Course Script:
Elementary Algebra, Functions, and Statistics
Vol. 2: Applications
2nd Edition
by Prof. Dr. Dietrich Ohse
ProCredit Academy GmbH
Hammelbacher Strasse 2
64658 Fürth-Weschnitz
Phone +49 6253 20080
© 2009 ProCredit Academy GmbH
Foreword
P a g e | III
Foreword
Mathematics plays a crucial role in banks because all banking operations
depend on accurate calculations and precisely defined methods. It is
therefore not nearly enough to rely on our pocket calculator, our desktop
PC or the server on our network. We need to develop an understanding of
the operations and processes that are fundamental to all areas of our
banks’ business. For this reason, a decision has been taken to introduce a
Mathematics Training Programme at all of the banks and academies
operated by ProCredit Holding. This Reader forms the basis for all of
these courses.
The course material is divided into two main parts:
Vol. 1
comprises the absolutely fundamental basics of algebra.
Confidence in applying these principles is essential. Numbers,
variables, calculations, expressions, equations and simple
functions are the basic components of mathematics and its
applications. A bank cannot afford to show any weakness in any
of these areas.
Vol. 2
addresses rather more advanced and in particular more bankspecific applications. In particular, it covers those functions that
are used in the financial sector, as well as the main themes of
financial mathematics, and finally the basics of statistics. This
material will be taught in its full scope primarily at the Regional
Academies; however, individual aspects will also be included in
the bank-level training courses.
In line with this division into two parts, the material is also presented in
two readers, the second of which builds upon the first. Accordingly, this
second volume contains applications related to banking problems and
some relevant topics of descriptive statistics.
The Reader is intended to define the subject matter to be covered during
the training. It thus forms a shared platform for both trainers and course
participants. For the training itself, the instructors are expected to
produce an overview of the sections and pages to be worked on during
each session. The overview, which should be handed out to the
participants, could look something like this:
IV | P a g e
Foreword
Block #
1
Structure and Topics
Volume 1 (revisited)
Introduction and Learning Targets
2
1 - 10
Special Functions
Exponential Functions
3
Pages
11 - 37
etc.
It is up to the trainer to select the sections that he or she will cover,
depending on the prior knowledge of the participants and how quickly
they are learning. It would therefore make no sense to try to offer a
universal recommendation here. Experience to date has shown that it is
better to cover the basics more slowly but therefore more thoroughly,
rather than trying to pack as much as possible into a single course at all
costs.
The main purpose of including references to specific sections and pages
in the textbook is to allow the participants to use the Reader for their own
independent study. The material has been developed very carefully and
comprehensively, so that it can be followed without the help of a trainer.
Numerous examples have been included in each section to illustrate the
steps that have been described. All of the examples are highlighted in the
text.
At the end of each section, we have set numerous practice exercises.
These exercises are an absolutely indispensable part of any training
course in mathematics. In fact, doing exercises is really the only way to
learn mathematics. Students will always be able to follow a teacher who
is good at motivating and explaining. But this is a very long way from
saying that they will automatically be able, independently, to apply what
they have understood. On the contrary: this ability only comes through
independent practice with the exercises.
In the book, the exercises are immediately followed by an answers
section, allowing the students to self-check their work. This section
contains only the final answers, without explaining in detail how these
results were arrived at. If you get stuck, please go back to the
Foreword
Page |V
corresponding unit and try to find the reason why you failed. Studying
means reading repeatedly.
Each chapter ends with progress tests designed to enable the students to
monitor their own progress. Students should compare their answers
against the ones given in the section immediately following the test.
References to the sections of the Reader that are relevant to specific parts
of the solution are designed to encourage students to go back and revise
these sections in order to acquire the skills needed to solve problems of
this particular type. However, the Reader does not contain any detailed
explanations of the solutions to the progress tests.
We hope you enjoy working with this Reader.
VI | P a g e
Contents
Contents
The numbering of the chapters continues from the first reader, i.e. the
first main chapter after description of the content of part 1 is the fifth
chapter of the course material as a whole. The following sections are
numbered accordingly.
1.
Volume 1 (revisited) ......................................... 1
1.1
Abstract of Volume 1 .................................................2
1.2
Contents of Volume 1 ................................................6
1.3
Learning Targets .......................................................8
5.
Special Functions ............................................ 11
5.1
Exponential Functions ............................................13
5.1.1
Graph of the Exponential Function ............................... 20
5.1.2
General Properties of Exponential Functions ................ 22
5.1.3
Application of the Exponential Function ...................... 23
5.1.4
Base e Exponential Function ......................................... 27
Exercise 5.1: Exponential Functions .................................. 32
Answers 5.1: Exponential Functions .................................. 34
5.1.5
5.2
Progress Test “Exponential Functions” ......................... 37
Logarithmic Functions ............................................39
5.2.1
Logarithmic Operations ................................................. 43
Exercise 5.2: Logarithmic Functions.................................. 46
Answers 5.2: Logarithmic Functions .................................. 47
5.2.2
5.3
Progress Test “Logarithmic Functions” ........................ 48
Answers to the Progress Tests ................................50
5.3.1
Answers to PT “Exponential Functions” ....................... 50
5.3.2
Answers to PT “Logarithmic Functions” ...................... 52
Contents
6.
P a g e | VII
Time Value of Money ..................................... 55
6.1
Interest .....................................................................58
6.1.1
Simple Interest ............................................................... 60
6.1.2
Compound Interest ........................................................ 62
6.1.3
Effective Interest ........................................................... 69
Exercise 6.1: Interest ......................................................... 78
Answers 6.1: Interest .......................................................... 80
6.1.4
6.2
Progress Test “Interest” ................................................. 81
Deposits ...................................................................83
6.2.1
Savings .......................................................................... 84
Single Payments ..................................................................... 84
Successive Payments .............................................................. 86
6.2.2
Regular Equal payments ................................................ 89
Exercise 6.2: Deposits ....................................................... 96
Answers 6.2: Deposits......................................................... 98
6.2.3
6.3
Progress Test “Deposits” ............................................... 99
Loans .....................................................................101
6.3.1
Repayments ................................................................. 102
Single Repayment................................................................. 102
Several Repayments ............................................................. 103
Repayment Plan ................................................................... 104
6.3.2
Instalments ................................................................... 106
Outstanding Balance ........................................................... 112
6.3.3
Effective Interest Rate of Loans .................................. 114
Periodic Interest Rate .......................................................... 115
Repayment Plan and Effective Interest Rate........................ 116
Floating Interest Rates......................................................... 120
Fees ...................................................................................... 122
VIII | P a g e
Contents
Exercise 6.3: Loans.......................................................... 129
Answers 6.3: Loans ........................................................... 132
6.3.4
6.4
7.
Progress Test “Loans” ................................................. 136
Answers to Progress Tests ....................................138
6.4.1
Answers to Progress Test “Interest” ............................ 138
6.4.2
Answers to Progress Test “Deposits” .......................... 139
6.4.3
Answers to Progress Test “Loans” ............................. 140
Basics of Statistics......................................... 143
7.0
Sum Symbol ...........................................................147
Exercise 7.0: Sum Symbol................................................. 152
Answers 7.0: Sum Symbol ................................................. 154
7.1
Presentation Techniques .......................................157
7.1.1
Tables .......................................................................... 158
7.1.2
Diagrams...................................................................... 165
7.1.3
Sum Curves ................................................................. 168
7.1.4
Base Problem ............................................................... 171
Exercise 7.1: Presentation Techniques ............................. 175
Answers 7.1: Presentation Techniques ............................. 178
7.1.5
7.2
Progress Test “Presentation Techniques” .................... 185
Key Figures: Centre ..............................................187
7.2.1
Mean ............................................................................ 189
7.2.2
Median ......................................................................... 196
7.2.3
Mode ............................................................................ 202
Exercise 7.2: Key Figures: Centre ................................... 205
Answers 7.2: Key Figures: Centre .................................... 208
7.2.4
7.3
Progress Test “Key Figures: Centre” .......................... 210
Key Figures: Spread .............................................212
Contents
P a g e | IX
7.3.1
Variance ....................................................................... 215
7.3.2
Standard Deviation ...................................................... 226
7.3.3
Normal Curve .............................................................. 230
Exercise 7.3: Key Figures: Spread ................................... 236
Answers 7.3: Key Figures: Spread ................................... 239
7.3.4
7.4
Progress Test “Key Figures: Spread” .......................... 241
Answers to Progress Tests ....................................243
7.4.1
Answers to PT “Presentation Techniques”.................. 243
7.4.2
Answers to PT “Key Figures: Centre” ........................ 246
7.4.3
Answers to PT “Key Figures: Spread” ........................ 247
Index ........................................................................ 249
X|Page
Tables
Figures
Figure 5-1:
Estimated population growth ........................................ 15
Figure 5-2:
Growth function of water hyacinths .............................. 18
Figure 5-3:
Different exponential functions with base a > 1 ........... 21
Figure 5-4:
Different exponential functions with base a < 1 ........... 21
Figure 5-5:
Inverse function of the exponential function ................. 40
Figure 6-1:
Time line with relevant data .......................................... 57
Figure 6-2:
Simple interest ............................................................... 61
Figure 6-3:
Compound interest......................................................... 63
Figure 6-4:
Future value of money ................................................... 64
Figure 6-5:
Present value of money .................................................. 66
Figure 6-6:
Interest compounded p times per year........................... 71
Figure 6-7:
Compounding periodically ............................................ 73
Figure 6-8:
Savings account ............................................................. 85
Figure 6-9:
Successive payments ...................................................... 87
Figure 6-10:
Positioning "today" ....................................................... 88
Figure 6-11:
Regular equal payments ................................................ 90
Figure 6-12:
Single repayment ......................................................... 103
Figure 6-13:
Several repayments ...................................................... 104
Figure 6-14:
Graph of the repayment plan ....................................... 106
Figure 6-15:
Instalments ................................................................... 107
Figure 6-16:
Calculation of outstanding amount ............................. 108
Figure 6-17:
Floating yearly interest rates ...................................... 120
Figure 6-18:
Floating periodical interest rates ................................ 121
Figure 6-19:
Fees and one repayment .............................................. 123
Figure 6-20:
Fees and instalments ................................................... 125
Tables
P a g e | XI
Figure 7-1:
Column diagram of grades .......................................... 165
Figure 7-2:
Column Diagram of a frequency table ........................ 166
Figure 7-3:
Column diagram of a grouped frequency table ........... 168
Figure 7-4:
Column diagram of the frequency sum of grades ........ 169
Figure 7-5:
Frequency sum of loans as a column diagram ............ 170
Figure 7-6:
Cumulative sum curve and distribution function......... 171
Figure 7-7:
Statistical figures in the Excel status bar .................... 191
Figure 7-8:
Histogram balanced at the average ............................ 197
Figure 7-9:
Right-tail distribution .................................................. 200
Figure 7-10:
Left-tail distribution..................................................... 201
Figure 7-11:
The tails of frequency distributions ............................. 201
Figure 7-12:
Narrowly spread attribute values ................................ 214
Figure 7-13:
Widely spread attribute values .................................... 214
Figure 7-14:
Spread around mean value .......................................... 228
Figure 7-15:
Distribution of the sizes of newborn babies ................ 231
Figure 7-16:
Histogram and fitted normal curve ............................. 232
Figure 7-17:
The former DM 10 bill with normal murve ................. 233
Figure 7-18:
A normal-shaped histogram ........................................ 234
XII | P a g e
Tables
Tables
Table 5.1:
Estimated world population (in thousands) .................. 14
Table 5.2:
Growth of water hyacinths ............................................ 17
Table 5.3:
Area Covered by water hyacinths.................................. 18
Table 5.4:
Limit process for number e ............................................ 29
Table 6.1:
Future values ................................................................. 71
Table 6.2:
Effective interest ............................................................ 72
Table 6.3:
Effective interest rates ................................................... 76
Table 6.4:
Repayment plan ........................................................... 105
Table 6.5:
Repayment plan with monthly instalments .................. 117
Table 6.6:
Repayment plan with quarterly instalments ................ 118
Table 6.7:
Repayment plan with floating interest rate ................. 119
Table 7.1:
List of grades ............................................................... 159
Table 7.2:
Frequency table of grades ........................................... 159
Table 7.3:
Frequencies of grouped data ....................................... 161
Table 7.4:
List of data ................................................................... 163
Table 7.5:
Grouped frequencies ................................................... 164
Table 7.6:
Frequency density of grouped data ............................. 167
Table 7.7:
“Base-Problem” .......................................................... 174
Table 7.8:
Grouped frequencies (Copy of Tab. 7.5) ..................... 195
Table 7.9:
List of 60 numbers ....................................................... 196
Table 7.10:
Number list .................................................................. 215
Table 7.11:
Calculating the variance ............................................. 216
Table 7.12:
Variance of grouped data ............................................ 225
1. Volume 1(revisited)
1.
Page |1
Volume 1 (revisited)
Because of its sheer size, this course, which was originally conceived of
as a single unit, has been divided into two parts and now appears in the
form of two separate readers. The first part mainly concerned itself with
the basics of algebra and simple functions. Most of this material is
normally taught in high school, so the first part can be regarded as an
introductory refresher course.
In contrast, this second part is devoted more to the applications of
mathematics, though naturally our selection is biased towards those areas
that play a role in everyday banking business. The mathematical
foundations underlying credit operations form one focal point,
descriptive statistics another.
Needless to say, this second part is based on the assumption that the
student has understood the basics discussed in the first volume. In order
to form a bridge from the first part and from earlier training courses that
you may have attended, this first introductory chapter gives a very brief
review of the topics covered in Volume 1. We have also included the
table of contents for Volume 1 to make it easier for you to refer back to
specific topics. Please note that, of course, the page numbers in this table
of contents refer to the first volume of the Reader. However, the
numbering of the chapters – with the exception of this first introductory
chapter – follows on consecutively from the first volume. Hence, the next
chapter is numbered 5 because the first volume ended with Chapter 4.
Finally, we re-iterate the training objectives, which naturally also apply
to this part and thus to the course as a whole.
2|Page
1. Volume 1 (revisited)
1.1 Abstract of Volume 1
The introductory Chapter 1 of Volume 1 focused primarily on the role of
mathematics as a language used to “translate real-world problems into
models which can be processed and solved using mathematical
methods”. With reference to examples, the transformation process from
real problem to solution was described.
The array of mathematical instruments which are available for solving
real-world problems are drawn initially from the basic instruments of
algebra. They were covered in Chapter 2.
Algebra could be regarded as the area of mathematics that deals in the
broadest sense of the term with “calculation”. At school we first learn to
calculate with numbers, that is, to apply certain operations such as
addition, subtraction, multiplication or division. Learning to calculate
with whole, natural numbers usually presents no problem. However,
people often start to make mistakes when they try to apply the same
operations to fractions and decimals. Fractions and decimals play an
important role in the banking sector, especially in connection with
percentages. A strong command of fractions in all their variations is
therefore indispensable.
That said, it would be inaccurate to equate algebra or indeed mathematics
with “calculation”, just as it would be a ridiculous understatement to
describe the literature of Shakespeare or Goethe as “letters”. One of the
main characteristics of mathematics is that its arguments consist not only
of numbers but also and above all of symbols, which stand for general
parameters or unknown values. Parameters and variables are
representatives of numbers. We can “calculate” with them in just the
same way as we do with known values, and in particular we can use them
to model real problems. It is only when we start to use variables in more
complex mathematical expressions and in functions that mathematics
becomes a powerful instrument that helps us generalise simple
relationships. Variables and parameters allow us to take the decisive step
from the specific to the general, and to analyse relationships that would
otherwise appear too confusing.
In the calculation of compound interest, algebraic expressions in which
numbers and variables are combined with exponents play an important
role in banking-specific applications. A discussion of the basics would
1. Volume 1(revisited)
Page |3
therefore not be complete without a section devoted to the calculation
rules that apply to integral and fractional exponents. As they are not
always covered sufficiently intensively at school, revision of the methods
for dealing with them are an important part of our course.
Through the combined application of mathematical operations on
numbers and variables, we can form complex expressions (indeed, there
is no limit to the complexity). One of the defining characteristics of
mathematics is that when performing calculations with these expressions
exactly the same rules must be applied as when calculating with
numbers. We might sometimes tend to forget this when the expressions
get really complicated. Yet it remains true to say: no matter how
complicated the expression is, the rules of algebra always remain in
force. In order to avoid confusion, however, it is usually necessary, and
certainly recommended, to use brackets to remove any ambiguity from an
expression. In other words, when solving expressions, it is better to use
too many brackets rather than one set of brackets too few.
Equations are the most frequently used relations in mathematics. In
mathematical models, i.e., in representations of real-world problems, they
provide the framework. They are created when accounting equations are
performed, physical laws are formulated or certain quantities for the
model are defined. The first volume of the Reader therefore included a
special section devoted mainly to explaining how equations are used in
the modelling process.
If the modelling process gives rise to an equation, then the primary goal
is usually to solve it. “Solve” means to determine the values of the
variables in the equation that satisfy the conditions of that equation.
Frequently the model will contain several equations, so-called systems of
equations. In that case, the solutions must satisfy the conditions of all
equations simultaneously. Ideally, they will then represent the solution to
the original real-world problem.
Linear and quadratic equations are, on the one hand, the simplest forms
of equations. At the same time, however, they are also the forms most
frequently used in economics. They were discussed in detail in two
sections of Chapter 3.
The fourth chapter then dealt for the first time with functions.
In our daily life we quite often encounter the verb “to function”. When
we turn the ignition key and the engine of our car starts, we could say
4|Page
1. Volume 1 (revisited)
that the starter motor is functioning correctly. When we switch on the
light and it becomes bright in the room, we would agree that the switch is
functioning correctly. Similarly, we could say that an organisation
functions if everything goes smoothly. In all cases, we say “it functions”
when an action leads to an expected and desired reaction.
If we want to describe the actions and reactions mathematically, starting
a car or switching on a light are not very interesting examples because
they are singular events. However, think about utility bills, in which the
final amount depends directly on the usage. In this case the action is the
consumption and the reaction is the amount shown on the invoice; thus,
there exists an almost infinite number of possible pairings of:
Action → Reaction
Both sides of this relation can be described in terms of numbers. Hence,
mathematics could serve very effectively as a language to describe the
dependencies between consumption and bill.
In fact, this is one of the major concepts in mathematics: to describe, by
means of equations, the relationships between factors which influence
each other, and the rules which govern how they influence each other. An
equation states the relationship between different variables. If we want to
describe the reaction of a factor (given by a variable, say y) to the action
of some other factor (expressed by another variable, say x), we could
write:
Action x
Rule orRelation
Reaction y
In mathematics the relation between different variables is called a
“function”. Functions form a key “vocabulary” in the language
“mathematics”, because practically all dependencies on real problems
must be described in terms of functions in order to apply mathematics to
solve them. That is, a model basically consists of a set of functions. If we
want to use models we have to know what to do with them.
Almost everything in mathematics is related to the concept of functions.
However, that does not mean that we have to learn everything about
them. Fortunately, we can easily restrict ourselves to
•
the most basic functions,
•
their representation,
1. Volume 1(revisited)
•
their properties, and
•
their use.
Page |5
Following the methodological approach of this whole reader, we began
with the general properties of functions, which are more or less valid for
all functions. In addition, we discussed general approaches to analysing
functions based on a number of fundamental properties such as zero
points, y-intercepts, slopes, and maximal or minimal points.
Very many functional relationships between economic quantities are
linear in nature, meaning that the resultant quantity is directly
proportional to the quantity that caused it. For example, sales revenues
are proportional to the quantity and the price of the goods or services
sold. Linear functions with one independent variable are represented
graphically by straight lines on the Cartesian plane. Based on this
simplest of all equations it is possible to explain the most important
properties and analytical steps in a particularly descriptive way.
Another section of Volume 1 was devoted to a discussion of quadratic
functions. In the case of a single independent variable, they are
represented by parabola. As the simplest type of non-linear function, they
provide a very descriptive example which can be referred to when
discussing more complex functions, such as the exponential and
logarithmic functions that are so important in banking. These special
functions are covered in this volume in the next chapter, Chapter 5.
6|Page
1. Volume 1 (revisited)
1.2 Contents of Volume 1
Volume 1 of the reader contained the basics of algebra and functions.
The page numbers shown here refer to Volume 1.
1.
Introduction ......................................................... 1
1.1 The Language Mathematics ......................................2
1.2 How to Apply Mathematics .......................................4
1.3 Learning Targets .......................................................8
2.
Basic Algebra ..................................................... 11
2.1 Numbers ..................................................................14
2.1.1 Numbers and Operations .............................................15
2.1.2 Fractions and Decimals................................................25
2.1.3 Percentages ..................................................................42
2.2 Exponents ................................................................58
2.2.1 Integer Exponents ........................................................61
2.2.2 Fractional Exponents ...................................................71
2.2.3 Radicals........................................................................80
2.3 Expressions..............................................................93
2.3.1 Integer Expressions ......................................................95
2.3.2 Fractional Expressions ...............................................104
Contents of Volume 1
3.
Page |7
Equations.......................................................... 123
3.1 Use of Equations ...................................................125
3.1.1 Modelling with Equations .........................................128
3.1.2 Solution ......................................................................132
3.2 Linear Equations ...................................................147
3.2.1 Normal Form of Linear Equations .............................148
3.2.2 Solution ......................................................................149
3.3 Quadratic Equations .............................................158
3.3.1 Forms of Quadratic Equations ...................................159
3.3.2 Solution ......................................................................160
4. Basic Functions ................................................... 177
4.1 Properties of Functions .........................................184
4.1.1 Characteristics of the Graph ......................................185
4.1.2 Inverse Functions .......................................................189
4.2 Linear Functions ...................................................203
4.2.1 Graph of a Linear Function........................................205
4.2.2 Properties of Linear Functions...................................208
4.3 Quadratic Functions .............................................216
4.3.1 Completion of the Square ..........................................218
4.3.2 Graph of the Quadratic Function ...............................220
4.3.3 Properties of the Quadratic Function .........................227
8|Page
1. Volume 1 (revisited)
1.3 Learning Targets
ProCredit is a bank. We aim to advise our customers, and above all we
want to give them correct and transparent advice. In order to do so, a
basic knowledge of algebra and some functions is indispensable.
Many of you will say: “But we’ve got computers and pocket calculators.”
That is a valid point, insofar as the software can ultimately calculate the
exact amount down to the last cent. However, there are various sources
of error: for example, you might enter the wrong figure. In this case, you
should be capable of judging whether the result is plausible or not.
You might find yourself in a situation in which you have to advise
someone and you don’t have your calculator or laptop with you. What
would you say to your friend, for example, if he stopped you on the street
and asked how much a loan would cost, or what the effective interest rate
on a savings deposit would be?
In our daily work, with or without customers, with or without colleagues,
with or without staff members: the essential details of banking business
are based on handling quantitative variables, figures and operations. It is
therefore essential to know
•
the basics of algebra,
•
some of the functions that are important for banking, and their
properties,
•
certain aspects of statistics, and
•
the background to the deposit-taking and lending business.
That is why in this course we will start by learning a sufficient amount of
vocabulary. We will familiarise ourselves with the absolutely
fundamental concepts of algebra, such as numbers and operators, and in
particular also fractions and exponents or indices, as well as how they are
combined to form algebraic expressions.
In our daily lives and in our work, quantities are very often compared and
weighed against each other. In mathematics, this generally leads to
equations that one would like to solve.
1.3 Learning Targets
Page |9
The most important form of an equation in mathematics is the function,
through which causal relationships in the real world can be described in
the mathematical model.
Basics of Algebra,
Functions, and
Statistics
1. Introduction
2. Basic Algebra
3. Equations
Vol. 2: Advanced
4. Basic Functions
5. Special Functions
6. Time Value
of Money
7. Basic
Statistics
In the bank, interest rates play an all-important role: calculating the
return on equity requires us to focus particularly on exponents when
working with algebraic expressions. Growth processes resulting from
compound interest are described by means of so-called exponential
functions, which naturally play an important role in describing problems
in the field of financial mathematics.
Because of the return on capital, the value of capital is time-dependent,
and because of this, financial-mathematical applications are often
referred to as the “time value of money”. In these course units, which are
10 | P a g e
1. Volume 1 (revisited)
of direct relevance to applications in the bank, the indispensable
fundamental principles of interest, deposits and loans will be described.
The above graphic shows the topics covered by this reader in their
context. They are designed to be taught in an advanced course, which
covers topics more closely related to applications in the banks.
The numbering of the chapters continues from the first reader, i.e. the
first main chapter after this introduction is the fifth chapter of the course
material as a whole. The following sections are numbered accordingly.
Each section contains numerous exercises. To allow you to check your
results, the correct answers are given at the end of each chapter. In most
cases, only the final answer is shown, but not the method of calculation
in all its details. We deliberately chose not to show the recommended
method because we wanted to encourage you to consider alternative
methods if you get the wrong answer.
5. Special Functions
5.
P a g e | 11
Special Functions
Basics of Algebra,
Functions, and
Statistics
1. Introduction
2. Basic Algebra
3. Equations
Vol. 2: Advanced
4. Basic Functions
Prerequisites:
5. Special Functions
6. Time Value
of Money
7. Basic
Statistics
Functions are one of the most important fields of
mathematics. In the first part of the course we
learned about the general properties of functions,
and focused on linear and quadratic functions. To
follow the discussion on exponential functions,
which are so important for the banking and financial
sector, familiarity with exponents is a basic
requirement.
Learning Targets: Exponential functions and their inverses, known as
logarithmic functions, together play a major role in
12 | P a g e
5. Special Functions
the banking sector because they are needed to
calculate interest charges on capital and evaluate
cash flows.
All of the functions we have considered so far have been polynomial
functions. In this reader we will discuss neither the rather large class of
rational functions – those which contain the operation “division” – nor
the functions containing roots. However, there is another class of
functions which cannot be described by means of the algebraic
operations addition, subtraction, multiplication, division, and taking of
powers and roots on variables. These are called “transcendental”
functions. Examples include the so-called trigonometric functions (sin,
cos, tan, ctg), which are extremely useful for technical applications.
Given our interest in economics and especially in banking, we will define
and investigate the properties of two other types of functions which
belong to the “transcendental” class: exponential functions and
logarithmic functions. They are used in describing and solving a wide
variety of real-world problems, including the growth of populations of
humans, animals, and bacteria; radioactive decay (negative growth);
normal distribution in statistics (see: Chapter 7); the growth of money at
compound interest; the absorption (negative growth) of light as it passes
through air, water, or glass; and the magnitude of sounds and
earthquakes. We will consider applications in these areas in more detail
in the chapters that follow (6. Time Value of Money, and 7. Basic
Statistics).
P a g e | 13
5.1 Exponential Functions
5.1 Exponential Functions
5. Special
Functions
5.1 Exponential
Functions
5.2 Logarithmic
Functions
Graph
Graph
Properties
Properties
Prerequisites:
Exponential functions occur if the variable is part of
the exponent. Therefore confidence in handling
exponential expressions is a very basic precondition
for understanding these functions.
Learning Targets: In this chapter we define and investigate the
properties of an important class of functions, called
exponential functions. They are used
use in describing
and solving a wide variety of real-world
real
problems,
especially processes of growth and decay. With
respect to banking,
banking exponential functions describe
the continuous compounding of interest.
14 | P a g e
5. Special Functions
Before we start to define some new mathematical concept we should
spend some time looking at the growth of the world’s population. The
following Tab. 5.1 shows the historical development of the estimated
number of people living on Earth (see: Wikipedia).
Year
Population
Year
Population
70,000 BC
2,000
1,000 BC
50,000
1960
2,981,659
10,000 BC
1,000
500 BC
100,000
1965
3,334,874
9,000 BC
3,000
1
200,000
1970
3,692,492
8,000 BC
5,000
1000
310,000
1975
4,068,109
7,000 BC
7,000
1750
791,000
1980
4,434,682
6,000 BC
10,000
1800
978,000
1985
4,830,979
5,000 BC
15,000
1850
1,262,000
1990
5,263,593
4,000 BC
20,000
1900
1,650,000
1995
5,674,380
3,000 BC
25,000
1950
2,518,629
2000
6,070,581
2,000 BC
35,000
1955
2,755,823
2005
6,453,628
Table 5.1:
Year Population
Estimated world population (in thousands)
If we look at the data in the table first, we notice that the world’s
population grew very slowly up to the year 1800. Then, all of a sudden,
the population started to grow rapidly.
The world population is the total number of humans living on Earth at a
given time. As of September 2008, the world’s population is estimated to
be just over 6.721 billion. In line with population projections, this figure
continues to grow at rates that were unprecedented before the 20th
century, even though the rate of growth has almost halved since its peak
of 2.2% per year, which was reached in 1963. The world’s population, on
its current growth trajectory, is expected to reach nearly 9 billion by the
year 2042.
If we draw the graph of the function p = f(t) i.e. number of people as a
function of time for the domain we get the picture shown in Fig. 5-1
P a g e | 15
5.1 Exponential Functions
(Source: Population Growth, Wikipedia). It shows a very distinct curve,
which remained almost totally flat for a very long period. Apparently, the
growth rate increased from the year 1000 on, then accelerated
explosively from 1800 and especially in the 20th century.
6
World Population (Billion)
5
4
3
2
1
0
10000 BC
8000
6000
4000
2000
AD1
1000
2000
Year
Figure 5-1:
Estimated population growth
Let us look for additional examples. In the German newspaper
“Frankfurter Rundschau”, dated November 10th, 2008, there was an
article by A. Beutelspacher with the headline: “Out of control – Woe
betide us if exponential functions start to gain momentum”. He described
three situations.
16 | P a g e
1.
5. Special Functions
Folding a newspaper exactly in the middle you get two layers of
half the area. Folding the smaller area again exactly in the middle,
the result will be 2 ⋅ 2 = 4 layers. If you continue to fold it in the
same way, you get 8, then 16, then 32 layers.
Each step of the folding process doubles the number of layers and
decreases the area by half.
Now, imagine that you can continue with the folding process as
long as you want. Certainly, if you really try to do it you will
come to an end pretty soon, because you will find that you
physically cannot fold it any more. The area will be too small,
even if you begin with a very large-sized newspaper.
But assuming that we could theoretically continue folding the
pieces, we could pose the question: How often do I have to fold
the newspaper until the stack of layers is so high that it closes the
gap between the Earth and the moon, which is about 380,000 km?
Although this is a totally unrealistic problem, a mathematician is
always allowed to ask questions like this because, as we will see,
it is a nice example with which to illustrate the behaviour of
processes which might get out of control.
Before reading on, just pause for a moment and guess how often
the newspaper needs to be folded in order to reach the moon. The
solution will be given at the end of this introductory section.
2.
There is a famous story about the inventor of chess whose
monarch wanted to reward him for this wonderful game. The
monarch offered him one free wish. Without hesitating, the
inventor answered that he would like the monarch to put one grain
of rice on the first square of the chessboard, two on the second,
four on the third, etc., doubling the number of grains each time
until all of the squares had been filled.
It does not sound as though the total amount of rice would be very
large, does it? However, taking into consideration that the
chessboard has 64 squares, the number of grains of rice on the last
square alone would be 263 which is almost the number 1 followed
by 19 zeros. This is many times the total amount of rice grown on
Earth in a year.
P a g e | 17
5.1 Exponential Functions
3.
While the first two examples are just amusing puzzles, the last
one is real and dramatic. Lake Victoria in East Africa, between
Tanzania, Uganda and Kenya, is the second largest freshwater
lake on Earth, and provides the local population with huge
quantities of fish. Or at least, it used to – until the arrival of the
water hyacinth. This pretty, decorative flower is in fact turning
out to be a kind of monster because of the phenomenal speed at
which it is spreading over the surface of the lake: when conditions
are ideal for it, the area covered doubles every 15 days.
So it is just a question of time before the entire surface will be
covered by an impenetrable carpet of plants which destroys the
shore, blocks the waterways, and prevents the infiltration of
oxygen, thus killing off the fish.
Recent explorations have shown that the carpet is growing at a
rate of 2000 hectares per week. If you measure every week, you
may find out that the growth per week depends on the area
covered the week before. The rate might be, say, 4%. That means,
if in one week the area covered is a then in the next week the
area will be approximately:
a + 0.04a = a ⋅ (1 + 0.04) = a ⋅1.04
During the following week, the covered area grows to:
1.4a ⋅ (1 + 0.04) = a ⋅1.042
because the rate of growth depends on the area covered in the
previous time period. It is a natural consequence of organic
growth due to cell division.
Starting the observation with the area a in week no. 0 you may get
the figures in Tab. 5.2 for the following 8 weeks.
Week
0
1
2
…
7
8
Area
a
a ⋅ 1.04
a ⋅1.042
…
a ⋅1.047
a ⋅1.048
Table 5.2:
Growth of water hyacinths
18 | P a g e
5. Special Functions
The area covered is obviously a function of time t and it is
certainly not too difficult to work out the structure of the
corresponding function y = f (t ) = a ⋅1.04 where t denotes the
number of weeks after the observations started.
t
Since a is simply a constant, it might be more interesting to study
the function:
y = f (t ) = 1.04t
Using the following value table the graph of the above function is
drawn in Fig. 5-2.
Year
0
1
Week
0
52
Area
1
2
3
4
4.3
4.6
5
208
225
242
260
7.7 54.6 419 3490
6800
13245
29022
104 156
Table 5.3:
Area Covered by water hyacinths
Area covered (hectares)
30000 a
20000 a
10000 a
5000 a
0
0
Figure 5-2:
1 year
2 year
3 year
4 year
Growth function of water hyacinths
5 year
P a g e | 19
5.1 Exponential Functions
Comparing the two curves in figs. 5-1 and 5-2 we may discover
similarities: Both functions are continuously increasing. In both
cases, the rate of growth starts relatively slowly, and accelerates
with increasing time. From a certain point on the gradient of the
function becomes very steep – it goes “out of control”.
Both functions are very realistic examples of the exponential
functions which we will study in this section.
We still have not answered one of the opening question: How do we have
to fold the newspaper until the stack of paper reaches the moon? Folding
a newspaper ten times leads to a stack of 210 = 1024 ≈ 103 pages. We can
assume that this stack is about 10 cm high (comparing it with two reams
of copy paper).
10 packs reach the height of 1 m
→
104 pages
103 packs are 1 km high
→
107 pages
The distance to the moon is about 384,000 km ≈ 4 ⋅105 km; therefore we
need:
4 ⋅105 ⋅107 = 4 ⋅1012 pages
Now let’s calculate how many pages we produce by folding the
1
newspaper. Folded once, we get 2 = 2 pages; folding twice, we obtain
22 = 4 pages, folding it n-times, we obtain 2n pages.
The final question is now: For what value of n is 2n = 4 ⋅1012 ?
The amazing result is that the above relation is true, when:
242 ≈ 4.4 ⋅1012
Did your guess come close to that surprisingly small number of fold?
Most probably you didn’t not because you likely didn’t take into account
that you were confronted with the special properties of an exponential
function.
20 | P a g e
5. Special Functions
5.1.1 Graph of the Exponential Function
Let us start by noting that the two functions f and g given by
f ( x) = 2 x and g ( x) = x 2
are not the same function – indeed, they are not even similar or related to
each other. The latter is called a power function, which is a special case
of a polynomial function. The variable x is raised to the power of 2.
In the function f ( x ) = 2 x the variable appears in the exponent. It is
called the exponential function.
EXPONENTIAL FUNCTION
The equation:
y = f ( x) = a x with a > 0, a ≠ 1
defines an exponential function for each different a, called the base. The
independent variable x appears in the exponent and may be any real
value.
The domain of these functions is the set of real numbers. It can be shown
that the range of all exponential functions is the set of all positive real
numbers. The base is required to be positive to avoid imaginary numbers
1
such as (−2) 2 = −2 .
Graphs of exponential functions for different bases a > 1 are shown in
Fig. 5-3. From left to right the curves in the first quadrant show the
functions:
y = f ( x) = 5x ; y = g ( x) = 2x ; y = h( x) = 1.2x
Notice that, in the upper right (first) quadrant, the greater the bases of the
function are, the steeper the curves are. In the upper left (fourth) quadrant
however, exactly the opposite is true: The functions become flatter with
increasing bases.
P a g e | 21
5.1 Exponential Functions
y
y = f ( x) = 5 x
26
24
22
20
18
y = g ( x) = 2 x
16
14
12
10
8
y = h( x ) = 1.2 x
6
4
2
-14
-12
-10
-8
-6
-4
-2
0
x
2
4
6
8
10
12
14
-2
-4
-6
Figure 5-3:
Different exponential functions with base a > 1
y = f ( x) = 15
x y
26
24
22
20
y = g ( x) =
1x
2
18
16
14
12
10
8
6
1
y = h( x) = 1.2
x
4
2
-14
-12
-10
-8
-6
-4
-2
0
x
2
4
6
8
10
12
14
-2
-4
-6
Figure 5-4:
Different exponential functions with base a < 1
22 | P a g e
5. Special Functions
The bases for the four functions shown in Fig. 5-3 were all greater than 1.
If we now choose different bases less than 1 we will get curves like the
ones in Fig. 5-4 for the exponential functions. Now the functions are
shown in the fourth quadrant from right to left in this order:
y = f ( x) =
1x
5
; y = g ( x) =
1x;
2
1
y = h( x) = 1.2
x
The graphs in Figs. 5-3 and 5-4 suggest the following important general
properties of exponential functions, which we state without a proof.
5.1.2 General Properties of Exponential Functions
•
•
•
•
•
•
The domain is the set of real numbers.
The range is the set of positive real numbers.
All graphs are continuous.
For all functions the y-intercept is y = 1.
The exponential functions have no zeros (see: point 2.).
The curves approach the x-axis on one side asymptotically, i.e. they
come infinitely close but never touch it.
•
If a > 1, then ax increases as x increases.
•
•
If a < 1, then ax decreases as x increases
The exponential function is a one-to-one function.
EXPONENTIAL OPERATIONS
Exponential functions obey the familiar laws of exponents we discussed
earlier for integer exponents (see: Volume 1; Section 2.2.1), which are
also valid for rational (fractional) exponents.
5.1 Exponential Functions
P a g e | 23
For exponential functions with the same base a > 0 and x,y real numbers
the following rules hold:
•
a x ⋅ a y = a x+ y
•
 ax
 y
a
•
a0 = 1
•
1
a
x

x− y
 = a

= a− x
•
(a )
•
n
•
ax = a y
x
y
= a x⋅ y
x
ax = a n
↔
x= y
For exponential functions with different bases a,b > 0 and the same
exponent x (real number) we obtain:
•
a x ⋅ b x = (a ⋅ b) x
•
 ax
 b x

•
a x = b x , for x ≠ 0
  a x
=
  b 

→ a=b
For examples, see Volume 1, Section 2.2.1 of the reader.
5.1.3 Application of the Exponential Function
We now consider three applications in which exponential functions are
utilised for analytical purposes: population growth, radioactive decay,
and compound interest.
24 | P a g e
5. Special Functions
POPULATION GROWTH
Our first example involves the growth of populations of organisms, such
as humans, animals, insects, and bacteria. Populations tend to grow
exponentially due to cell division. This means that their growth depends
at each point in time on the number of active cells. Assuming that in the
growth phase all cells will double within a certain time – the doubling
time – this parameter is a convenient and easily understood measure of
the growth rate. It is the time taken for the population to exactly double.
Over shorter time periods t the population growth can be described by
the model:
t
P = f (t ) = P0 ⋅ 2 d
where P is the population at time t, P0 is the population at time t = 0, and
d is the doubling time.
Note that when t = d:
d
P = f ( d ) = P0 ⋅ 2 d = P0 ⋅ 2
The population is double the original, as would be expected.
EXAMPLE
Mexico has a population of around 100 million, and it is estimated
that it doubles within 21 years. Assuming that the population grows
constantly at the same rate, what will be the population
a) 15 years from now?
b) 30 years from now?
We use the above model with the parameters: P0 = 100 million and
d = 21 .
15
a)
P(t ) = P(15) = 100 ⋅ 2 21 = 164.067 million
b)
P(t ) = P(30) = 100 ⋅ 2 21 = 269.18 million
30
P a g e | 25
5.1 Exponential Functions
RADIOACTIVE DECAY
Our second example deals with radioactive decay, which can simply be
regarded as a negative growth phenomenon. Decay means that if you
start with a certain amount A of a radioactive isotope, the amount
declines exponentially over time. The so-called half-life is a property of
every radioactive isotope. It defines the rate of decay in terms of the time
taken for the material to decay to half of its original amount.
Over any other time period t the decay can be described by the model:
A = f (t ) = A0 ⋅
( )
1
2
t
h
where A is the amount at time t, A0 is the amount at time t = 0, and h is
the half-life.
Note that when t = h:
A = f ( h) = A0 ⋅
( )
1
2
h
h
= A0 ⋅ 12
The amount is half the original material, as would be expected.
EXAMPLE
The radioactive isotope gallium-67 has a half-life of 46.5 hours. If we
start with 100 milligrams of the isotope, how many milligrams will be
left after
a) 24 hours?
b) 1 week?
We use the above model of radioactive decay with the parameters:
A0 = 100 mg; h = 46.5 and 1 week = 7 ⋅ 24 = 168 hours.
( 12 )
a)
A = f (t = 24) = 100 ⋅
b)
A = f (t = 168) = 100 ⋅
24
46.5
( )
1
2
= 69.925 milligrams
168
46.5
= 8.173 milligrams
26 | P a g e
5. Special Functions
COMPOUDED INTEREST
The third application is directly connected with your work in the
ProCredit Bank and is absolutely fundamental to many topics in banking
and finance. It relates to interest, which is generally calculated as a
percentage of the money saved or borrowed (principal) over a period of
time. Most often the percentage is the amount due for one year, but, as
you are all aware, the interest rate can also be defined for other periods.
We will discuss many topics relating to interest in much greater detail in
the sixth chapter, “Time Value of Money” (see: Page 55). However, for
the time being, let us simply consider the basic idea of interest and why it
is a nice example of the use of exponential functions.
If at the end of an interest period the interest earned on a savings deposit
is added to the balance, then in the next period interest is earned not only
on the original deposit but also on the interest earned during the previous
period. In other words, interest is paid on interest. This is called
compound interest.
Assume the interest rate is i per year, compounded annually. This means
that interest is capitalised (added to the balance) at the end of the year.
The balance of a deposit D after one year is:
B1 = D + i ⋅ D = D ⋅ (1 + i)
In the second year, interest is payable on the balance B1 :
B2 = B1 + i ⋅ B1 = B1 ⋅ (1 + i) = D ⋅ (1+ i)2
If we let B1, B2 , B3 , B4 ,... be the balances after 1, 2, 3, 4, … years,
respectively, then the following model describes the balance after n
interest periods (years):
Bn = D ⋅ (1 + i)n = D ⋅ qn
where:
Bn
=
balance after n interest periods
D
=
deposit at period n = 0
i
=
interest rate per interest period (year)
q = 1+i =
interest factor (balance plus interest)
P a g e | 27
5.1 Exponential Functions
EXAMPLE
Suppose you deposit D = EUR 1000 in a savings account that pays
i = 6% interest, compounded annually.
a) How much will the balance be after 10 years?
D = 1000;
i = 0.06;
q = 1.06;
n = 10;
B10 = ?
B10 = D ⋅ q10 = 1000 ⋅1.0610 = 1790.85 €
b) If you want the balance to have doubled 10 years from now, how
high must the interest rate be?
D = 1000;
i = 0.06;
n = 10;
B10 = 2000 ; q = ?
From the above model we obtain:
Bn = P ⋅ (1 + i ) n → (1 + i ) n =
i=n
Bn
P
→ 1+ i = n
Bn
P
Bn
−1
P
1
i = 10 2 −1 = 210 −1 = 1.07177 −1 = 0.07177
To double a principal in 10 interest periods, the interest must be
i = 7.177%.
5.1.4 Base e Exponential Function
When introducing numbers we mentioned in Volume 1, Section 2.1.1 the
three most commonly used irrational numbers: 2 , π , and e. The
constant π is a so-called natural constant which is well known from the
formulas used to calculate the circumference ( 2π ⋅ r ) and the area ( π ⋅ r )
of a circle (with radius r). The third irrational constant is at least as
important as the number π. It is the number e = 2.71828182846… which
is quite frequently the base of an exponential function describing many
natural growth phenomena. The irrational number e is named after the
great Swiss mathematician Leonard Euler (1707 – 1783).
2
28 | P a g e
5. Special Functions
The constant e occurs naturally when we study compound interest.
Although we are going to study the effects of interest in more detail in
the sixth chapter – as already mentioned – it is a good opportunity to
discuss some aspects of compound interest in connection with the socalled e-function, which is a synonym for the exponential function with
base e:
y = f ( x) = ex where e = 2.71828...
In the model of compound interest:
Bn = D ⋅ (1 + i)n = D ⋅ qn
we assumed that interest is paid (compounded) annually. If interest i (the
so-called nominal yearly rate) is compounded quarterly, monthly, or even
daily, the frequency of capitalisation is 4, 12, or even 360 times,
respectively, within one year. The corresponding formulas are:
(
)
•
B1 = D ⋅ 1 +
•
B1 = D ⋅ 1 + 12i
•
B1 = D ⋅ 1 +
i
4
4
(
)
(
i
360
when compounding quarterly
12
when compounding monthly
)
360
when compounding daily
Generalising the formulas for compounding p times within a year we
obtain:
•
(
B1 = D ⋅ 1 +
i
p
)
p
when compounding p times.
The balance after t years will be:
•
(

Bt = D ⋅  1 +

Substituting
i
p
=
the same formula:
•
1
x
i
p
)
p t
→
(
 = D ⋅ 1+

i
p
)
p⋅t
p = i ⋅ x provides a slightly different form of
P a g e | 29
5.1 Exponential Functions
•
( )
Bt = D ⋅ 1 +
i
p
p⋅t
(
= D ⋅ 1 + 1x
)
x⋅i⋅t
(
= D ⋅  1 + 1x

)
x  i⋅t

So far we have only done some simple algebraic manipulations.
However, if we look more carefully at the expression within the square
brackets we will realise that it tends to approach a limit.
1
 1
f ( x) = 1 + 
x

2
10
2.59374246
50
2.69158880
100
2.704813829
500
2.715568520
1000
2.716923932
5000
2.718010050
10000
2.71814592
100000
2.71826824
x
Table 5.4:
x
Limit process for number e
Although this series of numbers is not a proof, it is very obviously
approaching the irrational number e. Mathematicians use the limit
notation for describing approaches like this:
(
x→∞
lim 1 + 1x
)
x
= e = 2.7182818...
From the substitution p = i ⋅ x we should recognise that increasing x
means that the number of compounding periods increases equally.
Therefore the limiting procedure simply means that we divide the year
into an increasing number of interest periods. Hence lim describes
x→∞
infinitely many periods, which means continuous compounding. The
deposit is compounded every moment of time.
30 | P a g e
5. Special Functions
P a g e | 31
5.1 Exponential Functions
CONTINUOUS COMPOUND INTEREST
If a deposit D is compounded continuously at a yearly rate of i, in
nominal terms, then the balance Bt after the time t is given by:
Bt = f (t) = D ⋅ ei⋅t
EXAMPLE
A deposit of EUR 1000 will be compounded for t = 5 years at a
nominal interest rate i = 6%:
a) annually
b) quarterly
c) monthly
d) daily
e) continuously
How high will the balances be in each case?
a) p = 1:
B5 = 1000 ⋅ (1 + 0.06)5 = 1338.23 EUR
b) p = 4:
B5 = 1000 ⋅ (1 + 0.06
)20 = 1346.86 EUR
4
c) p = 12:
B5 = 1000 ⋅ (1 + 0.06
)60 = 1348.85 EUR
12
d) p = 360: B5 = 1000 ⋅ (1 + 0.06
)1800 = 1349.83 EUR
360
e) p = ∞:
B5 = 1000 ⋅ e0.06⋅5 = 1000 ⋅ e0.06⋅5 = 1349.86 EUR
32 | P a g e
5. Special Functions
EXERCISE 5.1:
EXPONENTIAL FUNCTIONS
You will find the answers to the exercise on the pages directly following
the problems. They consist only of the final answer, allowing you to
compare your answer with ours.
We deliberately chose not to show the recommended method because we
wanted to encourage you to consider alternative methods if you get the
wrong answer.
1. Sketch the following functions:
a)
f ( x) = 3 x
b) g ( x ) = ( 13 )
c)
h( x) = −(2 x )
d) k ( x ) = 2 − x
x
2. Simplify the following expressions:
a)
5
7
4
3
a ⋅a ⋅a
1
7
10
b)
a2
1
a3
3
2
1
c)
x ⋅ x5 ⋅ x3
d) (9a 2 ) 2
e)
34 x+1 ⋅ 32−3x
f)
( ) ⋅( )
2a
3b
2
2b
3a
3. Which values solve the following equations?
a)
2 x−3 = 23 x+1
c)
5 x −1 ⋅ 53− 2 x =
e)
2 x−5
−10x−3 = 0
b) 10
52 x −1
(2 x + 1) 4 = 9 2
5
x−2
d) 42 x −1 = 22 x + 4
P a g e | 33
5.1 Exponential Functions
4. The capital D = EUR 2000 was deposited at the beginning of
2004. What will be the balance at the end of 2007 at the nominal
interest rate i = 5% when
a) compounded annually?
b) compounded monthly?
c) compounded continuously?
5. A country has a population of 28 million. During the past 17 years
the population has doubled. What will be the estimated population
a) 10 years from now?
b) 20 years from now?
c) 30 years from now?
6. A radioactive isotope has a half-life of 7 hours. How much of 25
milligrams will be left:
a) after 36 minutes?
b) after 3 hours?
c) after 2 days?
d) How much was there 22 hours ago?
34 | P a g e
5. Special Functions
ANSWERS 5.1:
EXPONENTIAL FUNCTIONS
f ( x) = 3 x
1. a)
y
8
7
6
5
4
3
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
x
1
2
3
4
5
6
7
8
-1
-2
-3
-4
-5
b) g ( x) =
( 13 )
x
y
8
7
6
5
4
3
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
-5
x
1
2
3
4
5
6
7
8
P a g e | 35
5.1 Exponential Functions
( )
h( x) = − 2 x
c)
y
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
x
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
k ( x) = 2− x
d)
y
12
11
10
9
8
7
6
5
4
3
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
-2
x
36 | P a g e
5. Special Functions
577
1
2. a)
a 210
b)
e)
3x+3
f)
3. a) x = –2
e)
34
a6
c)
x 15
c)
x=
d) 3a
2 2 a +b
32b+ a
b) x = 2
1
2
d) x = 3
x =1
4. a) EUR 2431.01
b) EUR 2441.79
c) EUR 2442.81
5. a) 42.1 million
b) 63.3 million
c) 95.1 million
6. a) 23.56 milligrams
c) 0.2156 milligrams
b) 18.57 milligrams
d) 220.82 milligrams
P a g e | 37
5.1 Exponential Functions
5.1.5 Progress Test “Exponential Functions”
You should assign yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Don’t use the reader to look
for the solution. The aim of the test is to give you feedback on how much
you know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. Otherwise
(if you got the wrong answer or no answer at all) you will be given
advice on what to do. In most cases you will be directed to the section in
the reader which you should repeat in order to close the gap.
1. Sketch:
a)
y = f ( x) = 2x + 2
b)
y=
c)
y = ( −2 )
( 14 )
x
x
2. Simplify the following expressions:
a)
1
2
x ⋅x
− 23
5
=
b)
x4
x
c)
42 x −1 ⋅ 4 2+3 x
43 x − 2
1
7
=
=
3. Which values solve the following equations?
a)
3x−2 = 32 x+3
b)
4 2 x −1
2
2( x − 2)
= 16 x −1
38 | P a g e
c)
5. Special Functions
e x−1 = 1
4. At the beginning of the year EUR 400 was deposited in an
account paying 2% interest, compounded monthly. How much is
in the account at the end of the year?
5. A radioactive isotope has a half-life of 16 hours. How much of
124 mg will be left after
a)
24 hours?
b) 48 hours?
c)
1 week?
6. A radioactive isotope has a half-life of 8 hours.
a)
How much was there 24 hours ago, if there is 45 mg left?
b)
After which time will 10 mg of 50 mg be left?
P a g e | 39
5.2 Logarithmic Functions
5.2 Logarithmic Functions
5. Advanced
Functions
5.1 Exponential
Functions
5.2 Logarithmic
Functions
Graph
Graph
Properties
Properties
Prerequisites:
Since the logarithmic function is defined as the
inverse function to the exponential function,
function it is
helpful to review the general properties of functions,
which were explained in Volume 1, Section 4.1, and
especially the discussion of the inverse function.
Learning Targets: The discussion of the logarithmic function is very
strongly connected to the discussion of its inverse
function,, the exponential function. Since exponential
functions are one-to-one,
one
their inverse exists. We
will discuss how the general concept of inverse
functions helps to define and understand the
properties of logarithmic functions.
40 | P a g e
5. Special Functions
In Volume 1, Section 4.1, we extensively discussed one-to-one functions
and the existence of the inverses to one-to-one functions. In addition we
discussed a procedure for determining the inverse function. However,
this procedure is only applicable to functions whose variables can be
isolated.
The exponential function f ( x) = a x is clearly a one-to-one function. It is
relatively easy to construct its inverse as a mirror at the 45° line: y = x
(see: Fig. 5-5).
y
y = f ( x) = a x
16
14
12
10
8
6
y = g ( x ) = f −1 ( x) = log a x
4
2
x
-8
-6
-4
-2
0
2
4
6
8
10
-2
-4
Figure 5-5:
Inverse function of the exponential function
However, there is no algebraic procedure for solving the exponential
function f ( x) = a x for x in order to determine the function f −1 ( x ) = ?
As early as the 8th century an Indian mathematician described values for
the inverse functions of base 3 and 4. In the 13th century the first tables
were produced by Arabian mathematicians and in 1614 the Scottish
5.2 Logarithmic Functions
P a g e | 41
mathematician John Napier (1550 – 1617) published a book which made
him the “inventor of the logarithm”, although the Swiss watchmaker Jost
Bürgi (1552-1632) also independently developed a system for calculating
logarithms. Hence, the name “logarithm” was created about 400 years
ago to describe what is simply the inverse to the exponential function.
LOGARITHMIC FUNCTION
The inverse function of the exponential function y = a x with a > 0 is
called the logarithmic function for base a and is denoted by:
y = log a x where a > 0
The inverse function of the exponential function y = e x , i.e. to base e, is
called the natural logarithmic function and is denoted by:
y = ln x
The inverse function of the exponential function y = 10 x , i.e. to base 10,
is called the common logarithmic function (also decade logarithm) and is
denoted by:
y = log x (sometimes also: y = lg x )
Although the world had been given a new function, there was no
algebraic representation of it. Until the rise of electronic calculators, the
only way to use the function was by means of the graph and table values.
In fact, until the late 1960s logarithmic tables were an important part of
school textbooks and technical libraries.
DEFINITION OF THE LOGARITHMIC FUNCTION
As the inverse of the exponential function, the logarithmic function is
defined by:
y = loga x ↔ x = a y
with a > 0 and a ≠ 1
42 | P a g e
5. Special Functions
This means that the log to the base a is the exponent to which a must
be raised to obtain x. Or in other words, the log reverses the exponential,
i.e. executed consecutively they cancel each other out:
log a a x = x and a log a x = x
log10 x = x and 10log x = x
ln e x = x and a ln x = x
For some very obvious exponential numbers it is easy to deduce the
logarithm:
log 2 8 = 3 because 2 3 = 8
log 3 81 = 4 because 34 = 81
log100000 = 5 because 105 = 100000 (Remember: log = log10 is the
common logarithm.)
EXAMPLES
The following relationships between exponential functions and
logarithmic functions hold:
1.
25 = 32
↔
2.
3.43 = 39.304
3.
16 = 16 2 = 4
1
log2 32 = 5
↔ log3.4 39.304 = 3
↔ log16 4 =
1
2
4.
log4 64 = 3
↔ 43 = 64
5.
log5 22.8 = 1.9428 ↔ 51.9428 = 22.8
6.
log7 0.64 = −0.2293
↔ 7−0.2293 = 0.64
5.2 Logarithmic Functions
P a g e | 43
Similar to the properties of exponential functions (see: Section 5.1.2;
Page 22) there are a set of rules for logarithms which are the exact
inversions of the rules for exponential functions.
5.2.1 Logarithmic Operations
For logarithmic functions with the same base a>0 and x,y real numbers
the following rules hold:
•
log a ( x ⋅ y ) = log a x + log a y
•
x
log a   = log a x − log a y
 y
•
log a 1 = 0
•
1
log a   = log a 1 − log a x = − log a x
 x
•
log a ( x y ) = y ⋅ log a x
•
log a n x = log a x n = 1n ⋅ log a x
•
log a x = log a y ↔ x = y
1
For logarithmic functions with different bases a,b > 0 and the same
exponent x (real number) we get:
•
log a ( y x ⋅ z x ) = log a ( y ⋅ z ) x = x ⋅ ( log a y + log a z )
•
x
 yx 
 y
log a  x  = log a   = x ⋅ ( log a y − log a z )
z 
z
 
•
log a x = log b x ↔ a = b
44 | P a g e
5. Special Functions
EXAMPLES
The operations are illustrated by means of the following examples where
the logarithms are finally determined with a calculator. We use the
common logarithm; the formulas are the same for other logarithms with
different bases.
1.
log 200 = log ( 2 ⋅100 ) = log 2 + log100 = 0.30103 + 2 = 2.30103
2.
log 47849 = log(4.7849 ⋅10000) = log 4.7849 + log10000
log 47849 = 0.67987 + 4 = 4.67987
3.
3.53 = log 3.53 − log1000 = 0.54777 − 3
log 0.00353 = log 1000
log 0.00353 = −2.45223
4.
5.
(
)
log ( 4.6− ) = −0.2 ⋅ log 4.6 = −0.2 ⋅ 0.66276 = −0.13255
log 23.40.68 = 0.68 ⋅ log 23.4 = 0.68 ⋅1.3922 = 0.93107
0.2
If you look at a scientific electronic calculator you will notice that there
are (usually) only two logarithmic functions, namely the natural
logarithm (ln x ) and the common (decade) logarithm (log x is sometimes
also denoted by lg x). How can we calculate the general logarithm to
every base a > 0?
Going back to the definition and applying the above operations will help
us:
y = loga x ↔ a y = x with a > 0 and a ≠ 1
Applying some known logarithm (for instance the natural logarithm) to
both sides of the equation gives:
( )
ln a y = ln x →
y ⋅ ln a = ln x
→
y=
ln x
ln a
or, if you prefer to use the common logarithm:
( )
log a y = log x → y ⋅ log a = log x → y =
log x
log a
P a g e | 45
5.2 Logarithmic Functions
EXAMPLES
Again, the final logarithms must be determined with a calculator:
ln 349 5.8522
=
= 3.3624
ln 5.7 1.7405
1.
log 5.7 349 =
2.
log0.34 16.9 =
↔
5.7 3.3624 = 349
log16.9 1.2279
=
= −2.6208
log 0.34 −0.4685
↔ 0.34−2.6208 = 16.9
3.
log5 106 =
log106
6
=
= 8.584
log 5
0.699
↔
58.584 = 106
46 | P a g e
5. Special Functions
EXERCISE 5.2:
LOGARITHMIC FUNCTIONS
You will find the answers to the exercise on the pages directly following
the problems. They consist only of the final answer, allowing you to
compare your answer with ours.
We deliberately chose not to show the recommended method because we
wanted to encourage you to consider alternative methods if you get the
wrong answer.
1. Determine x without using a calculator:
a)
x = log 3 81
b)
x = log1.1 1.21
d)
x = log 25 5
e)
x = log 0.01
c)
x = log 0.5 8
c)
log x = 4
2. Solve the equations for x:
a)
log 2 x = 5
b)
log 3 x = 0.5
d)
log 4 x = 2
e)
log5 x = 0
3. Determine the unknown x in the following expressions:
a)
log 4 x = 3
b)
log 2 8 = x
d)
log10 −6 = x
e)
log 25
c)
log x 256 = 4
( 15 ) = x
4. Use your calculator to finally solve the problems:
a)
x = log 2.4
d)
x = log(2.4 ⋅1012 )
b)
x = log 0.02
c)
x = ln 0.00002
5. Use the logarithm to solve the following equations:
a)
e x = 100
b)
103 x +1 = 200
c)
2.43 x = 0.45
P a g e | 47
5.2 Logarithmic Functions
ANSWERS 5.2:
1.
2.
3.
4.
LOGARITHMIC FUNCTIONS
a) x = 4
b)
x=2
d) x =
e)
x = −2
a) x = 32
b)
x= 3
d) x = 16
e)
x=1
a) x = 64
b)
x =3
d) x = −6
e)
−0.2153
1
2
c)
x = −3
c)
x = 10000
c)
x=4
c)
x = −0.3040
a) x = log 2.4 = 0.3802
b) x = log 0.02 = −1.699
5.
c)
x = −10.8198
d)
x = 12.3802
a) x = 4.6052
b)
x = 0.4337
48 | P a g e
5. Special Functions
5.2.2 Progress Test “Logarithmic Functions”
You should assign yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Do not use the reader to look
for the solution. The aim of the test is to give you feedback on how much
you know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. Otherwise
(if you got the wrong answer or no answer at all) you should repeat the
corresponding units in order to close the gap.
1. Sketch the logarithmic functions by using the property that it is
the inverse to a certain exponential function:
a)
y = f ( x) = log 2 x is the inverse of: y = 2 x
b)
y = g ( x) = log0.5 x is the inverse of y =
c)
x
y = h( x ) = ln 2 x is the inverse of y = 12 e
( 12 )
x
2. Apply the calculation rule for logarithms to find simplified ways
of performing these calculations:
a)
log( x ⋅ 2y )
b)
 xn 
log  m 
y 


2
−2
c)
ln e x
ln e4.5
c)
log0.1 103
log5 x = 10
c)
log 2.5 x = 0
3. Calculate without a calculator:
a)
log5 125
b)
4. Solve the equations for x:
a)
log2 x = 16
b)
P a g e | 49
Progress Test
5. Determine the unknown x in the following expressions:
a)
log3 x = 1
b)
log10 − 3 = x
c)
ln 2 x = 1
6. Use the logarithm to solve the following equations:
a)
e2 x = 40
b) 102 x−1 = 1000
c)
23x+1 = 128
c)
log7.45 0.94
7. Use your calculator to find:
a)
log3.4 57
b)
log25.6 12.9
d)
log0.45 6.77
e)
log0.33 0.79
50 | P a g e
5. Special Functions
5.3 Answers to the Progress Tests
5.3.1 Answers to PT “Exponential Functions”
You should check your answers. If they are correct, you may continue
and start the next chapter. If not (if you got the wrong answer or no
answer at all) repeat the topics of the corresponding sections in order to
close the gap.
1. a)
b)
y = f ( x) = 2x + 2
y=
( 14 )
x
c) Not defined
P a g e | 51
5.3 Answers to Progress Tests
2. a)
1
2
x ⋅x
− 23
=x
− 16
5
b)
x4
1
31
= x 28
x7
c)
42 x −1 ⋅ 42+3 x
= 4 2 x +3
3 x −2
4
3. a) x = −5
b) x = 3
c)
x=1
5. a) 43.84 mg
b) 15.5 mg
c)
0.0856 mg
6. a) 360 mg
b) 18.57 hours
4. EUR 507.30
52 | P a g e
5. Special Functions
5.3.2 Answers to PT “Logarithmic Functions”
You should check your answers. If they are correct, you may continue
and start the next chapter. If not (if you got the wrong answer or no
answer at all) repeat the topics of the corresponding sections in order to
close the gap.
1. a)
y = f ( x) = log 2 x is the inverse of y = 2 x
y = 2x
y = f ( x) = log 2 x
b)
y = g ( x) = log0.5 x is the inverse of y =
y=
( 12 )
( 12 )
x
y = g ( x ) = log 0.5 x
x
P a g e | 53
5.3 Answers to Progress Tests
c)
x
y = h( x ) = ln 2 x is the inverse of y = 12 e
y = 12 e x
y = h( x) = ln 2 x
2. a)
log( x ⋅ 2y ) = log x + log 2 − log y
b)
 xn 
log  m  = n ⋅ log x − m ⋅ log y
y 


c)
ln e x
3. a)
2
−2
= x2 − 2
log5 125 = 3
b)
ln e4.5 = 4.5
c)
log0.1 103 = −3
4. a) x = 65,536
b) x = 9,765,625
c) not feasible
5. a) x = 3
b) x = −3
c) x = 1.359141
6. a) x = 1.844
b) x = 2
c) x = 2
log3.4 57 = 3.3038
b)
log25.6 12.9 = 0.7877
c)
log7.45 0.94 = −0.00308
d)
log0.45 6.77 = −2.395
e)
log0.33 0.79 = 0.2126
7. a)
P a g e | 55
6. Time Value of Money
6.
Time Value of Money
Basics of Algebra,
Functions, and
Statistics
1. Introduction
2. Basic Algebra
3. Equations
Vol. 2: Advanced
4. Basic Functions
Prerequisites:
5. Special Functions
6. Time Value
of Money
7. Basic
Statistics
It is required that one can deal competently with
equations and functions. In particular, exponential
and logarithmic functions will be applied.
Learning Targets: In order to understand the procedures in a bank, it
must be understood that the value of capital always
depends on the variable of time.
In every
calculation of cash flows, one must be able to
determine and understand the value of time.
56 | P a g e
6. Time Value of Money
Some of the most fundamental roles of a bank are:
collecting deposits, managing accounts and
providing loans. All these processes involve
charging interest on capital. This chapter will cover
the basics of the associated calculations.
Today, most calculations are performed with
computers. However, although it is possible to make
such calculations with computers, accurate results
depend on the correct entry of the figures. Therefore,
it is important to develop a feeling for the
plausibility of the calculated results. Only then is it
possible to be sure that accurate and meaningful
information is communicated to the customer.
The value of money is dependent on time. This implies neither that the
value printed on the bill changes, nor that the value decreases due to
inflation, but rather that the value changes when the money is invested.
The main concern is how the value of investments (in-payments) and
loans are affected by interest rates at different times. If we have an
investment at a certain time and repayment at another time, the values of
these two payments differ because of the possibility of earning interest.
In a situation in which all cash flows are certain – which we will assume
throughout this chapter – the rate of interest can be used to express the
"time value of money. The rate of interest will allow us to adjust the
value of cash flows, whenever they occur, at a particular point in time. It
will thus be necessary to position time-adjusted cash flows at a single
point in time.
A convenient way to represent cash flows at several points in time is to
use a time line. It is always very helpful to think of the time line as some
imaginary account where, over time, we can mark all relevant payments
with arrows (see: Fig. 6-1). Throughout this course we will represent inpayments with arrows towards the time line, indicating that there is cash
flow into the account, and mark out-payments with arrows pointing
downwards from the timeline, indicating that there is a cash flow out of
the account. Points in time (dates) are represented by points on the scale.
Since cash flow changes with time, we can only add, subtract, compare,
and balance cash flows when all payments are adjusted to a single point
P a g e | 57
6. Time Value of Money
in time. This point is usually marked extra as the reference point or often
simply by the word “today”.
in-payment
i.e. D = 200 EUR
compounding
time period
date
i.e. 5.5.2009
t (time line)
reference point
today
discounting
compounding
interest rate, i.e. i = 5%
out-payment
i.e. W = 100 EUR
Figure 6-1:
Balance
Time line with relevant data
The time line already contains all the relevant information. We will make
extensive use of it to structure real-world situations graphically. Once
you have configured a graphical model with all the relevant information
about a problem, it is much easier to translate it into the “language” of
mathematics, and then solve it using known mathematical procedures.
58 | P a g e
6. Time Value of Money
6.1 Interest
6. Time Value of
Money
6.1 Interest
6.2 Deposits
6.3 Loans
Simple
Interest
Savings
Repayments
Compound
Interest
Saving
Rates
Instalments
Effective
Interest
Prerequisites:
Effective
Interest
Interest is the reason for the growth of money. If
interest is paid on interest, then we must use
exponential expressions,
expression an indispensible part of
interest procedures.
Learning Targets: Interest is the reason why the value of money
depends on the time. This key fact is fundamental in
understanding financial operations,
operations and therefore the
basis of banking. We will discuss various aspects of
interest, like simple and compound interest. In order
to compare retail products,
product an understanding of
effective interest is essential.
6.1 Interest
P a g e | 59
INTEREST IS THE PRICE PAID FOR THE USE OF SAVINGS OVER A GIVEN
PERIOD OF TIME.
From Wikipedia:
"In the Middle Ages, time was considered to be property of God.
Therefore, to charge interest was considered to commerce with God's
property. Also, St. Thomas Aquinas, the leading theologian of the
Catholic Church, argued that the charging of interest is wrong because it
amounts to "double charging", charging for both the thing and the use of
the thing. The church regarded this as a sin of usury; nevertheless, this
rule was never strictly obeyed and eroded gradually until it disappeared
during the industrial revolution.
Usury has always been viewed negatively by the Roman Catholic
Church. The Second Lateran Council condemned any repayment of a
debt with more money than was originally loaned, the Council of Vienna
explicitly prohibited usury and declared any legislation tolerant of usury
to be heretical, and the first scholastics reproved the charging of interest.
In the medieval economy, loans were entirely a consequence of necessity
(bad harvests, fire in a workplace) and, under those conditions, it was
considered morally reproachable to charge interest.
In the Renaissance era, greater mobility of people facilitated an increase
in commerce and the appearance of appropriate conditions for
entrepreneurs to start new, lucrative businesses. Given that borrowed
money was no longer strictly for consumption but for production as well,
it could not be viewed in the same manner. The School of Salamanca
elaborated various reasons that justified the charging of interest. The
person who received a loan benefited and one could consider interest as a
premium paid for the risk taken by the loaning party. There was also the
question of opportunity cost, in that the loaning party lost other
possibilities of utilizing the loaned money. Finally and perhaps most
originally was the consideration of money itself as merchandise, and the
use of one's money as something for which one should receive a benefit
in the form of interest. Martín de Azpilcueta also considered the effect of
time. Other things being equal, one would prefer to receive a given good
now rather than in the future. This preference indicates greater value.
Interest, under this theory, is the payment for the time the loaning
individual is deprived of the money.
60 | P a g e
6. Time Value of Money
Economically, the interest rate is the cost of capital and is subject to the
laws of supply and demand of the money supply. The first attempt to
control interest rates through manipulation of the money supply was
made by the French Central Bank in 1847.
The first formal studies of interest rates and their impact on society were
conducted by Adam Smith, Jeremy Bentham, and Mirabeau during the
birth of classic economic thought. In the early 20th century, Irving Fisher
made a major breakthrough in the economic analysis of interest rates by
distinguishing nominal interest from real interest."
6.1.1 Simple Interest
Simple interest is interest that is paid (earned) on only the original
amount, or principal, that was borrowed (lent). The amount (in euro) of
simple interest is a function of three variables:
1. the original amount borrowed (lent), or principal,
2. the interest rate per interest period, and
3. the number of periods for which the principal is borrowed (lent).
The interest period is the time which elapses between successive dates
at which interest payments are due. Often, the interest period is one year.
However, we are going to discuss shorter periods as well, e.g., months or
days.
Simple interest is interest that is paid (earned) on the principal only. For
the deposit D, the interest earned at a rate of i after one period is:
I = D⋅i
P a g e | 61
6.1 Interest
D
t
I = D ⋅i
I = D ⋅i
.....
I = D ⋅i
interest rate i
Figure 6-2:
Simple interest
After n periods the total interest paid is:
In = D ⋅ i ⋅ n
The balance of the account is still D. In Germany it is possible to buy
certain types of bonds which are provided with simple interest.
EXAMPLE
For the deposit D = 200 EUR and an interest rate of i = 5% per year
with the interest paid at the end of each year:
I = D ⋅ i = 200 ⋅ 0.05 = 10 EUR
Simple interest will usually be applied to savings accounts for payments
within the year. Assume that on 12 February EUR 200 is deposited into a
savings account with 6% interest. If you look at the account on 7 July
nothing will have happened. The balance is still EUR 200. However, if
you close the account on that day, you get back your original EUR 200
plus simple interest for the time you left the money in the bank.
Simple interest is usually calculated on a daily basis. In Germany, the
banking year has 360 days and 12 months, meaning that every month has
the same number of 30 days. Another regulation is that either the day of
in-payment or the day of withdrawal can be counted as an interest day,
but not both.
62 | P a g e
6. Time Value of Money
Between 12 February and 7 July we can therefore count 145 interest
i
0.06
days. For each interest day
=
= 0.0001667 = 0.01667% interest
360 360
is paid, i.e. for 200 EUR and 145 days: 4.83 EUR.
We will refer to simple interest again in Section 6.2, when we discuss
savings.
6.1.2 Compound Interest
In Volume 1, Section 4.4.3 we learned about compound interest in
connection with the exponential function. Compound interest is the
concept of adding accumulated interest back to the principal, so that
interest can also be earned on interest from that moment on. Declaring
that interest is part of the principal is called compounding (i.e. interest is
compounded). The interest may be compounded every year, but also in
shorter (or even longer time periods). If compounded quarterly, the
interest is added to the principal every quarter (= 3 months); monthly
(one month); or even daily. We can also say that interest is capitalised,
i.e. added to the balance.
Interest rates must be comparable in order to be useful, and in order to be
comparable, the interest rate and the compounding frequency must be
known. Because most people think of rates as a yearly percentage, many
governments require financial institutions to disclose a comparable
yearly interest rate on deposits or advances. Compound interest rates may
be referred to as effective interest (also: Annual Percentage Rate,
Effective Annual Rate, and other terms). When a fee is charged up front
to obtain a loan, effective interest usually counts that cost as well as the
compound interest when calculating the converted rate. These
government requirements assist consumers in comparing the actual cost
of borrowing.
P a g e | 63
6.1 Interest
D
interest
period
t
.....
balance after 1 period:
B1 = D + D ⋅ i = D ⋅ (1 + i )
balance after 2 periods:
B2 = B1 + B1 ⋅ i = B1 ⋅ (1 + i )
balance after n periods:
interest rate i
Bn = Bn−1 ⋅ (1 + i )
Figure 6-3:
Compound interest
Suppose a principal (or deposit) of D (in euro) yields interest at a rate i %
per period. After the first period the balance will be:
B1 = D ⋅ (1 + i ) with the interest rate i
The interest rate may either be given as a percentage of the principal, i.e.
i = 5%, or – as is more common in practice and for most of this chapter –
as a decimal, i.e. i = 0.05.
After another period, capital and interest will be:
B2 = B1 ⋅ (1 + i) = D ⋅ (1 + i)2
After n periods the balance is:
Bn = D ⋅ (1 + i ) n
This is the formula for compound interest. The term implies that interest
paid (earned) on a deposit is periodically added to the principal. The term
q = 1+ i
is called the interest factor or often the compound factor. For each
interest period the capital will be multiplied by q.
64 | P a g e
6. Time Value of Money
FUTURE VALUE
A principal D, compounded with rate i will increase within n interest
periods to:
Bn = D ⋅ q n with q = 1+i being the interest factor
Bn is also called the future value of the principal D after n interest
periods.
Both formulas (that of simple interest and that of compound interest)
contain four basic financial terms:
•
D = principal, deposit, or present value
•
i = interest rate per time period (for instance one year), or in many
cases the interest factor q
•
n = number of periods (e.g. years)
•
Bn = balance after n interest periods, or future value
These four basic variables will occur in some form in every model. Three
of the four terms are usually given, and the fourth is then found.
D
today
eoy 1
eoy 2
...
eoy n
eoy n-1
Bn = ?
interest rate i
interest factor q = 1 + i
eoy = end of year
Figure 6-4:
Future value of money
t
6.1 Interest
P a g e | 65
We will begin with the most obvious case: Looking for the future value
of a principal D deposited n interest periods ago, we can calculate the
future value of the deposit using the known interest rate and number of
periods. Positioning "today" at the point in time for which we want to
know the balance (see Fig. 6-4) produces the standard compound interest
formula:
Bn = D ⋅ q n
EXAMPLE
Anna deposited D = 1000 EUR in a saving account with an interest
rate of 6%. What is the balance after 8 years?
B8 = D ⋅ (1 + i)8 = D ⋅ q8 = 1000 ⋅1.068 = 1593.85 EUR
As in all problems of financial mathematics, it is advisable to check
whether the result is roughly comprehensible or not. This is not only a
nice way of verifying whether the calculations are correct or not, but is
also effective training for your daily work.
Cross check
Anna will get at least EUR 60 annually. That means that in 8 years she
will earn at least EUR 480. Interest on interest will add some
additional amount to the balance. Therefore the result is probably
correct.
We all realise that 1 euro today is worth more than 1 euro to be received
one or two years from now, assuming that interest can be earned.
Calculating the present value of future cash flows allows us to place all
cash flows on one level field, or reference point, such that comparisons
can be made about present values.
66 | P a g e
6. Time Value of Money
D =?
today
eoy 2
eoy 1
eoy 8
...
t
eoy 7
B8 = 2500 EUR
interest rate i = 0.06
eoy=end of year
Figure 6-5:
Present value of money
PRESENT VALUE
Finding the present value is simply the reverse of compounding:
D = Bn ⋅
1
(1 + i )
n
= Bn ⋅
Note that the term
1
1+i
1
q
=
n
1
q
= Bn ⋅ v n
is the reciprocal of the interest factor q. It is
often called the discount factor, denoted by v =
1
q
= 1+1 i . Therefore,
finding the present value (or discounting) of a given future value consists
of multiplying by the discount factor for every interest period.
EXAMPLE
How much does Anna have to deposit if she wants to have a balance
of B8 = 2500 EUR after 8 years, assuming a constant interest rate of
6%?
D=
B8
(1 + i )
8
= B8 ⋅ v8 = 2500 ⋅
( 1.061 )
8
= 2500 ⋅ 0.9433968 =
D = B8 ⋅ v8 = 2500 ⋅ 0.62741 = 1568.53 EUR
P a g e | 67
6.1 Interest
Sometimes we are faced with a situation in which we know the future
and present values as well as the number of periods involved. What is
unknown, however, is the compound interest rate i, implicitly supplied in
the described situation. Again, three of the basic terms are known
( D, Bn , and n ); the fourth, the interest rate i, is unknown and should be
calculated.
Solving the formula of compounding interest for i results in:
1
Bn = D ⋅ (1 + i ) = D ⋅ q
n
n
B
→ q = n
D
n
→ q = 1+ i = n
Bn  Bn  n
→
=
D  D 
1
 B n
i =  n  −1
 D
INTEREST RATE
The interest rate i, given the present and the future value as well as the
number n of interest periods, is:
i=n
Bn
D
−1
EXAMPLE
We assume that Anna has only EUR 1400 to deposit. She wants to
know what the interest rate must to be in order to reach a balance of
EUR 2500 after 8 years?
1
2500
 2500  8
i=8
−1 = 
 − 1 = 1.075168 − 1 = 0.075168
1400
 1400 
Hence, the interest rate must be 7.5168%. Now she has to find a bank
which offers this rate of interest.
68 | P a g e
6. Time Value of Money
Cross check
The interest rate must definitely be greater than 6%, based on Anna’s
deposit. 7.5% is a realistic rate.
At times we may need to know how long it takes for an amount deposited
today to grow to a certain future value at a particular compound interest
rate. In this case the time is being calculated, in particular, the number of
periods an amount has to be compounded in order to reach a certain
balance.
This time, the three basic terms given are: D, Bn , and i , and we are
looking for the number of interest periods n until the required balance is
reached.
Now we have to solve the compound interest formula for n. Since n
occurs in the exponent we have to apply the logarithm (see: Section 5.2,
Page 43) in order to invert the term, such that we can isolate the
unknown. The sub index n of the symbol Bn is certainly not the variable
n but only a part of the balance name.
Bn = D ⋅ q n with q = 1 + i
qn =
Bn
D
→ n ⋅ ln q = ln
Bn
= ln Bn − ln D
D
n ⋅ ln q = ln Bn − ln D → n =
ln Bn − ln D
ln q
NUMBER OF INTEREST PERIODS
The number of interest periods n to reach a given future value is:
n=
ln Bn − ln D
ln q
It should be clear that instead of the natural logarithm "ln", any other
logarithm could also have been applied.
6.1 Interest
P a g e | 69
EXAMPLE
Anna was not successful in finding another bank which offers her the
7.5% interest rate she needed to reach a balance of EUR 2500 within 8
years. Therefore, at an interest rate of 6%, she has to wait longer than
8 years until her deposit of EUR 1400 grows to EUR 2500. How long
will it take?
D = 1400; Bn = 2500 ; q = 1.06
n=
ln Bn − ln D ln 2500 − ln1400
=
= 9.95 years
ln q
ln1.06
That means Anna has to wait almost 10 years.
Cross check
Anna did not find a bank offering a higher interest rate. The time until
the final amount is reached must definitely be longer than 8 years.
Since she has to earn about EUR 160 in interest, “two additional years
at the rate of 6% for EUR 1400” sounds plausible.
ATTENTION
Please note, that in all calculations the intermediate results shown in the
formulas are always displayed with a limited number of decimals,
although the maximal number of decimals on the calculator are always
used for calculations and final results. Final result and calculations with
fewer decimals may therefore be slightly different.
6.1.3 Effective Interest
Up to now, we have assumed that interest is paid annually, as this
assumption makes it easiest to get a basic understanding of the time value
of money. However, we are quite regularly confronted with
compounding periods shorter than a year. What is the relationship
between future value and interest rates in these situations?
70 | P a g e
6. Time Value of Money
To begin with, suppose that the interest is paid semi-annually, that is,
twice a year. We have already discussed the situation where
compounding periods are less than a year and the number of
compounding operations increases as time intervals become shorter (see
Section 5.1.4; Page 28).
It is now very important to have a clear understanding of the relationship
between time intervals, the number of compounding operations and the
applied interest rate.
•
If nothing else is mentioned, the interest period is generally a
year. Any other period must explicitly be stated!
•
The interest rate stated is usually the nominal interest rate,
implicitly assuming an interest period of one year.
•
In case of shorter compounding periods, the nominal interest rate
has to be adapted accordingly.
If you deposit D = 100 EUR in a savings account at a nominal 9%
interest rate, the future value at the end of the first half a year would be:
B 1 = D ⋅ (1 + 2i ) = 100 ⋅ (1 + 0.045) = 100 ⋅ (1.045) = 104.50 EUR
2
At the end of a full year, the future value would be:
B1 = B1 ⋅ (1.045) = 104.50 ⋅ (1.045) = 109.20 EUR
2
This amount has to be compared with Bɶ1 = 109 EUR, in which interest
was paid only once a year. The difference of ∆B = B1 − Bɶ1 = 0.20 EUR is
caused by interest earned in the second half of the year on the interest of
4.50 EUR that was earned at the end of the first half a year. This is the
effect of the compounding within the year.
We will now generalise the above problem by compounding p times in a
year – which we already did in Section 5.1.4 when discussing the
exponential function. We therefore divide the year into p periods. If we
assume a nominal interest rate i and interest paid periodically at a rate
i p = pi → q p = 1 + pi , the balance after p periods (= one year) is:
B1 = D ⋅ (1 + i p ) p
P a g e | 71
6.1 Interest
D
today
...
eop 2
eop 1
eop p
t
eop p-1
B1
interest rate per period i p
eop=end of period
Figure 6-6:
Interest compounded p times per year
The following Tab. 6.1 contains the future values of the deposit D = 100
EUR at the end of the first year, with different compounding periods, and
continuous compounding at a nominal interest rate of 8% (please refer to
Volume 1, Section 4.4.4):
Periods p
per year
1
B1 in EUR
108
3
6
12
360
108.22 108.27 108.30 108.33
Table 6.1:
continuous
108.33
Future values
The shorter the compounding period, the more periods there are and the
higher the future value; the maximum possible value can be reached with
continuous compounding − which does not differ from daily
compounding inside the first two decimals.
A consumer who wants to take out a loan may be faced with several
offers from competing financial institutions. It is therefore of
considerable importance to compare the various offers, especially if they
are based on different assumptions. The effective interest rate is often
used in making such comparisons.
For instance, comparing the effective annual interest rates applied to
Table 6.1, the interest paid (or earned) is:
72 | P a g e
Periods p per
year
Effectively paid
interest
6. Time Value of Money
1
8%
3
6
12
360
8.22% 8.27% 8.3% 8.33%
Table 6.2:
continuous
8.33%
Effective interest
It is clear that the yearly paid interest is higher when interest is
compounded more often within the year.
If we want to compare investments that have different compounding
periods, we need to state their interest using some common or
standardised form. This leads us to distinguish between nominal interest
rate and effective annual interest rate ieff .
In banking, and even more generally in the lending business, the concept
of effective interest rates is becoming more and more important.
EFFECTIVE INTEREST RATE
The effective interest rate ieff is always the interest rate per year taking
into account the total cost factors.
Basically, there are three factors that can create a difference between
nominal rate and the effective rate, though it should be obvious that in
credit the effective interest rate is always higher than the nominal rate:
1.
The compounding period may be less than a year, causing additional
interest on interest.
2.
Bank charges for services are not covered by the interest rate. These
fees are usually collected once, generally at the beginning of the
contract; sometimes just as a fixed cost, sometimes also depending
on the amount (of a loan, for example). In some cases, fees might
even be collected several times during the contract. Fees are costs for
the client and therefore increase the total cost of the money
borrowed.
P a g e | 73
6.1 Interest
3.
The nominal interest rate changes during the contract time (floating
interest). In this case, the effective interest rate is the average interest
rate for the entire contract time when compounding yearly. However,
“average” does not mean that the average is calculated as the
arithmetic mean of the interest rates!
We often see a combination of all three approaches.
The effective interest rate is always calculated as a yearly rate on the
basis of all costs charged to the client. In Germany, it is set in the law
that the effective interest rate must be explicitly stated on every contract
involving interest rates. As can be seen, calculating the effective interest
rate is a very important topic in banking and finance.
COMPOUNDING PERIODICALLY
If the interest is compounded periodically for p periods during a year at a
nominal rate i, the interest earned on interest influences the effective rate.
Shorter compounding interest periods increase the effective interest.
D
today
eop p
eop 1
eop 2
eop p-1
t
B1
interest rate per year (nominally) i
interest rate per period i p = pi
eop = end of period
Figure 6-7:
Compounding periodically
Let us assume that a deposit is compounded p times per year. After 1
year the balance will be:
74 | P a g e
6. Time Value of Money
( )
B1 = D ⋅ 1 +
i
p
p
Because this has to be the same amount if compounding yearly with the
effective interest rate, the following equation arises:
(
D ⋅ (1 + ieff ) = D ⋅ 1 +
i
p
)
p
D can be cancelled out, which means that the effective rate is
independent of the amount deposited.
EFFECTIVE INTEREST RATE: PERIODICALLY COMPOUNDING
The effective interest rate is independent of the amount D. For a given
nominal interest rate i, it increases with the number of periods p:
(
ieff = 1 +
i
p
)
p
−1
For continuous compounding, the effective interest rate is:
ieff = ei − 1
The same argumentation holds if we consider more than just one year –
which is commonly the case.
Let us assume again that a deposit is compounded p times per year. After
n years, the balance will be (see Volume 1, Section 2.2.1, or Volume 1,
Section 4.4.2):
(
Bn = D ⋅ (1 + pi ) p
)
n
= D ⋅ (1 + pi ) n⋅ p
This balance must also be the result when we compound the principal D
effectively over n years, because the effective interest rate is always the
correct rate (coving all adjustments) for exactly one year:
Bn = D ⋅ qeff n = D ⋅ (1 + ieff ) n
The right-hand sides of the last two equations must also be equal:
P a g e | 75
6.1 Interest
(
Bn = D ⋅ 1 +
i
p
)
n⋅ p
(
= D ⋅ 1 + ieff
)
n
Cancelling the deposit D (the effective interest rate should not depend on
the deposit!) and removing the n-th root from both sides will result in:
n
(1 + pi ) n⋅ p = n (1 + ieff ) n
→ (1 + pi ) p = 1 + ieff
The same result is found as before:
ieff = (1 + pi ) p − 1
EXAMPLES
1.
What is the effective yearly rate ieff corresponding to a nominal
annual interest rate of i = 10% with interest compounded:
a)
each quarter
b) each month?
a)
p = 4 → : ieff = 1 + 0.1
4
(
)
4
− 1 = 0.1038
effective interest rate is 10.38%
(
b) p = 12 → : ieff = 1 + 0.1
12
)
12
− 1 = 0.1047
effective interest rate is 10.47%
Cross check
In both cases, the effective interest rate must be higher than the
nominal 10% − which is the case. Also, and the rate in case b) must
be higher than in case a), because the compounding periods are
shorter. This is also true.
76 | P a g e
2.
6. Time Value of Money
The deposit of D = 200 EUR is compounded periodically by p=1,
3, 6, 12, 360 periods and continuously.
a)
Calculate the balance after 3 years at the nominal interest rate
of 10%.
b) Calculate the effective interest rates of the different
compounding periods: p = 3, 6, 12, 360, continuous
Periods p 1
per year
3
6
12
360
continuous
a) in EUR
266.20
268.65
269.31
269.64
269.96
269.97
b) ieff
0.1
0.1034
0.1043
0.1047
0.1052
0.1052
Table 6.3:
3.
Effective interest rates
The monthly interest for micro loans with 6 months maturity is
2%. What are the nominal and the effective interest rates for this
loan?
The nominal interest rate is: 12 ⋅ im = 12 ⋅ 2% = 24%
Despite a maturity of only 6 months, the effective interest rate is
always calculated on the basis of a complete year! That means that the
effective interest rate is:
ieff = (1 + im )12 − 1 = 1.0212 − 1 = 0.2682 = 26.82%
Cross check
The nominal interest rate is the yearly rate, i.e. the monthly rate times
12. The effective interest rate is always greater than the nominal rate:
26.8% is convincing.
Sometimes it is interesting to calculate the interest rate for a given period
so that a certain effective rate can be reached. Simply invert the question
and, consequently, the calculation: the result is the effective interest rate.
P a g e | 77
6.1 Interest
What should be the monthly (quarterly, or weekly) interest rate, so that
the corresponding compounding leads to the intended target value? In
order to answer this question we use the above formula:
ieff = (1 + pi ) p − 1 = (1 + i p ) p − 1
and solve it for i p :
(1 + i p ) p = 1 + ieff = qeff → 1 + i p = p qeff
→ i p = p qeff − 1
EXAMPLE
The effective rate of ieff = 8% is the result. What is the equivalent
interest rate if you compound
a)
quarterly? →
b) monthly? →
c)
daily?
→
i4 = 4 1.08 − 1 = 0.01943 , i.e. 1.943%
i12 = 12 1.08 − 1 = 0.006434 , i.e. 0.6434%
1
i360 = 360 1.08 − 1 = (1.08) 360 − 1 = 0.0002134 ,
i.e. 0.02134%
78 | P a g e
6. Time Value of Money
EXERCISE 6.1:
INTEREST
1. Assume the nominal interest rate is 8%. How much simple
interest will be paid on:
a)
EUR 200 for 5 months?
b) EUR 600 for 127 days?
c)
EUR 800 for the time between 18 March (day of deposit) and
10 October (day of closing the account)?
2. An investment of EUR 1500 is compounded 10 times at an
interest rate of 4.5%. What will the final balance be?
3. EUR 200 is deposited into a savings account at 5% interest. What
will the value of this money be after 6 years?
4. EUR 500 is invested in a savings account with 1% interest
compounded monthly. What will the balance of this account be
after 15 months?
5. The balance of an account is EUR 785.34. How much must have
been invested 3 years ago, if the interest rate was a constant 4%?
6. How much should you invest today if you need EUR 2000 in 5
years at a fixed interest rate of 7%?
7. Your account (0.75% interest compounded monthly) has a
balance of EUR 3211.56. What was the balance 15 months ago?
8. You have invested EUR 1250.
a)
Calculate the interest rate, so that the principal doubles within
10 years, without additional deposits.
b) Assuming monthly compounding, calculate the monthly
interest rate.
P a g e | 79
6.1 Interest
9. How long does it take to double your investment of EUR 1250 at
an interest rate of 6%? (Without additional clarification the
interest rate is understood to be a yearly compounding rate. Thus
the number of interest periods, and therefore the answer, should
be in years.)
10. The balance of your saving account is EUR 278.66. How long do
you have to wait until the balance is EUR 400, assuming a
monthly compounding interest rate of 0.8%?
11. Calculate the effective interest rate for:
a)
a monthly compounding rate of 1.5%
b) a quarterly compounding rate 0f 4.5%
c)
a bi-monthly rate of 4%
12. The nominal yearly interest rate is 9%. Calculate the effective
interest rate for:
a)
quarterly compounding
b) monthly compounding
c)
daily compounding, assuming that the banking year has 360
days
13. The effective interest rate is 8.5%. Calculate the interest rates for
the following compounding periods which lead to this given
effective rate:
a)
quarters
b) months
c)
days
80 | P a g e
6. Time Value of Money
ANSWERS 6.1:
1.
INTEREST
a)
I 5m = 6.67 EUR
c)
202 interest days → I 202 d = 35.91 EUR
2.
B10 = 2329.45 EUR
3.
B6 = 268.02 EUR
4.
B15 m = 580.48 EUR
5.
I = 698.16 EUR
6.
I = 1425.97 EUR
7.
B = 2871.05 EUR
8.
a)
9.
n = 11.9 years
i = 7.177%
b)
I127 d = 16.93
b)
im = 0.5793%
10. n = 45.36 months
11. a)
ieff = 19.56%
b)
ieff = 19.25%
c)
ieff = 26.53%
12. a)
ieff = 9.31%
b)
ieff = 9.38%
c)
ieff = 9.42%
13. a)
iq = 2.06%
b)
im = 0.682%
c)
ieff = 0.0227%
P a g e | 81
6.1.4 Progress Test "Interest"
6.1.4 Progress Test “Interest”
You should allot yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Do not use the reader to look
for the solution. The aim of the test is to get feedback on how much you
know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. In any other
case (you got the wrong answer or no answer at all) you should repeat the
corresponding units in order to close the gap.
1. Anna opens a savings account at 8% interest and deposits EUR
300 on 25 May. She closes the account on 6 November. How
much will she receive, provided simple interest is paid?
2. Anna invests EUR 500 at a nominal interest rate of 10%. What
will the balance be after 6 years, if interest is compounded:
a)
annually?
b) quarterly?
c)
monthly?
3. What will the effective interest rates be for Problem 2?
4. Anna would like to deposit a certain amount so that she has EUR
800 in her account after three years at a rate of 0.57%,
compounded monthly. How much does she have to deposit?
5. How long must she wait under the conditions of Problem 4, if she
has only EUR 550 for deposit?
6. Procredit agrees to increase the interest rate. What must the new
monthly interest rate be so that Anna receives EUR 800, although
she deposited only EUR 550 three years ago?
82 | P a g e
6. Time Value of Money
7. The effective interest rate is 14.5% for a ProCredit loan account.
What is the equivalent interest rate compounded:
a)
quarterly?
b) bi-monthly?
c)
monthly?
P a g e | 83
6.2 Deposits
6.2 Deposits
6. Time Value of
Money
6.1 Interest
6.2 Deposits
6.3 Loans
Simple
Interest
Savings
Repayments
Compound
Interest
Saving Rates
Instalments
Effective
Interest
Prerequisites:
Effective
Interest
The concept of interest and calculations of cash flow
as a function of time are essential for the calculation
of deposits.
Learning Targets: The basic rules of handling deposits will be
discussed. We will begin by calculating the futire
value of single investments add
a
then extend the
discussion to multiple deposits. Saving at fixed rates
leads to another important tool in financial
mathematics: the application of geometric series.
84 | P a g e
6. Time Value of Money
The two most basic operations of our banks are deposits and loans. Both
operations rely heavily on the concept of the “time value of money”, i.e.
interest. In the case of deposits, the client provides the money and the
bank owes it and thus pays interest. In the case of loans it is just the
opposite, the bank provides (disburses) the money and the client owes it
and pays interest for this service. In principle there is no difference when
analysing the processes. The procedures of dealing with interest is
identical in both situations – the distinction is only in perspective: in the
case of deposits, the client is the owner and the bank owes money. In the
case of loans, it is just opposite.
In this section we will cover the basic operations of saving. Many aspects
were already discussed in the last section. In order to avoid too much
redundancies, we refer often to the results of the former analyses.
6.2.1 Savings
Savings is a deposit in a specially designed account. Money is deposited
into the account, and beginning the next day, interest is calculated. The
client knows the nominal rate. In most cases this is equal to the effective
interest rate, because in most countries the interest is paid (capitalised =
added to the balance) on a yearly basis at the end of the calendar year,
and there are usually no fees on saving accounts.
SINGLE PAYMENTS
For deposits within a year, simple interest is paid (see Section 6.1.1; Page
61) only if the account is closed. In this case, the balance B is paid out
using the deposit D plus the interest earned for the days between inpayment (not included) and the day of out-payment (included). The
difference represents the number of interest days d.
P a g e | 85
6.2 Deposits
BALANCE AT SIMPLE INTEREST
Simple interest is paid for the deposit D at the nominal interest rate i for
d interest days:
(
i = D ⋅ 1+ d ⋅ i
Bd = D + D ⋅ d ⋅ 360
360
)
D
today
d interest days
Bd
interest rate i
interest factor q = 1 + i
Figure 6-8:
Savings account
EXAMPLE
Anna opens an account with a 5% interest rate by depositing EUR 400
on 25 May, and then close it on 12 October. What will the balance be?
On the basis that each month has 30 days, day of in-payment
excluded, day of out-payment included, we obtain 5 days in May + (4
months =) 120 days + 12 days in October = 137 interest days.
(
)
B137 = 400 ⋅ 1 + 137 ⋅ 0.05
= 407.61 EUR
360
Cross check
5% for the whole year would be EUR 20. Four months is a third of the
year, i.e. roughly EUR 7. Therefore the result is probably correct.
If the account exists beyond the end of the year, the interest earned
during the old year is capitalised on 31 December and then added to the
account. From the next day on, interest is paid also for the added interest.
t
86 | P a g e
6. Time Value of Money
The capital at the end of the old year with d o interest days is:
d ⋅i
o
Bo = D ⋅ (1 + 360
)
In the new year, interest is paid on Bo for d n interest days:
d ⋅i
d ⋅i
d ⋅i
n
0
n
Bn = Bo ⋅ (1 + 360
) = D ⋅ (1 + 360
) ⋅ (1 + 360
)
EXAMPLE
Anna opened an account with a 5% interest rate by depositing EUR
400 on 25 May 2007, and then close it on 18 April 2008. What will
the balance be?
Number of interest days in 2007: 5 days in May + (7 months =) 210
days in 2007 = 215 days
Number of days in 2008: (3 months =) 90 days + 18 days in April =
108 days
⋅0.05 ) ⋅ (1 + 108⋅0.05 ) = 418.12 EUR
B = 400 ⋅ (1 + 215360
360
Cross check
As in the previous example, interest for one complete year will be
EUR 20. More than 7 months, roughly three qaurters, i.e. EUR 15
interest, plus interest on interest. EUR 18 is probably correct.
SUCCESSIVE PAYMENTS
Suppose that three successive deposits are made for the amounts
D1 = 400 EUR at the beginning of the first year, then D2 = 700 EUR one
year later, and D3 = 300 EUR another year later. How much will be in
the account at the end of year 3, given an interest rate of 6% per year?
According to our notation, the three deposits are positioned in Fig. 6-9 as
arrows from above. We now have multiple payments plus balancing.
P a g e | 87
6.2 Deposits
D1 = 400
D 2 = 700
D 3 = 300
today
t
B3
interest rate per year i
Figure 6-9:
Successive payments
Since the cash flow changes with time, we can only add, subtract,
compare, and balance cash flows when all the payments are adjusted to
one point in time. This date is usually marked extra as the reference
point, or often simply by the word “today”.
If we want to know the account balance after 3 years, we must position
the reference point at the end of the third year. The only present value is
the balance B3 . The three deposits are past values.
BASIC RULES FOR BALANCING
There are two basic rules for calculations in finance, which always have
to be taken into consideration:
1.
All payments have to be adjusted to one point in time. After this
point in time is fixed, all past payments have to be compounded, all
future payments have to be discounted, and today’s payments are
present values.
2.
The sum of all compounded (or discounted) in-payments must be
equal to the sum of all compounded (discounted) out-payments.
Payments in the above sense are not necessarily physical cash flows. For
instance, if there are many deposits to an account and we ask for the
88 | P a g e
6. Time Value of Money
balance at a certain time, then we assume a withdrawal will comply with
Rule 2 above (balancing being an out-payment).
In the above example, we refer all payments to “today” (Rule 1):
Sum of deposits: B3 (present value)
Sum of payments:
D3 compounded one year →
compound factor
D3 ⋅ q
with
q = 1 + i being the
D2 compounded two years → D2 ⋅ q 2
D1 compounded three years → D1 ⋅ q 3
Rule 2 states that the sum of all in-payments and the sum of all outpayments should be the same:
B3 = D3 ⋅ q + D2 ⋅ q 2 + D1 ⋅ q 3
Thus, with q =1.06 the balance after 3 years is:
B3 = 318 + 786.52 + 476.41 = 1580.93 EUR
If we position our reference point differently (for instance in Fig. 6-10 at
the first deposit) then the calculation will be different; however, the result
will be the same.
D1 = 400
D 2 = 700
D3 = 300
t
today
B3
interest rate per year i
Figure 6-10: Positioning "today"
P a g e | 89
6.2 Deposits
If we balance all payments to the point of the first deposit, we obtain:
D1 + D2 ⋅ v + D3 ⋅ v 2 = B3 ⋅ v 3 with v =
1
q
= 1+1 i being the discount factor
Multiplying this equation by q 3 and bearing in mind that v = 1q , we can
prove that the balance B3 is the same as before.
6.2.2 Regular Equal payments
If all payments are equal ( D1 = D2 = D3 = a ) then the value of all
deposits can be calculated by applying the geometric series. Since equal
rates of payment and deposits occur quite often, we will discuss this in
more detail.
EXCURSION
The geometric series is often applied in financial mathematics. It is the
sum of terms, starting with the number 1 and always increasing by a
constant factor q. The geometric series with the constant factor q and n
terms is:
1 + q + q 2 + q 3 + ... + q n −1
Note that there are n term although the last exponent is n − 1 because we
start with 1 = q 0 . This sum occurs very often, with q being the interest
factor. There is a very convenient formula that avoids calculating the
sum:
1 + q + q + q + ... + q
2
3
n−1
1 − qn qn −1
=
=
1− q
q −1
Arriving at this formula is not difficult but is, however, beyond the scope
of this course.
An annuity is a sequence of equal payments made at set intervals over
some time span (although the word “annuity” suggests annual payments,
90 | P a g e
6. Time Value of Money
this is not necessarily the case). Let us assume there are n such inpayments at the beginning of each year. What will the balance be at the
end of the nth year?
a
a
a
....
a
today
t
Bn
interest rate per year i
interest factor per year q =1+ i
Figure 6-11: Regular equal payments
Assuming n regular deposits of value a at the start of each year, we are
interested in the balance at the end of year n. The interest rate i is fixed.
Balancing all in-payments requires that we set a reference point in time.
Since we are interested in the final balance, it is quite natural to select the
end of year n as the reference point, and adjust all payments to this point
in time. Because all payments are past payments, they have to be
compounded.
•
The last payment is one year old; it has to be compounded once:
→ a⋅q
•
The preceding payment is two years old; it has to be compounded
twice: → a ⋅ q 2
•
The preceding payment is three years old; it has to be
compounded three times: → a ⋅ q 3
•
… and so forth …
•
The first payment is n years old; it has to be compounded n times:
→ a ⋅ qn
Now we have adjusted all in-payments and their sum will be the balance:
P a g e | 91
6.2 Deposits
{
}
Bn = a ⋅ q + a ⋅ q 2 + a ⋅ q 3 + ... + a ⋅ q n = a ⋅ q ⋅ 1 + q + q 2 + ... + q n−1
The common factor a ⋅ q can be excluded. The sum in the brackets is the
geometric series resulting in:
{
}
Bn = a ⋅ q + a ⋅ q 2 + ... + a ⋅ q n = a ⋅ q ⋅ 1 + q + q 2 + ... + q n−1
Bn = a ⋅ q ⋅
qn − 1
qn − 1
= a⋅q⋅
q −1
i
This formula is one of the most used calculations in financial
mathematics. It occurs regularly whenever constant payments at regular
intervals arise: savings, loans, insurance, pensions, etc.
BALANCE AFTER N INSTALMENTS
Assuming regular payments of a per period and an interest rate i (→
compounding factor q = 1 + i ), the balance after n periods is:
Bn = a ⋅ q ⋅
qn − 1
qn −1
= a⋅q⋅
q −1
i
EXAMPLES
1.
Anna saves EUR 500 every year (in-payment at the beginning of
the year) in an account with 6% interest. What is the balance after
8 years?
a = 500; n = 8; i = 0.06 → q = 1.06
Balance: B8 = 500 ⋅1.06 ⋅
1.068 − 1
= 5245.66 EUR
0.06
Cross check
Sum of all in-payments:
8 ⋅ 500 = 4000 EUR
92 | P a g e
6. Time Value of Money
6% per year for in the average EUR 2000:
0.06 ⋅ 2000 ⋅ 8 = 960 EUR
Taking into account the compounding effect, the result is plausible.
2.
Anna saves EUR 10 monthly in your child’s account. How much
will be in the account when your child turns 18, assuming a
constant interest rate of 0.5% for the duration of the account?
a = 5; n = 18 times 12 months = 216 months; i = 0.005
→ q = 1.005: Bn = ?
B18 y = B216m = 10 ⋅1.005 ⋅
1.005216 − 1
= 3881.28 EUR
0.005
Cross check
Yearly investment: EUR 120 times 18 years:
EUR 2160 in deposits
Average balance of the account: EUR 1080
Interest per year: at least 6% per year → interest at least
1080 ⋅ 0.06 ⋅18 = 1166 EUR
The balance must be in the range of 3326 EUR, which is the case.
qn −1
The formula Bn = a ⋅ q ⋅
for regular successive payments once
i
more shows the four basic financial items:
• the future value:
the balance Bn
• the present value:
the payment a
• the time:
the number of periods n
P a g e | 93
6.2 Deposits
• the interest rate:
the interest rate i, or the compound factor q,
respectively
Again, three must be given in order to calculate the fourth.
The balance is already isolated in the above formula.
If we want to know how much we should pay periodically in order to
obtain a certain balance given the time n and the interest rate i , we have
to solve the formula for a:
a=
Bn ⋅ i
q ⋅ (qn − 1)
EXAMPLE
How much does Anna have to save monthly if the monthly
compounding interest rate is 0.5% and she wants to reach EUR 2000
within 18 months?
Bn = 2000; n = 18; i = 0.005; q = 1.005 : a = ?
→ a=
2000 ⋅ 0.005
(
)
1.005 ⋅ 1.00518 − 1
= 105.93 EUR
Cross check
Saving EUR 2000 within 18 months requires monthly in-payments of
approximately EUR 100, resulting in EUR 1800 plus interest. The
above result makes sense.
Assume Anna cannot afford the EUR 105.93 monthly but instead EUR
85. How long does she have to deposit that regular monthly amount in
order to reach the EUR 2000, provided the interest rate does not change?
Now the present and the future values are given, as well as the interest
rate. We are looking for the time, and have to solve the formula for n:
94 | P a g e
6. Time Value of Money
B ⋅i
qn −1
Bn = a ⋅ q ⋅
→ n + 1 = qn
i
a⋅q
We need to solve this equation for n, which is the exponent of the
compound factor. We have to apply the inverse operation of the
exponential operation, and that is the logarithmic operation. We can
choose any logarithm, for instance:
 B ⋅i 
ln  n + 1 
a⋅q
 B ⋅i 
 B ⋅i 

ln q n = ln  n + 1  → n ln q = ln  n + 1 → n = 
ln q
 a⋅q

 a⋅q

ATTENTION
It is not possible to simplify the expression in the numerator of the above
fraction. A common mistake is to solve the logarithm in the numerator as
follows:
 B ⋅i 
 B ⋅i 
ln  n + 1 = ln  n  + ln1
 a⋅q

 a⋅q 
It is easy to realise that the above calculation must be wrong, because
ln1 = 0 and the remaining two expressions cannot be equal!
EXAMPLE
We will try to find a solution to Anna's problem:
a = 85; Bn = 2000; i = 0.005; q = 1.005 : n = ?
 2000 ⋅ 0.005 
ln 
+ 1
85 ⋅1.005

 = 0.110702 = 22.2 months
n=
ln1.005
0.0049875
6.2 Deposits
P a g e | 95
Cross check
Every month Anna deposits about 20% less; i.e. she has to wait about
20% longer. 20% of 18 months is about 4 mothns: The 22 months
seem to be correct.
Now we would expect to discuss the fourth problem, calculating the
interest rate, provided the present and future values as well as the time. In
this case, a polynomial equation of degree n + 1 has to be solved. This is
a very complex problem, and can only be accomplished by means of an
advanced approximation technique. Therefore, the discussion of this
problem must be omitted here.
96 | P a g e
6. Time Value of Money
EXERCISE 6.2:
DEPOSITS
1.
How many interest days have to be taken into account if the day
of deposit is 17 April 2007 and the day of repayment is 5
September 2007?
2.
Anna deposits EUR 370 into a savings account at 5% nominal
interest on 7 June, and then withdraws it on 6 December. How
much will she receive?
3.
An investment of EUR 500 made on 4 Februaray will become
EUR 530.56 within 188 interest days, provided simple interest is
paid. What is the nominal interest rate?
4.
Anna makes four payments: at the beginning of the 1st year:
EUR 300; the 2nd year EUR 600; the 4th year EUR 500, and the
6th year EUR 800. How much will be in the account at the end
of the 8th year, given an interest rate of 5% per year?
5.
Together with the payments from Problem 4, Anna withdraws
EUR 400 at the beginning of year 5 and EUR 300 at the
beginning of year 7. How much will now be in the account at
the end of year 8, given an interest rate of 5% per year?
6.
Use the payments from the following figure, given an interest
rate of 6%:
D0
D1 = 400
D4 = 480
eoy2
today
eoy4
W2 = 800
interest rate per year i=0.06
eoy = end of year
eoy5
eoy3
eoy1
W3 = 350
W5 = 660
t
P a g e | 97
6.2 Deposits
How much should be deposited "today" in order to cover all
payments?
7.
Use the formula for the geometric series to calculate the sums:
( 12 ) + ( 12 ) + ( 12 ) + ( 12 ) + ( 12 )
2
3
4
5
6
a)
s = 1 + 12 +
b)
s = 1 + 0.3 + 0.32 + 0.33 + 0.34 + 0.35 + 0.36 + 0.37 + 0.38
c)
s = 1 + 1.1 + 1.12 + 1.13 + 1.14 + 1.15
8.
How much does Anna save within 6 years if she deposits EUR
25 every month at an interest rate of 0.6%, compounded
monthly?
9.
Assume an effective interest rate of 8.5%. You save EUR 50
monthly. What will the balance of your account be after 5 years,
if the interest is compounded monthly?
10. How many months does Anna have to invest EUR 50 in order to
save EUR 2500 at a monthly compounding interest rate of
0.8%?
11. How much does Anna have to save monthly at a monthly
compounding interest rate of 0.8%, if she needs EUR 2500 after
three years?
12. How much do you have to save quarterly if you need EUR 5000
within 2 years? The effective interest rate is 8% and interest is
compounded quarterly.
98 | P a g e
ANSWERS 6.2:
6. Time Value of Money
DEPOSITS
1.
Total: 138 interest days
2.
Balance = 379.20 EUR
3.
Nominal interest rate: i = 0.11704 = 11.7%
4.
Balance: B = 2851.74 EUR
5.
Balance: B = 2034.79 EUR
6.
Deposit today: D0 = 741.49 EUR
7.
a) s = 1.9844
b) s = 1.4285
c) s = 7.7156
8.
Balance: B6 = 2256.58 EUR
9.
Balance: B5 = 3716.87 EUR
10. Time: n = 41.94 months
11. Monthly payment: rm = 59.72 EUR
12. Quarterly payment: rq = 572.61 EUR
6.2.3 Progress Test “Deposits”
P a g e | 99
6.2.3 Progress Test “Deposits”
You should allot yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Do not use the reader to look
for the solution. The aim of the test is to get a feedback on how much you
know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. In any other
case (you got the wrong answer or no answer at all) you should repeat the
corresponding units in order to close the gap.
1. What will the balance of Anna's savings account be at 6%
nominal interest at the end of year 8 if she
•
deposits EUR 550 at the beginning of year 1,
•
deposits EUR 300 at the beginning of year 2,
•
withdraws EUR 400 at the beginning of year 3,
•
deposits EUR 600 at the beginning of year 5,
•
withdraws EUR 300 at the beginning of year 7?
2. Anna expects a tax refund of EUR 350 on 1 September, which
she deposits in her account at 0.75% compounded interest. She
wants to buy a notebook for EUR 980 at the end of the year. How
much must she deposit on 1 January?
3. Beginning 1 January, Anna wants to save EUR 80 monthly at
0.75% compounded interest. What will the balance be at the end
of the year?
100 | P a g e
6. Time Value of Money
4. How much would Anna have to save monthly until the end of the
year with 0.75% compounded interest if the balance of her
account at the beginning of the year is EUR 352.12, and she does
not want to spend the tax refund (of problem 2) for purchasing the
notebook at the end of the year?
5. Anna opens a new account with a 9% effective interest rate. How
much must she save monthly so that she can buy the notebook at
the end of the year?
6. Anna cannot afford the monthly amount she would have to pay
according to Problem 5. She can save only EUR 50 per month.
How long must she wait for the notebook?
P a g e | 101
6.3 Loans
6.3 Loans
6. Time Value of
Money
6.1 Interest
6.2 Deposits
6.3 Loans
Simple
Interest
Savings
Repayments
Compound
Interest
Saving Rates
Instalments
Effective
Interest
Prerequisites:
Effective
Interest
The concept of interest and calculations of cash flow
as a function of time are also the basis
bas of loan
calculations.
Learning Targets: The basic rules of handling loans will be discussed.
We will begin with the repayment of the priciple all
at once, and then extend the discussion to several
repayments. The most common way of repaying
loans is with fixed amounts, which leads to the
calculation of instalments.
102 | P a g e
6. Time Value of Money
The fundamental result of the last chapter was that all cash flow has to be
adjusted to a single point in time. Therefore, all past payments have to be
compounded, and all future payments have to be discounted. These basic
rules do not differentiate between in-payments (deposits) and outpayments (withdrawals or credits). Therefore it may not be surprising
that the differences between deposits and credits are only minor. The
general ideas are identical.
A loan implies a disbursement (out-payment) at the beginning of an
agreement, followed by in-payments until the loan is repaid. Usually, a
loan is not repaid all at once, but in the form of instalments. Depending
on the contract, they might be identical for each period (year or month)
or differ periodically.
6.3.1 Repayments
Each repayment always consists of two parts:
•
The principal repayment, i.e. the money owed to the lender.
•
The interest, which is the cost of lending the money.
We will begin the discussion with the easiest case: the total principal is
repaid all at once as a single amount.
SINGLE REPAYMENT
The loan is disbursed at the beginning of the time interval. We assume
that the principal will be repaid after n periods, given the compound
interest rate i per period (see: Fig. 6-12).
P a g e | 103
6.3 Loans
R
eop 2
eop 1
....
eop n-1
eop n
today
t
P
interest rate per period i
eop = end of period
Figure 6-12: Single repayment
We set the reference point "today" at the moment of repayment. R is the
only in-payment and is a present value in this situation. The principle P is
the only out-payment, and is a past value, and therefore has to be
compounded. n periods have passed since the loan was disbursed.
Therefore:
R = P ⋅ (1 + i ) n = P ⋅ q n
If we compare this formula with the formula for the future value of some
deposit D (see: Section 6.1.2; Page 63), we see that they are practically
identical. The difference is simply the direction of the payments.
Again, we have the four basic financial entries: present value (P),
future value (R), interest rate (i), and time (number of periods n). We can
only calculate one of the items if the other three are given.
In practice, single repayments of loans are exceptional. Usually a loan
will be repaid in more than one – often many − repayments.
SEVERAL REPAYMENTS
Let us assume that the principal is repaid in three instalments: the first
R2 after two periods, the second R3 after three periods, and the last one
R4 after four periods. After this last repayment, the principal should be
completely repaid (see: Fig. 6-13). The interest rate i is compounded per
period.
104 | P a g e
6. Time Value of Money
R4
R2
R3
t
today
P
eop = end o period
interest rate per period i
Figure 6-13: Several repayments
We set the reference point at the time of maturity and balance the
payments. Except for R4 , all payments are past payments, which have to
be compounded. The loan is four periods old and has to be compounded
four times. The balance of all in-payments and all out-payments equals:
R4 + R3 ⋅ q + R2 ⋅ q 2 = P ⋅ q 4
Given the principal and, for instance, the first two repayments as well as
the interest rate, we could calculate the last repayment R4 . However, it is
not possible to calculate all three repayments knowing only the principal
and the interest rate. There are too many unknowns to solve the equation.
REPAYMENT PLAN
Each repayment consists in part of interest on the outstanding principal,
and of some principal repayment. The repayments are called instalments.
The interest portion is largest in the first instalment because interest has
to be paid on the whole disbursed amount. If the interest rate is fixed, the
interest portion of the instalment will decrease in the following periods.
At the end of the contract, the instalment consists mainly of principal
repayments and only a small amount of interest, because the principal
amount owed decreases as payments are made.
Instalment payments are prevalent in mortgage loans, auto loans,
consumer loans, and certain business loans. Amortisation of a loan
P a g e | 105
6.3 Loans
involves determining the periodic payment necessary to reduce the
principal amount to zero at maturity, while providing interest payments
on the unpaid principal.
During negotiations about a loan, the bank usually provides a complete
repayment plan, which displays all the relevant information about the
loan and its repayment: the period, the outstanding amount, the interest,
the principal repayment, and – the sum of interest and principal
repayment – the instalment (the latter four entries all in the corresponding
period).
A person borrows P = 50,000 EUR (= principal) at the beginning of a
year and is supposed to pay it off in 5 instalments at the end of each
following year, with interest at i = 10% compounded annually.
Most regular clients want equal payments over a specified number of
periods, even though the repayment plan would look essentially the same
if the instalments were not constant. In the above case, the yearly
instalment payment would be a = 13189.87 EUR.
The repayment plan is shown in Tab. 6.4. In fact, the principal will be
repaid in five equal instalments.
p
Year
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
3
4
5
50000.00
41810.13
32801.27
22891.53
11990.81
5000.00
4181.01
3280.13
2289.15
1199.08
8189.87
9008.86
9909.74
10900.72
11990.73
13189.87
13189.87
13189.87
13189.87
13189.81
Sum
159493.74
15.949.37
49999.92
65949.29
Table 6.4:
1
Repayment plan1
p=periods; OA(p)=Outstanding Amount in period p; Int(p)=Interest to be paid in
period p; PR(p)=Principle repayment in period p; Inst(p)=Instalment to be paid in
period p
106 | P a g e
6. Time Value of Money
Principal Repayment vs. Interest
14000
Payment
12000
8189
9009
9910
10901
11991
10000
8000
6000
4000
5000
4181
3280
2000
2289
1199
year 1
year 2
Principal repayment
year 3
year 4
year 5
Interest
Figure 6-14: Graph of the repayment plan
In Fig. 6-14, the client’s yearly payments are represented by columns of
equal height, because the instalments are constant. The portion of interest
in that amount is shaded and the principal repayment is the white part.
Because the interest rate is fixed, the interest portion decreases with
every principal repayment. Correspondingly, the interest increases over
time.
The example raises the question of how to calculate instalments, which
are also called annuities when we mean yearly instalments.
6.3.2 Instalments
The principal amount P is now supposed to be repaid in n equal
instalments at the end of each year, with an interest rate i compounded
annually.
P a g e | 107
6.3 Loans
a
a
a
pricipal
repayment
pricipal
repayment
interest
a
pricipal
repayment
pricipal
repayment
.........
interest
interest
interest
t
eoy 1
eoy 2
.........
eoy n-1
eoy n
B n=0
P
today
interest factor per year q=1+i
eoy = end of year
Figure 6-15: Instalments
Fig. 6-15 represents the disbursed principal as the only out-payment
(arrow downwards) P. The constant yearly payments a are in-payments,
represented by arrows from above. The interest payment each year is i on
the outstanding balance; it is the dotted decreasing part of the instalment.
The complementary part of the instalment (bold) is the principal
repayment in the corresponding year, which is subtracted from the
outstanding balance left over from the previous year. At the end
(maturity) the balance should be zero.
At the beginning of the first year, the outstanding amount is the principal.
However, at the end of that year, the first instalment a is paid. This
payment reduces the outstanding amount (balance B1 ) at the end of year 1
(= start of year 2) to:
B1 = P ⋅ q − a
Note that we simply balance all compounded in and out-payments.
Similarly, at the end of year 2 (= start of year 3), we balance:
B2 = B1 ⋅ q − a = P ⋅ q 2 − a ⋅ q − a
Thus, at year j the balance will be:
B j = B j −1 ⋅ q − a = P ⋅ q j − a ⋅ q j −1 − ... − a ⋅ q − a
108 | P a g e
6. Time Value of Money
a
a
a
a
int
int
int
int
rep
rep
rep
rep
...
eoy 2
eoy 1
...
eoy j
B1 = P ⋅ q − a
eoy n
today
t
B2 = B1 ⋅ q − a = P ⋅ q 2 − a ⋅ q − a
B j = B j −1 ⋅ q − a = P ⋅ q j − a ⋅ q j −1 − ... − a ⋅ q − a
P
Bn=0
interest factor per year q=1+i
eoy = end of year
Figure 6-16: Calculation of outstanding amount
If we continue until the last period, the balance at the end of year n is:
Bn = P ⋅ q n − a ⋅ q n−1 − a ⋅ q n−2 − ... − a ⋅ q − a =
{
}
Bn = P ⋅ q n − a ⋅ 1 + q + q 2 + ... + q n−1
We again see the geometric series in the brackets (see: Section 6.2.1;
Page 89), and can therefore simplify the formula to:
{
Bn = P ⋅ q − a ⋅ 1 + q + q + ... + q
n
2
}
n −1
qn −1
= P⋅q −a⋅
q −1
n
On the other hand, the balance of the outstanding amount must be zero at
maturity ( Bn = 0 ):
qn −1
Bn = 0 = P ⋅ q − a ⋅
q −1
n
→
a = P ⋅ qn.
q −1
qn −1
We can simplify this term by replacing q − 1 = i and, instead of
multiplying the numerator, can divide the denominator by q n :
a = P ⋅ qn .
q −1
q −1
n
= P⋅
i
q −1
qn
n
=
P ⋅i
1 − q−n
P a g e | 109
6.3 Loans
INSTALMENTS
The principal P will be amortised in instalments
a=
P ⋅i
1 − q−n
payable for n periods if the interest rate is i (compounded periodically).
The periods might be years, months, or quarters; the interest rate i
(interest factor q = 1 + i ) should be adjusted to the period.
EXAMPLES
1.
The loan P = 5000 EUR shall be amortised within 9 periods at an
interest rate of 2%. There are no fees. What will the instalments
be?
P = 5000; i = 0.02; q = 1.02; n = 9 : a = ?
a=
5000 ⋅ 0.02
1 − 1.02−9
= 612.58 EUR
Cross check
The sum of all payments are 9 times the instalment = 5,513.19
EUR. The principal plus 10% interest is about this amount, which
makes sense.
2.
A loan P = 4000 EUR should be repaid within 6 months. The
effective interest rate is ieff = 15% . There are no fees. What will
the instalments be?
The solution to this problem has two parts:
a)
We have to calculate the monthly compound interest rate im
resulting from the given effective interest rate (see: Section
6.1.3; Page 77):
110 | P a g e
6. Time Value of Money
im = 12 qeff − 1 = 12 1 + ieff − 1 = 12 1.15 − 1 = 0.0117
b) Keeping this monthly interest rate in the calculator memory,
we can continue:
P = 4000; n = 6; im ; qm = 1 + im : a = ?
a=
4000 ⋅ im
1 − qm−6
= 694.27
Cross check
The sum of all repayments is EUR 4,165.60; EUR 4,000 principal
repayment and EUR 165.60 interest for half a year at the effective
interest rate of 15% makes sense.
Calculation of the instalment a =
P ⋅i
1 − q−n
contains again the four well-
known financial entries:
•
the future value:
•
the present value: principal P
•
the time:
number of periods n
•
the interest rate:
rate i, or compound factor q, respectively
instalment a
The old rule remains true: Three must be known in order to calculate the
fourth. Of the above, we already discussed the first bullet.
If we want to calculate the principal P given the instalment, the time and
the interest rate, we have to solve the formula for P:
P=
a ⋅ (1 − q − n )
i
P a g e | 111
6.3 Loans
EXAMPLE
Anna can afford monthly instalments of EUR 300 for 12 months. She
has been informed by the bank that the monthly compounding interest
rate is 2%. Now she can calculate how much she could borrow:
a = 300; n = 12; i = 0.02; q = 1.02 : P = ?
P=
300 ⋅ (1 − 1.02−12 )
= 3172.60 EUR
0.02
Cross check
Anna will pay EUR 3,600 in total. This sum contains the principal and
the interest. Therefore we have to subtract roughly 28% interest
(effective!) from the average of EUR 1500, i.e. about EUR 420. This
matches the result almost perfectly.
In the third bullet we solve the equation for time n, because the other
three items are given:
a=
P⋅i
1 − q −n
→ 1 − q −n =
P⋅i
P ⋅i a − P ⋅i
→ q −n = 1 −
=
a
a
a
In order to isolate n we have to apply the logarithmic operation to both
sides:
q −n =
n=
a − P⋅i
a
↔ qn =
→ n ⋅ ln q = ln a − ln(a − P ⋅ i)
a
a − P ⋅i
ln a − ln(a − P ⋅ i)
ln q
ATTENTION
The logarithm ln( a − P ⋅ i ) in the numerator may not be split into the
difference of the two arguments!
112 | P a g e
6. Time Value of Money
EXAMPLE
How long does Anna have to repay the principal of EUR 4000 if she
can afford instalments of EUR 250 every month at n compound
interest rate of 1.5% monthly?
P = 4000; a = 250; i = 0.015; q = 1.015;
n=
n=?
ln 250 − ln(250 − 4000 ⋅ 0.015)
= 18.43 months
ln1.015
Cross check
The sum of all payments is 18 times the instalment = 4,600; the
principal plus EUR 600 interest for the average outstanding amount of
EUR 2000 for 1.5 years at an effective rate of 20% is sufficient.
The fourth bullet would involve calculating the interest rate from the
other three financial items. Again this is unsolvable for us, because it
requires solving a polynomial equation of degree n, which is beyond the
scope of this course.
OUTSTANDING BALANCE
The discussion of instalments and their calculation provided a convenient
side effect. Reviewing the process of deriving instalments, we calculated
the outstanding amounts for successive periods until we finally reached
maturity with the outstanding balance of 0 (see: Figure 6-16; Page 108).
In period j the outstanding amount was calculated as:
B j = P ⋅ q j − a ⋅ (1 + q + q 2 + ... + q j −1 ) = P ⋅ q j − a ⋅
q j −1
q j −1
= P⋅q j − a⋅
q −1
i
P a g e | 113
6.3 Loans
OUTSTANDING AMOUNT
The outstanding amount of the principal P at the end of period j is:
Bj = P ⋅ q j − a ⋅
q j −1
with the compounding interest rate i
i
We should realise that we first have to calculate the instalment a before
we can calculate the outstanding balance of the loan in question.
EXAMPLE
A loan of EUR 15,000 will be repaid within 18 months at an interest
rate of 1.5% compounded monthly. What is the outstanding amount
after 4 (8, 12) months?
P = 15000; n = 18; i = 0.015; q = 1.015 : a =
P ⋅i
1 − q−n
= 957.09 €; B j = ?
1.0154 − 1
B4 = 15000 ⋅1.015 − a ⋅
= 12005.10 EUR
0.015
4
B8 = 15000 ⋅1.0158 − a ⋅
1.0158 − 1
= 8826.43 EUR
0.015
B12 = 15000 ⋅1.01512 − a ⋅
1.01512 − 1
= 5452.70 EUR
0.015
Cross check
The total payments within 4 months are four times the instalment, i.e.
EUR 3,840. This sum contains effective interest of 20%. The average
deposit for the first 4 months is EUR 13,500, which includes interest
of roughly 7% for 4 months: EUR 940. Therefore the principal
repayment is about 3,840 – 940 = 2900 EUR. That is a relatively
precise estimate of the result.
Afterwards, the principal repayments
outstanding balance decreases:
increases
because
the
114 | P a g e
6. Time Value of Money
The principal repayment within the first 4 months:
15,000 – 12,005.10 = 2,994.90 EUR
The principal repayment during the next 4 months:
12,005.10 – 8,826.43 = 3,178.67 EUR
The principal repayment during the next 4 months:
8,826.43 – 5,452.70 = 3,373.73 EUR
The results are comprehensible.
6.3.3 Effective Interest Rate of Loans
A consumer who wants to take out a loan may be faced with several
offers from competing financial institutions. It is therefore of
considerable importance to compare the various offers, especially if they
are based on different assumptions. The only way to make such
comparisons is to use the effective interest rate.
If we want to compare investments that have different compounding
periods, we need to state their interest using some common or
standardised form. This leads us to distinguish between the nominal
interest rate and the effective interest rate ieff . This was discussed in
Section 6.1.3 (see: Page 72).
Basically, there are three factors that can create a difference between the
effective rate and the nominal rate:
1. The compounding period may be less than or greater than a year.
Since the effective interest rate is always an annual rate, it
generally differs from the nominal rate.
2. The nominal interest rate may change during the contract time. In
this case, the effective interest rate is the average interest rate for
the entire contract time when compounding each year. “Average”
does not mean that the average is calculated as the mean of the
interest rates!
3. Bank charges for services are not covered by the interest rate.
These fees are usually collected once, generally at the beginning
P a g e | 115
6.3 Loans
of the contract, sometimes as a fixed cost, sometimes depending
on the amount (of a loan, for example).
Often we will see a combination of all three factors.
The effective interest rate is always calculated from the client’s point of
view. In Germany, it is set in the law that the effective interest rate must
be explicitly stated on every contract involving interest rates. As can be
seen, calculating the effective interest rate is a very important topic in
banking and finance.
EFFECTIVE INTEREST RATE
The effective interest rate ieff is always the annual interest rate taking
into account costs.
We will divide the discussion by looking separately at each of the three
aforementioned factors affecting the interest rate.
PERIODIC INTEREST RATE
Compounding in periods of less than a year means that interest is paid
out within the year and additional interest is then earned on previously
earned interest. We have already discussed this in Section 6.1.3 (see:
Page 72). The following paragraph is just a reminder.
If the interest on deposit D is compounded periodically with the nominal
interest rate i, the balance of the account after p periods (= one year) will
be:
B1 = D ⋅ (1 + pi ) p
Since the effective interest rate is always the annual interest (paid or
earned), we have to determine how much interest must be paid yearly in
order to reach the same balance:
D ⋅ (1 + ieff ) = B1 = D ⋅ (1 + pi ) p
→
ieff = (1 + pi ) p − 1
This formula remains true for deposits and loans.
116 | P a g e
6. Time Value of Money
EXAMPLE
For monthly instalments, p = 12, and a nominal interest rate of i =
24%, or a monthly interest rate of im = pi = 2% , the effective interest
rate will be:
ieff = (1 + 0.02)12 − 1 = 1.0212 − 1 = 0.2682 = 26.82%
Cross check
The effective interest rate must be more than the nominal rate, which
is the case.
REPAYMENT PLAN AND EFFECTIVE INTEREST RATE
A general method of calculating the effective interest rate in the case of a
loan may be applied with reference to the repayment plan. The
periodically paid interest rate i p can always be calculated from the
repayment plan as follows :
ip =
∑ all interest paid
∑ all outstanding amounts
This relation is valid for any period: With p being some portion of a year
(p = 12 for monthly instalments, p = 4 for quarterly instalments, etc.),
the periodic interest rate can be calculated using the above relationship,
and the effective interest rate will be: ieff = (1 + i p ) p − 1 .
We will illustrate this procedural method by means of three Examples.
P a g e | 117
6.3 Loans
EXAMPLES
1. A typical repayment plan for a loan of EUR 1000, with six monthly
instalments of EUR 184.60 and an interest rate of i p = 3% , is:
p
Months
OA(p)
EUR
1
2
3
1000.00
686.17
522.15
4
5
6
Sum
PR(p)
EUR
Inst(p)
EUR
30.00
25.36
20.59
154.60
159.24
164.01
184.60
184.60
184.60
353.22
179.22
0.00
15.66
10.60
5.38
168.93
174.00
179.22
184.60
184.60
184.60
3586.17
107.59
1000.00
1107.59
Table 6.5:
Int(p)
EUR
Repayment plan with monthly instalments2
The interest paid per period (= month) is: i p =
107.59
= 0.03
3586.17
This is of course not surprising, because the instalment was
calculated with a monthly interest rate of 3%! However, the
calculation shows that the approach is correct in the above case.
In order to calculate the effective interest rate, which is always the
annual effective rate, we would have to apply the above formula:
ieff = (1 + i p ) p − 1 = 1.0312 − 1 = 42.58%
2. A loan of EUR 10000 has a maturity of 7 quarters and a quarterly
interest rate of i p = 8% . The quarterly instalments will be with
P=10000, i=0.08 → q=1.08, and n=7:
a=
P ⋅i
1 − q −n
=
10000 ⋅ 0.08
1 − 1.08−7
= 1920.72 EUR
The repayment plan for 7 quarters is:
2
Headlines: see page 103
118 | P a g e
6. Time Value of Money
p
Quarters
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
3
10000
8879.28
7668.90
800.00
710.34
613.51
1120.72
1210.38
1307.21
1920.72
1920.72
1920.72
4
5
6
6361.69
4949.91
3425.18
508.94
395.99
274.01
1411.78
1524.73
1646.71
1920.72
1920.72
1920.72
7
1778.48
142.28
1778.48
1920.76
Sum
43063.45
3445.08
10000.00
13445.08
Repayment plan with quarterly instalments3
Table 6.6:
3445.08
= 8% , i.e. again exactly the
43063.45
rate we started with, which didn’t change.
The interest per quarter is: i p =
The effective interest rate in this case is:
ieff = (1 + 0.08 ) − 1 = 1.084 − 1 = 36.05%
4
In the third example we drop the assumption that the interest rate is
constant during the contract.
3. We assume that the loan from EUR 1000 with a six-month
maturity, the instalments of monthly EUR 185 include the
following floating interest payments:
3
Headlines: see page 103
P a g e | 119
6.3 Loans
p
Months
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
3
1000.00
845.00
685.00
35.00
30.00
25.00
155.00
160.00
165.00
190.00
190.00
190.00
4
5
6
520.00
350.00
177.00
20.00
17.00
13.00
170.00
173.00
177.00
190.00
190.00
190.00
Sum
3577.00
140.00
1000.00
1140.00
Table 6.7:
Repayment plan with floating interest rate4
The interest rate per month is: i p =
140
= 0.03914
3577
In order to get the effective interest rate, we have to calculate the
yearly compounded rate:
(
ieff = 1 + i p
)
p
− 1 = 1.0391412 − 1 = 58.52%
A relatively small and seemingly harmless looking change of the
interest generated a substantially different effective interest rate.
Cross check
The interest rate for the first month is 3.5%; however, for the last
month the interest rate raised to 7.3%. An average rate of about 4%
seems plausible.
4
Headlines: see page 103
120 | P a g e
6. Time Value of Money
FLOATING INTEREST RATES
The interest rate may change during the contract time, as in the above
repayment schedule. In another example, it is possible to imagine an
investment with increasing interest rates. In Germany, there are treasury
bonds which are sold exactly under these conditions.
On the credit side, we can also imagine that there could be floating –
generally increasing – interest rates, especially if the maturity of a loan is
long-term. We start by discussing the less complex problem of repaying a
loan in one single payment. In Fig. 6-17, interest i j will be paid in year j
for the principal P, and the principal and all interest, together R, will be
repaid at the end of the agreement.
R
q1
q2
eoy 1
q4
q3
eoy 2
eoy 3
P
eoy 4
t
today
eoy = end of the year
Figure 6-17: Floating yearly interest rates
The corresponding interest factor is q j = 1 + i j , and the repayment R −
from the bank's point of view − will be:
R = P ⋅ q1 ⋅ q2 ⋅ q3 ⋅ q4 after four years
The effective interest rate is the true rate for the customer. From his (or
her) point of view, the repayment will be the investment compounded
effectively four times:
R = P ⋅ qeff 4 = P ⋅ (1 + ieff ) 4
The repayment is the same for both sides, thus:
R = P ⋅ q1 ⋅ q2 ⋅ q3 ⋅ q4 = P ⋅ (1 + ieff ) 4
P a g e | 121
6.3 Loans
The effective interest rate will be independent of the principal amount.
EFFECTIVE INTEREST RATE WITH FLOATING INTEREST
The effective interest rate for floating nominal interest rates is:
ieff = 4 q1 ⋅ q2 ⋅ q3 ⋅ q4 − 1
EXAMPLE
Let the interest rates be: i1 = 3% for the first year, and i2 = 5% ,
i3 = 6% , i4 = 7% , and i5 = 10% for the following years. The effective
interest rate for a loan over five years is:
ieff = 5 1.03 ⋅1.05 ⋅1.06 ⋅1.07 ⋅1.1 − 1 = 0.0617 = 6.17%
Cross check
The arithmetic mean of the five interest rates would be 6.2%. The
result is convincing.
In the case of floating interest rates for periods shorter than a year, we
have to combine the latter procedure with the first. The year will be
replaced by the more general term “period”:
R
q1
...
eop 1
qj
...
qn
...
eop j
eop n-1
today
P
eop = end of period
Figure 6-18: Floating periodical interest rates
t
122 | P a g e
6. Time Value of Money
Using the same argumentation as before we will obtain a single periodic
interest factor:
q p = n q1 ⋅ q2 ⋅ ... ⋅ qn
Since the effective interest rate is always the true yearly rate, and
assuming that the year is divided into p periods, we have to compound
the periodic rate p times:
ieff = (1 + i p ) p − 1
EXAMPLE
The monthly interest rates for a loan with 6 months maturity are
assumed to be: 2%, 2.5%, 3%, 3.3%, 3.8%, and 5%.
What is the effective interest rate?
The average monthly factor will be:
qm = 6 1.02 ⋅1.025 ⋅1.03 ⋅ 1.033 ⋅ 1.038 ⋅ 1.05 = 1.0326
The effective interest rate is:
ieff = qm12 − 1 = 0.4699 = 46.99%
Cross check
Again, we calculate the arithmetic mean of the interest rates, obtaining
3.27% per month. The exact result is very close.
FEES
Almost all banks charge fees when disbursing a loan, some even more
than once. Usually it is either a fixed percentage of the loan amount or a
fixed amount, paid once at the beginning of the contract. The fee f is
basically an in-payment by the customer (see: Fig. 6-19); it is deducted
from the principal amount P to be disbursed, so that the customer gets
only P− f.
P a g e | 123
6.3 Loans
We start by assuming that the principal plus interest will be repaid after
one year in one single payment R.
R
1 year
today
P-f
t
f
interest factor per year qnom=1+i
Figure 6-19: Fees and one repayment
The bank demands nominal interest for the full loan:
R = P ⋅ qnom = P ⋅ (1 + i)
However, the customer gets only P−f, yet he repays R; the true interest
must therefore be calculated on the basis of this reduced amount:
R = ( P − f ) ⋅ qeff = ( P − f ) ⋅ (1 + ieff )
Due to the equality of the repayments we obtain:
ieff =
P
(1 + i) −1
P− f
EXAMPLE
Let the loan be EUR 1000 and the fee 2% with a nominal interest rate
of 24%. The effective interest rate is:
ieff =
1000
⋅1.24 − 1 = 0.2653 = 26.53%
980
124 | P a g e
6. Time Value of Money
Cross check
The effective interest rate must be slightly higher than the sum of the
nominal rate and fee because the fee is paid one year before.
If the time between disbursement and repayment is less than a year – for
instance 4 months – the result of a similar argumentation will be an
average interest rate for the corresponding time interval.
The year may be divided into p periods. We obtain an average interest
factor per period:
q p = (1 + i p ) =
P
⋅ (1 + pi )
P− f
The effective interest rate will be yearly compounded interest rate:
p
ieff = q p
p
 P

−1 = 
⋅ (1 + pi )  − 1
 P− f

EXAMPLE
Let the loan be EUR 1000 and the fee 2% with a nominal interest rate
of 24%. Repayment after four months means p=3:
q p = (1 + 0.24
)=
3
1000
⋅1.08 = 1.1020
980
i p is the true interest rate for four months, ieff the effective rate
(annually!):
ieff = q p3 − 1 = 0.3384 = 33.84%
Cross check
Since a 2% fee is paid for considerably less time and the periodical
rate has to be compounded, the effective interest rate must be
significantly larger than the nominal interest rate.
P a g e | 125
6.3 Loans
Repaying the principal at once is rather rare. Generally, loans will be
repaid in the form of instalments. Therefore, we will now discuss the
most probable situation: A customer has to pay a disbursement fee and
repays the principal in monthly instalments.
For a loan P, with a nominal monthly interest rate im (interest factor
qm = 1 + im ) and maturity after 12 periods, the amount of instalment a
will be (see: Section 6.3.2; Page 108):
a=
P ⋅ im
1 − qm −12
Now we assume that there is a disbursement fee f (see: Fig. 6-20). Instead
of in-paying the fee it is usually subtracted from the principal amount
which is indicated in the figure by means of the reduced length of the
downward arrow.
a
a
a
a
12 times
...
today
P-f
t
f
interest factor per month qm = 1 + im
Figure 6-20: Fees and instalments
Besides the fee, after 12 months the bank receives the principal amount
plus interest, which is 12 times the compounded monthly interest.
Bank receives:
P ⋅ qm12
The customer received in fact only (P−f) for which he paid the effective
interest.
126 | P a g e
Customer pays:
6. Time Value of Money
( P − f ) ⋅ (1 + ieff ) for the year
Balancing both sides leads to:
 P

ieff = 
qm12  − 1
 P− f

EFFECTIVE INTEREST RATE FEE AND INSTALMENTS
The effective interest rate for the principal amount P with a monthly
interest factor qm , a fee f , and 12 months maturity is:
 P

ieff = 
qm12  − 1
 P− f

EXAMPLE
P = 1000 EUR; fee = 2%; monthly interest i p = 2% ; n = 12 months
maturity.
The instalments are: a = 94.56 EUR
 1000

⋅1.0212  − 1 = 29.41%
The effective interest rate is: ieff = 
 980

Cross check
Without a fee, we remember that the effective interest rate in the case
of monthly compounded interest of 2% was about 26%. Due to the
fee, the effective interest rate must be significantly higher. Almost
30% sounds reasonable.
If maturity is less than a year, which is often the case for micro-loans, the
effect of a fee will be correspondingly higher.
We now assume that the maturity of the loan is t months. Hence,
6.3 Loans
P a g e | 127
•
t < 12
→ maturity is less than one year,
•
t = 12
→ maturity is exactly one year, and
•
t > 12
→ maturity is more than one year.
After t months the bank receives:
P ⋅ qmt
The customer received only (P−f) and pays for that amount ( P − f ) ⋅ it .
( P − f ) ⋅ (1 + it )
Hence the customer pays:
Balancing both amounts leads to:
qt = 1 + it =
P
⋅ qmt
P− f
This is the interest rate which the customer truly has to pay for the t
months. The equivalent interest factor per month is therefore (see:
Section 6.1.3; Page 77):
q1m = qt = ( qt ) =
1
t
t
(
P
P− f
⋅ qmt
) = ( ) ⋅q
1
t
P
P− f
1
t
m
=t
P
P− f
.qm
This is the true monthly compounding factor. In order to obtain the
effective interest rate, we have to compound this factor 12 times:
ieff =
( )
P
P− f
12
t
⋅ qm12 − 1
The effective interest rate for a loan P with a fee f, t months maturity, and
monthly compounding interest rate im is:
ieff =
( )
P
P− f
12
t
⋅ qm12 − 1 , where qm = 1 + im
EXAMPLE
Let us now keep all the conditions the same, except that the loan has
to be repaid in t = 6 instalments, i.e.:
P = EUR 1000; f = 2% = 20; qm = 1.02 ; maturity t = 6 months
128 | P a g e
6. Time Value of Money
Although we don’t need it for calculating the effective interest rate,
bear in mind that repayment in less time certainly will imply a larger
instalment: a = 178.53 EUR
The effective interest rate is:
ieff =
( 1000
980 )
12
6
⋅1.0212 − 1 = 0.3205 = 32.05%
Cross check
Compared to the last example, the only change is the shorter maturity.
The same fee is therefore charged for less time and the effective
interest raises to 32% compared to 29.4%.
The same formula holds for t > 12, i.e. for maturity of more than one
year.
EXAMPLE
All data remain unchanged, except that the loan has to be repaid in t =
18 instalments, i.e.:
P = EUR 1000; f = 2% = 20; qm = 1.02 ; maturity t = 18 months
Instalment: a = EUR 66.70
The effective interest rate is:
ieff =
( 1000
980 )
12
18
⋅1.0212 − 1 = 0.2854 = 28.54%
Cross check
Compared to the two previous examples, maturity is now more than
one year. The same fee is therefore charged for more time and the
effective interest decreases, which was expected.
6.3 Loans
EXERCISE 6.3:
P a g e | 129
LOANS
1.
Anna receives a loan of EUR 500. She pays it back together
with the interest after 6 months. The interest rate is i = 1%
compounded monthly. How much must she pay after 6 months?
2.
Anna can pay EUR 200 after 2 months (without any fee) and the
remaining amount together with the interest at the end of the
sixth months. How much does she have to pay?
3.
For a loan of EUR 1800 Anna agrees on the following
repayment plan: payment of EUR 400 after 3 months, EUR 300
after 6 months, EUR 500 after 8 months, and the remaining
amount together with the interest after 12 months. The interest
rate is 1% compound monthly. How much is the last payment?
4.
Anna receives a loan of EUR 5,000.
a) Calculate the instalment at 1.5% monthly compounded
interest and a 12 month maturity.
b) Set up the repayment plan for Anna.
5.
Set up the repayment plan for a loan of EUR 2,000 and an
effective interest rate 16%, monthly instalments, and a 9 month
maturity.
6.
Anna receives a loan of EUR 3,000. The monthly interest rate is
1.2%. She wants to repay the loan in 8 equal instalments.
Calculate the instalments and the effective interest rate.
7.
How long must Anna repay the loan from Problem 6, if she can
afford only EUR 250 per month?
8.
Anna's financial situation allows her to repay EUR 350 for a
loan. Procredit offers her 1.2% monthly interest, if she repays
the loan in 12 instalments. What size loan can she get?
130 | P a g e
9.
6. Time Value of Money
What is the outstanding balance of Anna's loan from Problem 4
after 6 months?
10. Calculate Anna's outstanding balance from Problem 8 after 4
months.
11. What is the effective interest rate Anna has to pay for the loan in
problem 1?
12. Anna asks for a loan of EUR 1400 at a competitor of ProCredit
and was offered the follwing conditions:
5
•
Bi-monthly instalments, i.e. every two months.
•
3% interest per period (i.e. for two months) on the principal
amount
•
7 periods maturity (= 14 months)
p5
Months
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
1400.00
42.00
200.00
242.00
2
3
4
1200.00
1000.00
800.00
42.00
42.00
42.00
200.00
200.00
200.00
242.00
242.00
242.00
5
6
7
600.00
400.00
200.00
42.00
42.00
42.00
200.00
200.00
200.00
242.00
242.00
242.00
Sum
5600.00
294.00
1400.00
1694.00
p=periods; OA(p)=outstanding amount in period p; Int(p)=interest to be paid in period
p; PR(p) = principal repayment to be paid in period p; Inst(p) = instalment to be paid in
period p
6.3 Loans
P a g e | 131
Interest of 3% for two months signals a nominal rate of 18% or
an effective rate of 19.4%. But, is this true?
What will be the effective interest rate of this agreement, really?
13. On 1. January 2009 Anna got a loan of 2000 EUR under the
following conditions:
•
ProCredit has agreed that she will repay the principal and all
interest at once at the end of the fourth year.
•
However, Anna has to pay a service fee of 100 EUR when
the loan is disbursed.
•
For the first year Anna has to pay 8% interest, the next year
9%, the third year 11%, and the last year 15%.
a) How much has the Anna to pay at the end of year 2012?
b) What is the effective interest rate for Anna?
14. Anna receiced a loan of 2000 EUR . She has to repay the
principal in 15 months with a monthly compounding interest
rate of 1.5%. At the time of disbursement she has to pay a fee
which is 3% of the principal.
a)
Calculate the instalments.
b) Prepare the repayment plan for Anna.
c)
What is Anna’s effective interest rate?
132 | P a g e
6. Time Value of Money
ANSWERS 6.3:
LOANS
1.
Repayment is: R = 530.76 EUR
2.
Repayment is: R = 322.64 EUR
3.
Last Payment: R = 752.05 EUR
4.
a) Instalment: a = 458.40 EUR
b) Repayment plan6 :
6
p
Months
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
3
5,000.00
4,616.60
4,227.45
75.00
69.25
63.41
383.40
389.15
394.99
458.40
458.40
458.40
4
5
6
7
3,832.46
3,431.55
3,024.62
2,611.59
57.49
51.47
45.37
39.17
400.91
406.93
413.03
419.23
458.40
458.40
458.40
458.40
8
9
10
2,192.36
1,766.85
1,334.95
32.89
26.50
20.02
425.51
431.90
438.38
458.40
458.40
458.40
11
12
896.58
451.63
13.45
6.77
444.95
451.63
458.40
458.40
Headline: see footnote page 128
6.3 Loans
5.
P a g e | 133
Instalment is: a = 236.28 EUR
Repayment plan7:
6.
p
Months
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
3
2,000.00
1,788.61
1,574.59
24.89
22.26
19.60
211.39
214.02
216.68
236.28
236.28
236.28
4
5
6
1,357.91
1,138.53
916.41
16.90
14.17
11.40
219.38
222.11
224.88
236.28
236.28
236.28
7
8
9
691.54
463.87
233.36
8.61
5.77
2.90
227.67
230.51
233.38
236.28
236.28
236.28
Instalment a = 395.53 EUR; effective interest
ieff = 0.1539 = 15.39%
7
7.
Possible maturity: n = 13 months
8.
Possible principle is: P = 3,889.95 EUR
9.
Outstanding amount is: B6 = 2611.59 EUR
Headline: see footnote page 128
134 | P a g e
6. Time Value of Money
10. Outstanding amount is: B4 = 2654.65 EUR
11. ieff = 1.112 − 1 = 0.1268 = 12.68%
294 = 0.0525
12. True bi-monthly interest rate: i2m = 5600
Effective interest rate: ieff = 1.05256 − 1 = 0.3594 = 35.94%
13. a)
b)
14. a)
R = L ⋅ q1 ⋅ q2 ⋅ q3 ⋅ q4 = 3005.39 EUR
The effective rate is:
a=
P ⋅i
1− q
−n
=
ieff = 12.15%
2000 ⋅ 0.015
1 − 1.015−15
= 149.89 EUR
6.3 Loans
b)
P a g e | 135
Repayment Plan8
p
Months
OA(p)
EUR
8
PR(p)
EUR
Inst(p)
EUR
1
2
3
2000.00
1880.11
1758.42
30.00
28.20
26.38
119.89
121.69
123.51
149.89
149.89
149.89
4
5
6
1634.91
1509.54
1382.29
24.52
22.64
20.73
125.37
127.25
129.16
149.89
149.89
149.89
7
8
9
1253.14
1122.05
988.99
18.80
16.83
14.83
131.09
133.06
135.06
149.89
149.89
149.89
10
11
12
13
853.93
716.85
577.71
436.49
12.81
10.75
8.67
6.55
137.08
139.14
141.22
143.34
149.89
149.89
149.89
149.89
14
15
293.15
147.65
4.40
2.21
145.49
147.66
149.89
149.87
16555.23
248.33
2000.00
2248.32
Sum
c)
Int(p)
EUR
ieff =
(
2000
1940
)
12
15
⋅1.01512 − 1 = 1.2251 = 22.51%
Headline: see footnote page 128
136 | P a g e
6. Time Value of Money
6.3.4 Progress Test “Loans”
You should allot yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Do not use the reader to look
for the solution. The aim of the test is to get feedback on how much you
know or have learned up to now.
At the end of the chapter, you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. In any other
case (you got the wrong answer or no answer at all) you will find some
advice on what to do. In most cases, you will be directed to the section in
the reader which you should repeat in order to close the gap.
1. A loan of EUR 800 must be repaid in a single payment after 6
months. The monthly compounded interest rate is 1%. What is the
repayment amount?
2. How much would the monthly instalments be for Problem 1?
3. A client wants to compare the offer of a competitor with the
conditions at ProCredit for a loan of EUR 2,000 with a 9 month
maturity.
The competitor offers 15% nominal interest plus a service charge
and calculated monthly instalments of EUR 244.34.
ProCredit asks for 16 % effective interest. What is the instalment
amount for the ProCredit loan?
4. Which loan amount can you offer as a loan officer, if the client's
monthly instalment should not exceed EUR 140 with an effective
interest rate of 15% and a 9 month maturity?
6.3 Loans
P a g e | 137
5. Anna asks for a loan of EUR 1,000 because she wants to buy a
notebook immediately. ProCredit offers an interest rate of 1.5%,
compounded monthly with a 12 month maturity.
a)
Calculate the instalments.
b) Set up the repayment plan.
6. The client from Problem 4 asks for a loan of EUR 2,000. He
agrees to monthly instalments of EUR 150. Calculate the maturity
of the agreement.
7. A mortgage loan of EUR 200,000 should be repaid within 20
years at an effective interest rate of 8%.
a)
Calculate the annuity.
b) What is the outstanding amount after 10 years?
8. Anna and ProCredit signed the following savings plan:
•
Anna deposits EUR 50 every month.
•
The monthly compound interest rate rises from 1.5% in the
first and second year, to 1.8% during the third and fourth year,
and to 2.5% during the fifth year.
a) What will be the balance at the end of fifth year?
b) What is the effective interest?
9. Anna receives a loan of EUR 4000 with a nominal interest rate of
18%. Maturity is 18 months, and disbursement fee is 2.5%.
a)
Calculate the instalments.
b)
Calculate the effective interest rate.
138 | P a g e
6. Time Value of Money
6.4 Answers to Progress Tests
6.4.1 Answers to Progress Test “Interest”
You should check your answers. If they are correct, you may continue
and start the next chapter. In any other case (you got the wrong answer or
no answer at all), repeat the topics in the corresponding sections in order
to close the gap.
1. 141 interest days → 309.40 EUR
2. a)
EUR 885.78
b) EUR 904.36
c) EUR 908.80
3. a)
10%
b) 10.38%
c)
10.47%
b) 2.282%
c)
1.135%
4. EUR 651.97
5. 65.92 months
6. 1.046 % monthly
7. a)
3.443%
7.0 Sum Symbol
P a g e | 139
6.4.2 Answers to Progress Test “Deposits”
You should check your answers. If they are correct, you may continue
and start the next chapter. In any other case (you got the wrong answer or
no answer at all), repeat the topics in the recommended sections in order
to close the gap.
1. EUR 1,180.70
2. EUR 566.26
3. EUR 1,008.11
4. EUR 47.20
5. EUR 77.92
6. 18.28 months
140 | P a g e
6. Time Value of Money
6.4.3 Answers to Progress Test “Loans”
You should check your answers. If they are correct, you may continue
and start the next chapter. In any other case (you got the wrong answer or
no answer at all), repeat the topics in the corresponding sections in order
to close the gap.
1. EUR 849.22
2.
EUR 138.04
4. EUR 1,185.04
5.
a)
3.
EUR 236.28
EUR 91.68
b) Repayment Plan9
p
Months
9
OA(p)
EUR
Int(p)
EUR
PR(p)
EUR
Inst(p)
EUR
1
2
1,000.00
923.32
15.00
13.85
76.68
77.83
91.68
91.68
3
4
5
845.49
766.49
686.31
12.68
11.50
10.29
79.00
80.18
81.39
91.68
91.68
91.68
6
7
8
604.92
522.32
438.47
9.07
7.83
6.58
82.61
83.85
85.10
91.68
91.68
91.68
9
10
11
12
353.37
266.99
179.32
90.33
5.30
4.00
2.69
1.35
86.38
87.68
88.99
90.33
91.68
91.68
91.68
91.68
Headline: see footnote page 128
7.0 Sum Symbol
P a g e | 141
6. 67 months
7. a) EUR 2,0370.44
b) EUR 136,687.32
8. a) EUR 5738.25
b)
ieff = 24.15%
9. a) EUR 255.22
b)
ieff = 21.6%
142 | P a g e
6. Time Value of Money
7. Basics of Statistics
7.
P a g e | 143
Basics of Statistics
Basics of Algebra,
Functions, and
Statistics
1. Introduction
2. Basic Algebra
3. Equations
Part 2: Advanced
4. Basic Functions
Prerequisites:
5. Special Functions
6. Time Value
of Money
7. Basics of
Statistics
All calculations in statistics depend on the correct
application of algebra. Therefore it is necessary to be
able to deal confidently with the transformation and
evaluation of expressions and equations.
144 | P a g e
7. Basics of Statistics
Learning Targets: Statistics is the art of making numerical conjectures
about puzzling questions:
•
What are the effects of increasing the interest
rate?
•
How can a class’s exam performance be
measured?
•
How much profit does a casino make on its
roulette tables?
•
How many job applicants are actually hired?
•
What is the inflation rate in a given country?
These are complex issues, and statistical methods
can be very useful for analysing them. Statistical
methods have been developed over hundreds of
years, and weighty volumes have been written about
them.
Here we will study just some of the basic statistical
methods of descriptive statistics.
WHAT IS STATISTICS?
The word is Latin in origin (status → state, condition, situation, position,
shape): It incorporates all methods of analysing large sets of data by
summarising them in the form of concentrated information. Instead of
analysing all data – which might be very time consuming – we
summarise the data by classifying them in tables, representing them in
graphs, and creating key figures which reveal the basic properties of the
set of data. Of course, in each of these steps we lose some of the details
contained in the complete set of information, yet at the same time, by
summarising the data, we gain a better overview of the situation as a
whole. Key figures provide a much better insight into the conditions
reflected in the data. The shape of a column diagram or a curve can be
recognised at a glance. Instead of looking at the details we are able to see
the “big picture”. This is precisely what the concentration of information
is about: it is the intended purpose of statistics.
7. Basics of Statistics
P a g e | 145
Statistics can be subdivided into descriptive and conclusive statistics. The
former is the art of summarising and concentrating data. The latter tries
to deduce properties from a sample of the complete set of data and draw
conclusions regarding the whole using probability theory. In this course
we focus on the fundamentals of descriptive statistics. We will discuss
the collection, tabulation and systematic classification of quantitative
data with the aim of gaining an overview of the set of data in order to
draw conclusions about future trends. We will discuss the concentrated
representation of large sets of data in tables and in diagrams. Graphical
representations are ideal tools for providing a general picture of the
situation as a whole, which can then be easily visualised and analysed at
a glance. That is why they are frequently used in presentations, lectures,
speeches or academic papers. Summarising data by calculating key
figures, such as the average, minimum, maximum and median of the list
of observed numbers, is probably the most prominent part of descriptive
statistics. These numbers provide us with information about the
concentration, i.e. the focus or the centre of gravity, of the data. The final
topic under discussion here will be the spread of the data. While the
average value identifies the centre point of the data, the variance provides
information on how widely or narrowly the data are spread or distributed
around the centre. Only if we have both key figures, the centre and the
variance, will we have a complete picture of the whole set of data.
Statisticians usually have to handle huge amounts of data. If one were to
do everything manually – the calculations, the concentrations of data, and
the representations in suitable forms – the work would become tedious,
monotonous and tiresome. This is especially true of all banking
operations which are naturally based on large sets of data (account,
balances, loans, deposits, etc.). The use of suitable software is
indispensable in these areas. We will use Microsoft Excel, which offers a
broad variety of useful statistical functions and can generate nice-looking
diagrams. The screenshots in this chapter were made with the version of
Excel that forms part of MS Office 2007.
146 | P a g e
7. Basics of Statistics
ESTIMATION
Modern statistics distinguishes between the population and a sample. A
sample may be used to gain certain information about the – in its totality
– unknown population from which it was drawn. For the most part, the
data which mathematical statistics deals with consist of random samples
constructed in such a way that every item in the population has an equal
chance of being chosen in the sample.
In this course, in line with the needs of the ProCredit group, we will
mainly focus on descriptive statistics, i.e. the branch of statistics devoted
to summarising the large sets of data that make up the population. This
means that we will not discuss any probability-based statistical problems,
nor will we argue with samples. However, in some very rare cases –
when it comes to discussing variance, for example – we will have to
differentiate between population and sample. We will return to the
distinctions that need to be made in these cases when discussing the
topics that require us to be aware of them.
In economic statistics we generally do not know everything that we
would like to know about the population from which the data are a
sample. Therefore, we have to consider the data as a random sample of
the hypothetically infinite population. For example, if a set of prices for
some banking service is recorded it must be considered as a random
sample of all (unrecorded) service prices charged by all the other banks
in the market.
Statistical estimation is the method by which we extract information
about the population from the sample. This is the only object of
sampling. The estimates derived from the sample are functions of various
observations made on the sample. Since each observation is subject to a
certain probability, they are random variables. Consequently, the
estimates themselves are random variables as well.
One important property of estimation is that it is unbiased. An unbiased
estimate is defined as an estimate whose mean value is equal to the
corresponding value of the population. Since we usually use the
population for our statistical analysis, the distinctions between biased and
unbiased data are not really important for us.
7.0 Sum Symbol
P a g e | 147
7.0 Sum Symbol
7. Basics of Statistics
7.0 Sum
Symbol
7.1 Presentation
Techniques
7.2 Key Figures:
Centre
7.3 Key Figures:
Spread
Tables
Mean
Variance
Diagrams
Median
Standard
Deviation
Sum
Curves
Mode
Normal
Curve
In statistics we often have to add up large quantities of numbers, and we
quite regularly have to describe these sums of entries. In order to avoid
writing the complete sum, mathematicians use a short notation for sums.
sums
The so-called sum symbol is simply the Greek capital sigma: Σ . Because
it is used quite extensively in this chapter it is defined and explained in
this section.
In order to define how many terms should be added together, the sum
symbol is usually accompanied by a count index (often one of the letters
i, j, k)) with its starting and ending value. The starting value is written
under the sigma, while the ending value is given on top:
148 | P a g e
7. Basics of Statistics
10
The notation
∑
indicates that the sum has 10 terms.
j=1
The sum symbol is nothing more than an operator telling us that we have
to take the sum of a certain number of terms. The term is the operand
following the symbol. Generally it contains the count index j; however,
the term may also be independent of j (see problem 4 in the examples
below).
Very often the term is simply one variable with index j. In this case we
get:
n
∑ x j = x1 + x2 + ... + x j + ... + xn
j =1
However, the term may also be any algebraic term, containing the count
index as a parameter and/or index:
3
∑ a j ⋅ x j = ao ⋅ x0 + a1 ⋅ x1 + a2 ⋅ x2 + a3 ⋅ x3
j =0
Thus, we can write a general polynomial expression of degree 3 (see
Volume 1, Section 2.3.1) using the sum symbol.
SUM SYMBOL
∑
n
The
∑ t( j)
is a short notation for the sum of terms t ( j ) where j has to
j =k
be replaced by all consecutive integers starting from j=k (the starting
value or lower bound) and ending at j=n (the ending value or upper
bound). The starting value k cannot be greater than the ending value:
k≤n
7.0 Sum Symbol
P a g e | 149
EXAMPLES
1. The sum of all natural numbers from 1 to 5 can be written using
the sum symbol:
5
1+ 2 + 3 + 4 + 5 = ∑ j
j =1
2. The sum of the odd numbers from 1 to 21 can be denoted as:
11
1 + 3 + 5 + ... + 19 + 21 = ∑ (2 j − 1)
j =1
Cross check
We find that the first term will be 2 − 1 = 1 , and the last term in
the sum is 22 − 1 = 21 . Therefore, the representation with the sum
symbol is correct.
3. The geometric series of n terms with quotient q can also be
written using the sum symbol (see Section 6.2.1; Page 89):
n −1
1 + q + q 2 + ... + q n−1 = ∑ q j
j =0
ATTENTION
The sum has n terms if the counter index starts with 1 and ends with n.
However, if the counter index begins with j = 0 there is one term more.
Therefore, the last sum has n terms even though the counter index ends
with n-1.
8
4.
The sum
∑1 might look rather strange at first glance; however,
j=1
once you write the complete sum and apply the rules consistently
the above expression makes sense:
8
∑1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8
j=1
150 | P a g e
7. Basics of Statistics
CALCULATIONS USING THE SUM SYMBOL
You have to follow some very strict rules if the sum symbol is to be used
in algebraic calculation. Always remember and take care to ensure that
•
the sum symbol replaces brackets.
The impact of the sign is to place the entire sum inside a pair of implicit
brackets. This is extremely important if multiplicative factors are
involved:
3
c ⋅ ∑ x j = c ⋅ ( x1 + x2 + x3 ) or
j =1
3
∑ c ⋅ x j = ( c ⋅ x1 + c ⋅ x2 + c ⋅ x3 )
j =1
The fact that the sum sign implicitly replaces brackets means that the
range of that rule ends with
•
the term losing uniqueness when it is composed without brackets.
For instance:
3
∑ x j 2 + b = x12 + x22 + x32 + b , but
j =1
3
∑ ( x j 2 + b) = x12 + b + x22 + b + x32 + b = x12 + x22 + x32 + 3b
j =1
The brackets in the second expression are necessary if b should be part of
the summation.
A similar situation occurs if parts of the sum
•
are to be factored out.
If a symbol (number or variable) appears as a factor in every summand it
can be factored out. Instead of putting it in front of the brackets (replaced
by the sum symbol) it can be taken out of the sum and placed before the
sum sign:
7.0 Sum Symbol
n
n
j =1
j =1
P a g e | 151
∑c⋅ xj = c⋅∑ xj
If the term of the summation is a sum itself
•
the sum symbol and additive operators can be interchanged.
This means that we can split the term of the sum symbol into individual
sums.
n
n
n
n
n
n
j =1
j =1
j =1
j =1
j =1
j =1
∑ ( x j 2 + x j + 2) = ∑ x j 2 + ∑ x j + ∑ 2 = ∑ x j 2 + ∑ x j + 2n
152 | P a g e
7. Basics of Statistics
EXERCISE 7.0:
SUM SYMBOL
1. Simplify the expressions using the sum symbol and a count index:
a)
b)
10 + 20 + 30 + 40 + 50 =
1 + 2 + … 10 =
c)
3 + 6 + 9 + 12 + 15 + … + 27 + 30 =
d)
x6 + x7 + … + x19 + x20 =
e)
1 1 1 1
1
1
1+ + + + +…+ +
=
2 3 4 5
99 100
2. Write the sums given in symbolic form in explicit terms:
7
a)
2
∑ x2 j −1 =
b)
5
c)
∑
3
j ⋅ x 2 j −1 =
d)
j =1
xj
∑  j 2
j =1 
3
e)
∑ aj ⋅x
j
=
j =0

=


6
f)
∑ ( x j − x )2 =
k =3
10
g)
∑ x2 j =
j =−2
j =3
7
∑2 =
h)
j =1
∑ −2x 2 =
j =11
6
i)
∑qj =
j =0
3. Evaluate the sums as much as possible:
5
a)
∑ ( x j − 4)
j =1
2
=
5
b)
∑ 3( x j − 2) ⋅ ( x j + 2) =
j =1
7.0 Sum Symbol
c)
P a g e | 153
 2
  3

 ∑ ( x j − 4)  ⋅  ∑ x j  =
 j =1
 


  j =1 
4. Factor out as much as possible:
3
a)
∑ 22 j ⋅ x3 j =
j =1
5
c)
∑ (2 xk + 4 xk 2 − 6) =
k =3
3
b)
∑3⋅ x ⋅ x j =
j =1
154 | P a g e
7. Basics of Statistics
ANSWERS 7.0:
SUM SYMBOL
1. Simplify the expressions using the sum symbol and a count index:
5
a)
10 + 20 + 30 + 40 + 50 = ∑10 j
j =1
10
b)
1 + 2 + ... + 10 = ∑
j
j =1
10
c)
3 + 6 + 9 + ... + 27 + 30 = ∑ 3 j
j =1
d)
x6 + x7 + ... + x20 =
20
∑ xj
j =6
100
e)
1 =
1 + 12 + 13 + ... + 100
∑ 1j
j =1
2. Write the sums given in symbolic form in explicit terms:
7
a)
∑ x2 j −1 = x5 + x7 + x9 + ... + x13
j =3
2
b)
∑
j =−2
x2 j = x−4 + x−2 + x0 + x2 + x4
5
c)
∑ j⋅x
2 j −1
= x + 2 ⋅ x 3 + 3 ⋅ x5 + 4 ⋅ x 7 + 5 ⋅ x9
j =1
3
d)
∑ aj ⋅x
e)
xj
∑  j 2
j =1 
j =0
3
j
= a0 + a1 ⋅ x + a2 ⋅ x 2 + a3 ⋅ x3

x 2 x3
= x+
+

4
9

7.0 Sum Symbol
P a g e | 155
6
f)
∑ ( xk − x )2 = ( x3 − x )2 + ( x4 − x )2 + ( x5 − x )2 + ( x6 − x )2
k =3
10
g)
∑ 2 = 2 + 2 + 2 + ... + 2 = 20
j =1
7
∑ −2x 2 = not defined
h)
j =11
6
∑ q j = 1 + q + q 2 + q3 + q 4 + q5 + q6
i)
j =0
3. Evaluate the sums as much as possible:
5
a)
5
∑ ( x j − 4) =∑ x j
j =1
2
j =1
2
5
− 8∑ x j + 80
j =1
5
5
j =1
j =1
b)
∑ 3( x j − 2) ⋅ ( x j + 2) =3∑ x j 2 − 60
c)
 2
  2

 ∑ ( x j − 4)  ⋅  ∑ x j  = x12 + 2 x1 ⋅ x2 + x2 2 − 8 x1 − 8 x2
 j =1
 


  j =1 
ATTENTION
This answer would be wrong:
2
 2
  2
 2
 ∑ ( x j − 4)  ⋅  ∑ x j  ≠ ∑ ( x j − 4) ⋅ x j = ∑ x j 2 − 4 x j !
 j =1
 

j =1

  j =1  j =1
(
)
156 | P a g e
7. Basics of Statistics
4. Factor out as much as possible:
a)
b)
c)
3
3
j =1
j =1
∑ 3 x 3 j =3 ⋅ ∑ x 3 j
3
3
j =1
j =1
∑ 3x ⋅ x j =3x ⋅ ∑ x j
5
5
5
k =3
k =3
k =3
∑ (2 xk + 4 xk 2 − 6) = 2 ∑ xk + 4 ∑ xk 2 − 18
7.1 Presentation Techniques
P a g e | 157
7.1 Presentation Techniques
7. Basics of Statistics
7.0 Sum Symbol
7.1
Presentation
Techniques
7.2 Key Figures:
Centre
7.3 Key
Figures: Spread
Tables
Mean
Variance
Diagrams
Median
Standard
Deviation
Sum
Curves
Mode
Normal
Curve
Prerequisites:
There is no special mathematical knowledge
required except your readiness to find out about the
best way to represent large sets of data in order to
gain a qualitative overview of the whole.
Excel provides a wide range of tools which help to
generate tables and diagrams. Therefore it is a good
idea to study the options available in the “statistical
functions” section of the program. It is also very
helpful to know how different diagrams can be
produced using Excel or similar tools.
158 | P a g e
7. Basics of Statistics
Learning Targets: Descriptive statistics is the art of summarising large
sets of data.
In all our institutions we are dealing with thousands
of data records, such as disbursed loans, maturities,
account balances or average transactions. We are
interested in summarising the observed data in such
a way that the management, our colleagues and we
ourselves are able to grasp and understand the main
attributes of the complete list. Tables, diagrams and
curves are the main instruments for doing so.
7.1.1 Tables
A list of characteristic values or property values forms a statistical
sequence. Very often we get lists with just one entry; in this case the
attribute values form a one-dimensional list. The attributes may be:
•
Nominal: for instance → Name (John, Anna, Juan,...); Marital
status (single, married); Gender (female, male); etc.
•
Ordinal (can be ordered by connotation): for instance → grades
(A, B, C,...); dress size (S, M, L, XL,...); loan size range (micromicro, micro, small, medium,…); etc.
•
Cardinal (numbers): for instance → temperature (25°C), speed
(100 km/h); loan amount (EUR 2500); age (36 years); etc.
In many cases, especially in the banking sector, the attributes are cardinal
or ordinal. Therefore we assume throughout this reader that the attribute
values can be ordered by connotation or by size, i.e. that they are ordinal
or cardinal.
Let us consider an Example the grades attained by 82 students in their
final exam in statistics. The list of all students would look like Tab. 7.1,
which is not shown in its full length. If we want to know whether or not
the class as a whole was successful in the exam, the list with the
individual grades does not help us much because it is too detailed.
Therefore, we usually start by creating a tally list, which simply means
counting the frequencies of the different grades.
7.1 Presentation Techniques
No
1
2
3
4
5
Name
Alois
Juan
David
Kathrin
Annette
......
Pts
46
80
92
108
98
Grade
D
B
C
A
D
Table 7.1:
P a g e | 159
No
78
79
80
81
82
Name
......
Willi
Jan
Hans
Rolf
Pamela
Pts
Grade
113
78
29
97
100
A
B
F
A
A
List of grades
The result could look like Tab. 7.2, which provides a much better
summary of the class’s exam performance. With nine A grades, only 3 Fs
(failed), and a relatively large number of Bs and Cs, one could state that
the results are encouraging. Thus we can see that the frequency offers a
much clearer overall impression than the preceding detailed list.
Grade
Frequency
Percent
Relative
Frequency
Sum of Rel.
Frequencies
A
B
C
9
25
34
11.0%
30.5%
41.5%
0.110
0.305
0.415
0.110
0.415
0.829
D
F
11
3
13.4%
3.7%
0.134
0.037
0.963
1.000
Sum
82
100.0%
Table 7.2:
Frequency table of grades
The frequency table provides a good overview of the exam: We can see
at a glance that the overall result is rather good, because good grades are
more frequent than poor grades. Therefore, by summarising the long list
of individual grades we have achieved a much clearer impression of the
quality of the class’s exam performance at the price of losing the details
of the individual students’ grades.
160 | P a g e
7. Basics of Statistics
For each grade (each line) the percentage is the portion of students with a
9 = 11%
certain grade relative to all students: “9 out of 82” → 82
The relative frequency is basically the same thing expressed as a decimal
9 = 0.110
number: “9 out of 82” → 82
For a list of n items (= grades) the frequency of item k is commonly
f
denoted by f k , and for the relative frequency we use: hk = k
n
The percentage is simply the relative frequency multiplied by 100.
An additional source of information is the “sum of relative frequencies”
(the last column of Tab. 7.2). This is the sum of all relative frequencies
for all items less than or equal to the item in the corresponding line.
However, the sum of relative frequencies only makes sense for cardinal
or ordinal attribute values. It can be very useful to know the sum of
relative frequencies. For instance, if we know that grades A to D
represent a “pass”, the sum of the D grades tells us that 96.3% of all
students passed with grade D or better. In the following we usually refer
to this curve as the sum curve.
EXCEL:
In the past, tally lists of frequencies used to be created
very laboriously using pencil and paper. With software
like Excel it is now much easier to analyse even very long
lists using the function COUNTIF. In the above example,
COUNTIF counts the number of cells within a range that
meet a given condition. The range would be the list of all
grades; the condition would be the corresponding grade.
The result is the second column of Tab. 7.2.
GROUPED DATA
Suppose you are the statistician of your bank and you are asked to
provide an overview of the number of loans in the loan portfolio.
Obviously it would make no sense to use one category for every different
principal amount, because there would simply be too many items (=
different principals) in the list. Therefore, all of the banks in our group
classify their loans into different size categories by dividing the attribute
7.1 Presentation Techniques
P a g e | 161
value into intervals from “micro-micro” to “large” and assigning each
loan to the appropriate category.
In the following Tab. 7.3 we illustrate10 the classification scheme by
numbers. For the purposes of illustration, we have chosen to use only
four groups: Micro, Small, Medium and Large. “Left” (“Right”) denotes
the left (right) bound of the corresponding group in EUR thousands.
Thus, “Micro” ranges from 0 to EUR 4000, while “Medium” stands for
loans between EUR 10,000 and EUR 30,000. “Middle” refers to the
middle point in the corresponding interval.
The frequency in the category “Micro” means that we identified 377 loan
contracts with disbursed amounts smaller than or equal to EUR 4000.
The relative frequency shows that these make up 13.4% of all loans, and
the sum of relative frequencies tells us that, for instance, 87.4% of all
loans are smaller than EUR 30,000.11
If we have to summarise large sets of data with many different attribute
values, we have to categorise the list entries into different groups. That
means that we must further condense the large volume of information.
The grouped frequency tables with grouped data provide a very
comprehensive insight into the loan portfolio. However, we gain this at
the price of the loss of detailed information about the individual
contracts.
No.
Class
Left
Right
Middle
Freq.
Relative
Freq.
Sum of
Rel. Freq.
1
2
3
Micro
Small
Medium
0
4
10
4
10
30
2
7
20
377
1429
649
0.134
0.509
0.231
0.134
0.643
0.874
4
Large
30
90
60
355
0.126
1.000
2810
1
Sum
Table 7.3:
10
Frequencies of grouped data
Neither the numbers nor the scales are realistic. They have been chosen simply in
order to create tables and diagrams that clearly illustrate the intended purpose.
11
Again, the figures presented here are not realistic. The magnitudes have been chosen
in order to demonstrate the properties of the diagrams in an easily understandable way.
162 | P a g e
7. Basics of Statistics
In Tab. 7.3 the number of entries within each group and their relative
frequencies are listed. The relative frequency is the number of entries per
group divided by the number of entries in the list. The middle point of the
group is often used as the representative or mean value.
Let us denote:
k=
the count index for the class k
j=
the count index for the item j
m=
the number of attributes (groups) defining the length of the list
n=
the number of items (entries) in the statistical sequence
(population)
f k = the frequency of items in group k
hk = the relative frequency of items in group k
FREQUENCY AND RELATIVE FREQUENCY OF GROUPED DATA
Using the above symbols we can conclude:
m
n = ∑ fk →
The sum of all frequencies is equal to the size of the
k =1
population (the length of the list of all items).
hk =
fk
→
n
The relative frequency is given by the absolute frequency
divided by the size of the population.
m
∑ hk = 1
k =1
EXCEL:
Grouped data is a simple but extremely efficient way to
aggregate large sets of data and thus concentrate
information. However, it is also hard work to go through
long lists and count entries of different groups using a tally
list. Excel provides a helpful function to count
frequencies. In the “Statistical Functions” domain you will
7.1 Presentation Techniques
P a g e | 163
find the function FREQUENCY. However, it is slightly
complicated to apply:
•
Define the intervals of the groups; they can be
equidistant, meaning that the intervals all have the
same width. They may also have different widths. The
right endpoints are included; for instance, 15 will be
in group k = 1 of Tab. 7.5.
•
Include the function FREQUENCY in the first
position of column “absolute frequency” and define as
“Data Array” the list of entries which are to be
grouped.
•
Define “Bins-Array” by marking the array of Right
end points in the class list.
•
Pull formula into all class rows and keep them
marked.
•
Press key F2 and then CTRL+SHIFT+ENTER.
EXAMPLE
In Tab. 7.4 a list of 60 entries is given in the form of a matrix:
25
25
49
66
57
15
17
22
13
1
16
19
21
23
24
30
25
43
29
28
26
28
26
27
28
5
26
27
39
33
Table 7.4:
EXCEL:
20
52
19
29
27
30
26
29
28
29
24
30
32
35
38
37
26
27
27
26
21
21
30
40
63
25
45
12
48
41
List of data
Assuming that the data array is in the area D6:O10 and the
bins-array (the left value of the class range) is in the area
S5:S10, the Excel formula in the first line of column
“Abs.Freq.” has to be:
=FREQUENCY(D6:O10;S5:S10)
164 | P a g e
7. Basics of Statistics
This formula is tracked down the complete table; the
column stays marked. Then press F2 and afterwards
CTRL+SHIFT+ENTER. The result will be the frequencies
f k in column “Abs.Freq.”. Their sum is n = 60.
fk
and the sum of
n
the relative frequencies "Freq.Sum". The result is shown in Tab. 7.5.
Now you can calculate the relative frequencies: hk =
Class
k
Left
Right
Middle Abs.Freq.
lk
rk
mk
fk
hk
1
0
15
7.5
5
0.0833
0.0833
2
15
25
20
16
0.2667
0.3500
3
25
30
27.5
23
0.3833
0.7333
4
30
40
35
7
0.1167
0.8500
5
40
50
45
5
0.0833
0.9333
6
50
70
60
4
0.0667
1.0000
60
1
Table 7.5:
Rel.Freq.
Freq.Sum
Grouped frequencies
Just imagine: In your bank the list of items probably contains several
thousand entries. In this case it is very important
1.
to summarise the data in suitable frequency tables and
2.
to apply Excel in order to compile the list.
7.1 Presentation Techniques
P a g e | 165
7.1.2 Diagrams
Another good way to summarise large sets of data is the graphical
representation. Typically one first has to condense the data in frequency
tables which then serve as the sources of data for column, line, pie, bar or
area diagrams. The purpose of a diagram is to communicate the main
features of the information to the addressee. The properties of the
complete set of data can be grasped at a single glance.
Let us use as an example the frequency Tab. 7.2 showing the exam
results. One way to judge the quality of the students’ exam performance
would be to count the excellent and very good grades (A and B) and
compare them with the corresponding numbers in other categories (say,
D and F). In Fig. 7-1 we have created a column diagram based on the
frequency distribution. The heights of the columns communicate a clear
picture that the majority of the students achieved good grades because the
concentration of the columns towards the good grades (the left side) is
obvious.
Number of Students
Results of the Exam
40
35
30
25
20
15
10
5
0
A
B
C
D
F
Grades
Figure 7-1:
Column diagram of grades
In principle, we can handle tables with the attribute values fragmented in
intervals (grouped attribute values) in the same way. The result is shown
in Fig. 7-2. Again, we receive a clear picture: Almost 400 loans are in the
range 0 – EUR 4,000 (Micro), while about 1,400 loans of between EUR
4,000 and EUR 10,000 (Small) are outstanding. However, Fig. 7-2 differs
166 | P a g e
7. Basics of Statistics
from Fig. 7-1 in an important respect. In Fig. 7-1 the columns are
assigned to a single attribute only, while in Fig. 7-2 each column
represents the whole range of values of the corresponding class.
EXCEL:
Excel provides a large set of different diagram templates.
When you select Insert Chart you are prompted to choose
a template from the list. The chart wizard offers a lot of
support for creating the appropriate presentation.
Loan Portfolio by Class of Loans
Number of Loans
1600
1400
1200
1000
800
600
400
200
0
M ic r o
Sm all
M e d iu m
L ar ge
Categories
Figure 7-2:
Column Diagram of a frequency table
In the above column diagram, each column represents a whole class.
Consequently, statisticians often use a slightly modified form of the
column diagram.
•
They select a representative point for the class, which is in many
cases the middle point of the interval.
•
They close the gap between the columns, demonstrating that the
area of the column represents the frequencies of the group as a
whole.
•
They divide the height of each column, i.e. the frequency of the
class members, by the width of the interval. Thus, the area of the
7.1 Presentation Techniques
P a g e | 167
column, i.e. the height multiplied by the width, becomes
equivalent to the portion of the class as a whole.
In Tab. 7.3 we divided the frequencies (column "Freq.") as well as the
relative frequencies ("Rel.Freq.") by the width of the corresponding
class. In Tab. 7.6 we see the results in columns headed “Freq. Density”,
and “Rel.Freq. Densisty”, respectively.
L
R
M
Freq.
Micro
0
4
2
377
94.25
Small
Med.
Large
4 10 7
10 30 20
30 90 60
1429
649
355
238.17
32.45
5.92
Sum
2810
Table 7.6:
Rel.
Freq.
Density
Sum of
Rel.
Freq.
0.134
3.35%
0.134
0.509
0.231
0.126
5.09%
0.66%
0.18%
0.643
0.874
1.000
Freq.
Rel.
Density Freq.
Class
1
Frequency density of grouped data
Finally, we again produce a column diagram with the relative frequencies
(in percentages) as data: Fig. 7-3. This type of diagram is referred to as
the density distribution of the relative frequencies of the class attributes.
Statisticians call this type of column diagram a histogram. Because the
area of each column is exactly equivalent to its portion of the whole, the
sum of all areas must be 1.
168 | P a g e
7. Basics of Statistics
Density Distribution
Frequency Density
6 .0 %
5 .0 %
4 .0 %
3 .0 %
2 .0 %
1 .0 %
0 .0 %
M icro
Small
M e dium
L arge
Categories
Figure 7-3:
EXCEL:
Column diagram of a grouped frequency table
It is common sense to close the gaps between the different
columns in the diagram when representing densities. This
is because in the density distribution the entire area under
the curve will be normalised to 1. If the gaps are closed,
all the columns are connected.
If you want to close the gaps, just click on one column; all
columns should now be marked. Choose FORMAT
SECTION in the FORMAT menu. In the opened dialog
you can change the gap width to 0%.
7.1.3 Sum Curves
In Tab. 7.2 and in all the following frequency tables we created a column
for the “Sum of Relative Frequencies”. The idea behind taking the sum
was to obtain the frequency for all entries less than or equal to the
corresponding attribute value. For instance, we found that the frequency
of grades equal to or better than (= less than or equal to) grade B turned
out to be 0.415 = 41.5% (see Tab. 7.2), or that micro or small loans
accounted for 64.3% of the total loan portfolio based on Tab. 7.6. In
order find out how many numbers in the list of Tab. 7.4 will be between
16 and 40 we could either add the frequencies of the three classes k = 2,
7.1 Presentation Techniques
P a g e | 169
3 and 4 or simply subtract from the frequency sum for class k = 4 (0.85)
the frequency sum for class k = 2 (0.0833), giving 0.7667. Of course, in
short lists it can also be calculated by adding the frequencies of the
classes k = 2, 3 and 4. However, for very long lists it is much more
convenient to obtain the frequencies of intervals by simply subtracting
the frequency sums.
It is now quite straightforward to represent the frequency sum in a
column diagram too. The result for the data of Tab. 7.2 is given in Fig.
7-4. In order to accentuate continuity the columns of the sum frequencies
are usually sketched without gaps. In each column you will find the
relative frequency of each group on top of the cumulative relative
frequencies of the groups left of it. The dotted horizontal lines are added
in order to highlight this.
Sum of Relative Frequencies
Relative Frequencies
1.20
1.00
0.80
0.60
0.40
0.20
0.00
A
B
C
D
F
Grades
Figure 7-4:
Column diagram of the frequency sum of grades
If we illustrate the frequency sums of Tab. 7.6 in the same way, we
obtain a very similar diagram (see Fig. 7-5). Both diagrams have in
common that they cumulate towards the value 1, because the sum of the
relative frequencies is always 1.
170 | P a g e
7. Basics of Statistics
Sum Curve (Loan Portfolio)
Relative Frequencies
1 .2 0 0
1 .0 0 0
0 .8 0 0
63.4%
0 .6 0 0
0 .4 0 0
0 .2 0 0
0 .0 0 0
M ic r o
S m a ll
M e d iu m
L ar ge
Catagories
Figure 7-5:
Frequency sum of loans as a column diagram
We have already pointed out that the sum curve is very informative if we
are interested in the relative frequencies of more than one class. For
instance, the arrow on top of the second column at the value 0.643 on the
vertical axis shows that 64.3% of all loans are either micro or small.
However, if we were interested in estimating some cumulative value not
exactly on top of one the columns, the diagram would not help us. In this
case, however, a logical assumption may help us: If we assume that the
individual loans in each class are equally distributed, meaning that each
loan size within a class has the same probability, a diagonal line in each
column of relative frequencies would represent the distribution within the
class. If we now transfer the diagonals together with the columns into the
cumulative sum diagram, all pieces of the diagonals are connected to
each other. Together they form a so-called polygon of linear pieces of
lines whose form is usually s-shaped. This line is called the distribution
curve of the underlying statistic.
7.1 Presentation Techniques
P a g e | 171
Sum Curve (Loan Portfolio)
Relative Frequencies
1 .2 0 0
1 .0 0 0
0 .8 0 0
0 .6 0 0
0 .4 0 0
0 .2 0 0
0 .0 0 0
0
1000
M ic r o
8500
S m a ll
20000
10000
M e d iu m
30000
90000
L ar ge
Catagories
Figure 7-6:
Cumulative sum curve and distribution function
With reference to Fig. 7-6 we can now estimate – due to the many
assumptions the consequences are always only rough estimates and never
real results – that 50% of all loans in the portfolio may be smaller than or
equal to EUR 8,500. Or we may conclude that about 78% of all loans are
less than or equal to EUR 20,000. The difference of the two values,
namely 28, tells us that 28% of the loans are consequently between EUR
8,500 and EUR 20,000.
If we made the group sizes smaller, the distribution curve would be
smoother. In the limit the curve loses its property of being a linear
polygon. Instead it is a smooth s-shaped curve, which for continuous
distributions is finally the distribution function.
7.1.4 Base Problem
Statistics deals with large sets of entries. This makes it absolutely
necessary to use statistical software. For this reader we apply the broad
array of statistical functions available in Excel.
In the classroom we can discuss only small examples and illustrate the
topics using statistical lists with only a few entries. In order to provide
exercises which are more realistic, we will draw up a list with 500
entries. It will be generated by means of random numbers using Excel.
172 | P a g e
7. Basics of Statistics
For most of the following exercises we will formulate a special question
related to this problem. Due to the size of the data set, it can only be
answered using Excel.
This problem will be called the “Base-Problem”. Each student must
generate this example on his/her own computer. None of the problems
generated will be the same, but their properties will be statistically
identical.
In order to generate the problem, prepare a table with 20 columns and 10
rows. In the upper left corner assign the following Excel function to the
first coefficient:
= RANDBETWEEN(0;9)
This generates a random number between 0 and 9. Track the formula
over the complete 10×20 matrix, generating 10 random numbers in each
column. Place the sum of the ten random numbers in an extra row. We
refer to this row as the generating “base line” for the final list. The base
line changes its values randomly with every input in Excel.
Now we will set up the problem in a separate table of 25 rows and 20
columns.
•
COPY the base line.
•
PASTE it to the next empty line of the final list applying the
special paste PASTE VALUE.
•
Repeat this procedure for the 25 lines of the final table. The
easiest way to do this is to use function key F4 repeatedly for
each empty row.
The table is now complete. Save it under the name “Base-Problem”. We
will refer to this problem under this name regularly throughout this
chapter.
The following Tab. 7.7 contains the data generated as the sum of 10
random numbers in the range 0 to 9. It is inserted only to provide an
example with which you can compare your own list. The head row and
the head column (both bold and italic) are not part of the list. They are
added only for better orientation.
7.1 Presentation Techniques
P a g e | 173
ATTENTION
The problem you have generated will contain completely different
entries. Do not allow this to confuse you: the derived results (e.g.,
average, median, variance, standard deviation etc.) may also differ
slightly from the given values. However, the differences will be so minor
that they are insignificant. All lists generated using the above method
have the same statistical properties.
38
43
67
62
35
40
36
46
49
47
38
35
38
45
51
56
41
51
37
46
41
36
53
36
54
54
49
57
43
28
48
49
49
29
41
40
38
66
47
59
32
50
52
42
37
60
55
58
62
43
36
51
48
55
25
63
19
40
31
39
40
39
36
51
38
36
52
52
37
47
41
42
36
54
55
52
45
40
57
41
42
53
26
52
48
45
57
37
51
27
33
55
40
48
40
56
51
40
35
47
44
53
31
41
34
32
43
38
57
30
29
37
59
43
42
44
41
54
30
54
52
46
41
44
39
43
57
42
43
34
55
51
38
51
52
38
53
43
59
31
55
54
44
37
45
38
52
39
52
58
33
42
41
63
37
38
41
42
31
39
43
62
49
37
60
41
48
55
44
40
35
54
34
45
47
39
54
40
53
61
60
37
37
55
47
46
43
55
32
37
29
29
43
38
36
45
52
44
46
44
54
36
68
52
38
46
49
47
43
54
49
46
38
47
45
53
45
47
57
32
31
58
48
64
53
39
42
46
38
49
32
47
48
34
46
45
46
47
38
18
40
49
51
40
35
36
43
41
57
60
37
51
26
34
54
51
48
29
42
64
54
37
46
55
50
44
38
48
52
57
40
33
47
34
36
42
46
45
26
39
41
43
44
47
50
56
58
45
49
45
65
46
41
43
52
41
48
56
49
61
47
36
49
46
44
34
39
52
53
59
37
67
50
43
27
26
49
41
42
48
47
40
42
36
45
48
54
49
31
42
50
49
54
40
48
51
58
47
55
51
47
58
31
51
55
50
50
39
36
65
61
51
60
48
39
25
71
43
35
36
174 | P a g e
7. Basics of Statistics
45
70
43
36
42
55
42
42
56
61
40
32
54
27
42
47
57
49
35
40
24
46
56
37
40
41
39
49
39
51
49
50
47
60
28
57
46
46
47
66
46
43
61
49
50
42
38
40
40
43
27
57
33
60
47
43
44
47
46
38
48
51
46
28
40
40
54
32
33
31
42
38
40
44
52
52
27
43
38
44
45
47
40
40
50
43
28
53
44
62
52
52
49
46
53
57
48
39
50
38
50
44
33
47
33
43
55
54
52
52
45
47
35
42
41
57
41
27
44
58
53
40
17
47
47
41
40
40
34
55
48
49
54
29
62
54
46
32
40
60
Table 7.7:
“Base-Problem”
7.1 Presentation Techniques
EXERCISE 7.1:
P a g e | 175
PRESENTATION TECHNIQUES
1. After reweighing 40 0.5kg packages of sugar Anna found the
following values (in grams = g).
492 497 497 491 499 504 495 503 499 504
495 496 502 500 499 500 508 502 500 499
498 501 493 506 494 511 496 488 494 500
506 503 497 497 498 504 499 504 501 498
Set up frequency tables with:
a)
Class width = 2g
b)
Class width = 4g
2. The following list contains the grades scored by German students
in their maths exam. Set up a frequency table and estimate the
mean grade.
2
2
4
3
2
1
1
4
2
5
3
4
3
2
1
1
2
4
3
3
3
5
4
2
5
3
1
3
5
1
1
4
3
2
4
3
5
1
2
2
3
4
3
3
4
2
5
1
2
1
1
2
4
4
3
5
3
2
3
2
4
5
2
1
4
3
5
2
3
2
Use Excel – if available – in order to present the distribution as a
column diagram.
176 | P a g e
7. Basics of Statistics
3. Given the following list of 60 numbers:
25
6
16
30
26
5
20
30
24
37
35
17
19
18
33
26
45
16
30
26
49
22
21
48
26
27
19
29
32
7
12
13
23
19
27
39
29
28
45
37
34
1
24
28
28
33
17
29
38
36
21
25
21
45
20
12
40
48
22
41
a)
Set up a list with grouped data. Choose the class sizes (0-10,
10-20, 20-30, etc.) with the right values included, and the left
values excluded.
b) Use Excel to set up the table of grouped data.
c)
Calculate the sum of absolute frequencies and present them in
a column diagram. Close the gaps between the columns in the
diagram.
d) Add the s-shaped piecewise linear sum curve.
7.1 Presentation Techniques
P a g e | 177
4. In the last statistics exam the teacher set up the following
frequency table of marks:
Marks
from
to
0
21
51
20
50
75
50
90
170
76
100
90
Sum
a)
Frequency
400
Sketch the distribution function via relative frequencies, sum
of rel. frequencies and the sum curve.
b) Estimate the percentage of students who scored 60 or higher.
5. Use the “Base-Problem”:
a)
Guess the average value of the 500 numbers.
b) Define class widths of 10 (11-20, 21-30, etc.) and set up the
table of frequencies for grouped data.
c)
Create a histogram and estimate the mean value.
d) Choose the alternative class width of 5 and repeat b) and c).
Compare the results dependent on the class size.
6. Use the “Base-Problem” and the frequency tables of problem 5:
a)
Create a diagram of the sum of relative frequencies and the
sum curve.
b) Mark and estimate the percentage of numbers between 37 and
54.
c)
Repeat the process for the table with class size 5 and compare
the estimates.
178 | P a g e
ANSWERS 7.1:
1.
a)
7. Basics of Statistics
PRESENTATION TECHNIQUES
Min is 488 and Max is 511
Class
Freq.
Class
Freq.
488-489
1
500-501
6
490-491
1
502-503
4
492-493
2
504-505
4
494-495
4
506-507
2
496-497
6
508-509
1
498-499
8
510-511
1
Sum
40
b)
Class
Freq. Rel. Freq.
488-491
492-495
496-499
500-503
2
6
14
10
5%
15%
35%
25%
504-507
508-511
6
2
15%
5%
Sum
40
100%
7.1 Presentation Techniques
P a g e | 179
2.
Grade Freq.
Rel. Freq.
1
2
3
12
18
18
17.1%
25.7%
25.7%
4
5
13
9
18.6%
12.9%
Sum
70
1
Grade Distribution
Relative Frequencies
30%
25%
20%
15%
10%
5%
0%
1
2
3
4
Grades
3.
Groups Middle Abs.Freq. Rel.Freq.
Freq.Sum
1-10
5
4
0.0667
0.0667
11-20
15
13
0.2167
0.2833
21-30
25
25
0.4167
0.7000
31-40
35
11
0.1833
0.8833
41-50
45
5
0.0833
0.9667
51-60
55
2
0.0333
1.0000
60
1
Sum
5
180 | P a g e
7. Basics of Statistics
Sum of Relative Frequencies
Sum Curve
1.2
1.0
0.8
0.6
0.4
0.2
0.0
1-10
11-20
21-30
31-40
41-50
51-60
Groups
a) and b)
Marks in Statistics Exam
Relative Frequencies
4.
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0.0
0-20
21-50
Groups
51-75
76-100
7.1 Presentation Techniques
P a g e | 181
Sum of Relative Frequencies
Sum Curve of Marks
1.2
1.0
0.8
0.6
0.53
0.4
0.2
0.0
60
0-20
21-50
51-75
76-100
Groups
The estimated percentage of students with marks better than 60 is
100% - 53% = 47%.
5.
Class widths of 10:
Group
till
Freq.
Rel. Freq.
Freq. Sum
10-20
20
3
0.6%
1%
21-30
31-40
41-50
30
40
50
25
134
195
5.0%
26.8%
39.0%
6%
32%
71%
51-60
61-70
71-80
60
70
80
120
22
1
24.0%
4.4%
0.2%
95%
100%
100%
500
100%
182 | P a g e
7. Basics of Statistics
Relative Frequencies
Frequency Distribution (Width: 10)
45%
40%
35%
30%
25%
20%
15%
10%
5%
0%
45
1-10
21-30
31-40
41-50
51-60
Groups
Estimated mean is 45.
Class widths of 5:
Group
till
Freq.
Rel. Freq. Freq. Sum
16-20
21-25
20
25
3
3
0.6%
0.6%
0.6%
1.2%
26-30
31-35
36-40
30
35
40
22
39
95
4.4%
7.8%
19.0%
5.6%
13.4%
32.4%
41-45
46-50
51-55
45
50
55
96
99
82
19.2%
19.8%
16.4%
51.6%
71.4%
87.8%
56-60
61-65
66-70
60
65
70
38
16
6
7.6%
3.2%
1.2%
95.4%
98.6%
99.8%
71-75
75
1
0.2%
100.0%
500
100.0%
61-70
71-80
7.1 Presentation Techniques
P a g e | 183
Frequency Distribution (Width: 5)
Relative Frequencies
25%
20%
15%
10%
5%
0%
45
16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 71-75
Groups
Estimated mean is 45.
6.
Sum of relative frequencies
Sum Curve (width: 10)
120%
100%
80%
60%
40%
20%
0%
1-10
21-30
31-40
41-50
51-60
61-70
Groups
Estimated percentage of numbers between 37 and 54 is:
81% – 22% = 59%
71-80
184 | P a g e
7. Basics of Statistics
Sum of relative frequencies
Sum Curve (width: 5)
120%
100%
80%
60%
40%
20%
0%
16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 71-75
Groups
Estimated percentage of numbers between 37 and 54 is:
82% – 20% = 62%
7.1 Presentation Techniques
P a g e | 185
7.1.5 Progress Test “Presentation Techniques”
You should assign yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Don’t use the reader to look
for the solution. The aim of the test is to give you feedback on how much
you know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. Otherwise
(if you got the wrong answer or no answer at all) you will be given
advice on what to do. In most cases you will be directed to the section in
the reader which you should repeat in order to close the gap.
1. Evaluate the sum and simplify the expression:
n
∑ 2( x j − a)2 =
j =1
2. The following table shows the application data for six ProCredit
banks and the respective percentages of admissions:
Men
Admissions
Women
Applicants Admissions
Bank
Applicants
A
B
224
267
43%
54%
345
336
62%
57%
C
D
E
127
117
191
23%
37%
44%
287
256
312
34%
29%
59%
F
263
67%
211
43%
Calculate the weighted average of admissions for men and
women.
186 | P a g e
7. Basics of Statistics
3. The following table shows the number of disbursed loans per loan
officer.
22
17
19
21
22
26
23
21
18
25
29
33
21
26
28
22
31
33
35
27
24
22
23
24
29
30
28
20
31
24
25
26
37
27
34
23
24
36
23
33
a)
Set up a tally list of frequencies.
b) Draw a column diagram of the relative frequency
distribution.
4. Given the following list of 40 numbers.
22
17
19
21
22
26
23
21
18
25
29
33
21
26
28
22
31
33
35
27
24
22
23
24
29
30
28
20
31
24
25
26
37
27
34
23
24
36
23
33
Set up a tally list after grouping the data in steps of 3 (17-19, 2022, etc.) and draw a column diagram of the relative frequencies,
the frequency sum, and the sum curve.
5. Using the sum curve from problem 4, estimate the numbers
between 24 and 30. How many numbers really were within this
range?
7.2 Key Figures: Centre
P a g e | 187
7.2 Key Figures: Centre
7. Basics of Statistics
7.0 Sum Symbol
7.1 Presentation
Techniques
7.2 Key
Figures: Centre
7.3 Key Figures:
Spread
Tables
Mean
Variance
Diagrams
Median
Standard
Deviation
Sum
Curves
Mode
Normal
Curve
Prerequisites:
To work through this section it is necessary to be
familiar with the sum symbol introduced and
discussed in Section 7.0.1. Very obviously, when
calculating with large sets of numbers it is necessary
to be able to use notation which allows us to argue
without writing long expressions.
Again, we would like to remind the reader that Excel
provides a wide range of functions which can be
used to calculate the figures needed. In most cases
we will refer to these
the functions explicitly.
188 | P a g e
7. Basics of Statistics
Learning Targets: Descriptive statistics is the art of summarising large
sets of data.
Having summarised the data by means of graphical
tools, we now introduce key figures which also offer
certain concentrated information about the statistical
list and its attributes.
The figures will be calculated. We differentiate
between the various types of information about the
centre of the data. The catchwords are mean,
median, and mode.
By combining information about the centre of the
data with information about their concentration or
spread, which will be discussed in the next section,
we can obtain quite an accurate picture of the
distribution of data without drawing a diagram.
An appropriate and convincing method of analysing distributions of large
sets of data with cardinal attribute values is to represent them graphically.
This kind of visual analysis is especially useful in presentations.
However, the analysis of diagrams and curves is less effective when we
need concentrated information about data for further quantitative
analysis. For this purpose it is much more efficient to characterise
distributions by key figures = measurement figures. They are constructed
in such a way that they offer a comprehensive picture of certain central
properties of the distribution. In this section we will calculate figures
which characterise certain accentuated positions of the distribution. They
can be referred to as the “big M’s”:
•
Mean
•
Median
•
Mode
•
Minimum
•
Maximum
These figures already allow us to gain a very good understanding of the
distribution.
7.2 Key Figures: Centre
P a g e | 189
7.2.1 Mean
The most prominent figure for describing distributions is the statistical
mean (often also called: average, midpoint, average value, mean value,
or arithmetic mean). The many synonyms are a strong indicator that
almost everybody has a rough idea of how to calculate a mean, because it
is quite common in daily life to argue with averages. We calculate the
average maturities of loans, the average days of arrears, the average fuel
consumption of our car, the average speed of the finishers in a marathon,
the average family income, or the average tax.
How do we calculate these average values? Most of us are quite familiar
with the process of adding up all the attribute values of which we want to
take the average, and dividing the sum by their number.
EXAMPLE
We assume that the maturities of five loans are: 1 year, 6, 9, 8 and 6
months, respectively. What is the average maturity of the five loans?
In order to calculate the average maturity, we add up the attribute
values.
ATTENTION
When adding up attribute values, make sure that they all have the same
dimension. In other words, do not add apples and pears if you want to
know the number of apples in the basket.
That means, before adding together the maturities, they all have to be
transformed either into years or into months.
Using the dimension “months”, we get the sum:
m = 12 + 6 + 10 + 8 + 6 = 42 months
Dividing this by the number of loans gives: m =
42
= 8.4 months
5
The average maturity of the five loans is: 8.4 months.
190 | P a g e
7. Basics of Statistics
In the following discussion we refer to a statistical list consisting of a set
of entries having (cardinal) attribute values of the same dimension.
STATISTICAL LIST
If we have n observations x1, x2 ,…, xn we call L = { x1 , x2 ,… , xn } a set or
list of observations.
n is the number of entries in the list.
The sum of all observations or attribute values is:
n
x1 + x2 + ... + xn = ∑ x j
j =1
The sum symbol, introduced in Section 7.1.1 (see: Page 147), will now
be used extensively. If we divide the sum by the number of entries in the
list we will get the average value of all attribute values. We refer to it as
the statistical mean of the attribute values, or – even more precisely – as
the arithmetic mean of the values.
ARITHMETIC MEAN
The arithmetic mean (for short: mean or average) of n observations xi
from the list L = { x1 , x2 ,… , xn } is:
n
∑ xj
x=
j =1
n
=
1 n
∑ xj
n j =1
The mean is commonly denoted by the un-indexed symbol topped by a
bar ( x ).
EXCEL:
In Excel the function AVERAGE returns the average or
arithmetic mean of its arguments.
7.2 Key Figures: Centre
P a g e | 191
EXAMPLE
Use Excel to calculate the average of the numbers in Tab. 7.4 (see
Section 7.1.2; Page 163). The following Fig. 7-7 is a screenshot of the
table using MS Office 2007. The calculated value of the average is
29.1666667.
Figure 7-7:
EXCEL:
Statistical figures in the Excel status bar
Excel provides some additional statistical features which
are very useful for quick information: If you mark any
array in Excel, Office 2007 automatically calculates
different statistical figures for the marked numbers and
displays them in the status bar at the bottom of the screen
(see screenshot in Fig. 7-7):
Average: 29.16667
=
the arithmetic mean of the
numerical attributes.
Count: 60
=
the number of entries in the
list.
Numerical Count: 60
=
the number of numerical
entries in the list. If the list of
attributes contains numerical
and non-numerical values, the
other results are calculated on
192 | P a g e
7. Basics of Statistics
the basis of the numerical
values only.
Min: 1
=
the minimum numerical value
in the list.
Max: 66
=
the maximum numerical value
in the list.
Sum: 1750
=
the sum of all numerical
attributes.
If one or more attribute values appear more than once in the list, the
calculation can be simplified. We assume that the attribute xk appears in
the list with frequency f k . The list of entries now consists of the m
different attributes and the list of corresponding frequencies:
L = {x1, x2 ,..., xm} and F = { f1, f2 ,...., fm }
The total number of entries is in this case:
m
n = ∑ fk
k =1
Every attribute value appears in the sum of attributes as often as the
frequency indicates. Hence, the arithmetic mean is now:
m
∑ xk ⋅ fk
x=
k =1
m
∑ fk
=
1 m
∑ xk ⋅ fk
n k =1
k =1
In this formula we can multiply the sum by the constant factor
1
n
to get:
7.2 Key Figures: Centre
P a g e | 193
m
∑ xk ⋅ fk
x=
k =1
m
∑ fk
=
m
f
1 m
x
⋅
f
=
xk ⋅ k
∑
∑
k
k
n k =1
n
k =1
k =1
We defined the absolute frequency divided by the total number of entries
f
as the relative frequency hk = k (see: Section 7.1.2; Page 167). Thus,
n
we arrive at an alternative formula for the arithmetic mean:
x=
m
1 m
x
⋅
f
=
∑ k k ∑ xk ⋅ hk
n k =1
k =1
EXAMPLE
In Germany exams are graded using a number system. Grade 1 is the
best grade (“excellent”), while grade 5, the lowest, means “failed”.
The following table contains all grades for the 16 students in the
statistics course.
Student i
Grade g j
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2 2 4 3 2 1 1 4 2 5 3 4 3 2 1 3
Counting the frequencies and the relative frequencies of the grades we
obtain:
Grades k
Frequency f k
Rel. Frequency hk
1
3
2
5
0.1875 0.3125
3
4
0.25
4
3
5
1
Sum
16
0.1875 0.0625
Now we calculate the average grade using the three different formulas.
•
Using the grades only:
194 | P a g e
7. Basics of Statistics
1 n
1 ⋅ (2 + 2 + 4 + ... + 1 + 3) = 1 ⋅ 42 =
g = ∑ g j = 16
16
n j =1
•
= 2.625
Using the frequencies:
g=
•
21
8
m
1
n
∑ gk ⋅ f k = 161 ⋅ (1⋅ 3 + 2 ⋅ 5 + 3 ⋅ 4 + 4 ⋅ 3 + 5 ⋅1) = 1642 = 2.625
k =1
Using the relative frequencies:
m
g = ∑ g k ⋅hk = 1 ⋅ 0.1875 + 2 ⋅ 0.3125 + 3 ⋅ 0.25 + 4 ⋅ 0.1875 + 5 ⋅ 0.0625
k =1
m
g = ∑ g k ⋅hk = 2.625
k =1
We can see that the results are exactly identical, regardless of which
formula we use.
EXCEL:
In addition to the Excel function AVERAGE it is
sometimes very helpful to apply the function
SUMPRODUCT. This returns the sum of the product of
the coefficients of two congruent arrays. Congruent means
the arrays must be of identical size. Identically positioned
coefficients of the two arrays are multiplied and added.
For instance, we can apply this Excel function when
m
calculating
∑ gk ⋅ fk or
k =1
m
∑ gk ⋅ hk
k =1
7.2 Key Figures: Centre
P a g e | 195
ARITHMETIC MEAN IN THE CASE OF GROUPED DATA
If the entries and their frequencies are exactly enumerated, as was the
case in the above example, all three formulas yield identical exact results.
However, if the entries are grouped in classes with the middle point as
the reference value, the results differ slightly.
Class
k
Left
lk
1
2
3
4
5
6
0
15
25
30
40
50
Right Middle Abs.Freq.
rk
mk
fk
15
25
30
40
50
70
Sum
Table 7.8:
7.5
20
27.5
35
45
60
Rel.Freq.
hk
Freq.Sum
5
16
23
7
5
4
0.0833
0.2667
0.3833
0.1167
0.0833
0,0667
0.0833
0.3500
0.7333
0.8500
0.9333
1,0000
60
1
Grouped frequencies (Copy of Tab. 7.5)
For example, let us look again at the data of Tab. 7.4 which is also
presented in the screenshot in Fig. 7-7. The analysis is given in Tab. 7.8
which is a copy of Tab. 7.5. The calculated average of the 60 entries x j
is: x = 29.1667
We can apply the formula with the frequencies only if we have one
representative value for each group. A value which offers itself naturally
could be the middle point of the interval (group), which is also calculated
in Tab. 7.8. If we apply the frequency-related formula using the middle
points as representative values we obtain:
1 (7.5 ⋅ 5 + 20 ⋅16 + 27.5 ⋅ 23 + 35 ⋅ 7 + 45 ⋅ 5 + 60 ⋅ 4) = 1700 = 28.3333
x = 60
60
Obviously, there is a difference between the two results which derives
from the inaccuracy of the representative value. The middle point of the
interval is only in exceptional situations the true average of the entries in
the group.
196 | P a g e
7. Basics of Statistics
This might be especially true in the case of loans. We defined for
instance the group of micro loans as consisting of loans between 0 and
4000. In this group there will not be any entries 1, 2, or even 99. The
majority of all micro loans are probably close to or exactly equal to 4000.
Therefore, 2000 is never the representative value of the entries in this
group. It will be much closer to the right-hand interval bound. This is
probably also the case in the other groups; and if so, the arithmetic means
calculated on the basis of the middle points of each interval
underestimate the true average.
Applying the formula with the sum-product of middle point times
relative frequencies gives the same result as the last line:
6
x = ∑ mk ⋅ hk = 28.3333
k =1
7.2.2 Median
Let us have a look at the following list of 60 numbers in Tab. 7.9:
List of Entries
2
7
2
9
3
4
1
4
9
4
3
8
1
9
8
0
3
4
0
2
3
9
4
0
3
1
4
7
9
2
0
4
1
0
2
4
1
0
2
2
9
3
3
8
9
9
8
9
3
5
1
0
9
3
9
2
4
1
8
8
Table 7.9:
List of 60 numbers
The arithmetic mean of these 60 numbers is 4.2. It is probably natural to
guess that 50% of the entries in the last list were above average, and 50%
were below. However, this guess is not correct. In fact, only 33.3% of the
entries were above, and about 6.67% were below average.
7.2 Key Figures: Centre
P a g e | 197
To understand this apparent contradiction better, let us start with a
hypothetical list of data:
L = {1, 9, 1, 9, 1, 1, 1, 9, 1, 1}
34
10
The average of the entries is
= 3.4 ; i.e. 7 entries in the list are below
and only 3 entries are above average. If we arrange the entries in
ascending order and divide them into two parts
L = {1, 1, 1, 1, 1 | 1, 1, 9, 9, 9}
with the same number of entries, the divider lies between two entries
“1”. The value where the entries are divided by size into two equal halves
is called the median. In the example the median is therefore 1, the
average is 3.4.
(a) List: 1, 2, 2, 3; Average 2
Median = Average
50%
25%
0%
0
1
2
3
4
(b) List: 1, 2, 2, 5; Average 2.5
Average
Median
50%
25%
0%
0
1
2 2,5 3
4
5
6
4
5
6
(c) List: 1, 2, 2, 7; Average 3
50%
Median
Average
2
3
25%
0%
0
Figure 7-8:
1
7
8
Histogram balanced at the average
198 | P a g e
7. Basics of Statistics
The histograms in Fig.7-8 explain what is happening. Histogram (a) is
derived from the ordered list La = {1, 2, 2,3} ; it is symmetric around the
average (of 2), and exactly one half of the area lies on either side of the
dotted line.
What happens when the value 3 in the list is increased, say to 5 (the list
will become Lb = {1, 2, 2,5} ) or to 7 (with the list Lc = {1, 2, 2,7} )? The
average moves off-centre to the right if one of the numbers is increased
in value.
If the histogram is made of wooden blocks attached to a stiff board, the
board will always be in balance when supported at the average.
However, although the average moves to the right if the right weight is
moved to the right, the area is still divided into two parts with two blocks
on either side at line 2. The value dividing the area into two equal parts is
called the median. It is another statistical number used to summarise
data.
MEDIAN
The median xmed is the value with half of the entries of a list to the left =
less or equal, and the other half to the right = greater or equal. Since we
divide the list of entries by size, we argue with a sorted list where the
entries are sorted in ascending order.
If the number of entries n is odd, xmed is exactly the middle element and
therefore an element of the list. In a list with an even number of entries
the median lies between two entries in the middle of the list. In this case
the median is the average of the two middle entries. If they are equal, the
median is this value. If they are different, the median is the arithmetic
mean of the two adjacent numbers. In this case the median is not an entry
of the list.
EXCEL:
In the division “statistical functions” Excel provides the
function MEDIAN which returns the median, or the
7.2 Key Figures: Centre
P a g e | 199
number in the middle of the set of given numbers. The
operand of this function can be any array.
EXAMPLES
1.
xj
Let L1 be a list of random numbers x j with average x = 2.95 :
7 3 2 3 5 2 1 3 0 5 1 6 2 1 4 9 1 0 2 1 4
In order to illustrate the median we arrange the numbers in ascending
order:
xj
0 0 1 1 1 1 1 2 2 2 2 3 3 3 4 4 5 5 6 7 9
The list contains an odd number (21) of entries; hence, the median
is the element in the middle of the list, which is the shaded
number:
xmed = 2
In this example the median is smaller than the average: Therefore,
the smaller entries in the distribution “outweigh” the larger ones.
The following frequency distribution of the numbers in Fig. 7-9
shows the left-sided concentration of the numbers. This type of
distribution is called a right-tail distribution.
200 | P a g e
7. Basics of Statistics
Frequency Distribution
Frequencies
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
10
Numbers
Figure 7-9:
Right-tail distribution
Now look at the list L2 with 20 entries; their average is x = 6.15 .
xj 7 6 2 3 5 6 8 7 8 5 8 6 8 7 4 9 8 9 0 7
Again, we arrange the numbers in the list in ascending order:
xj 0 2 3 4 5 5 6 6 6 7 7 7 7 8 8 8 8 8 9 9
The first half with the smaller values is shaded. Since we have an even
number of entries the median lies between 7 and 7, the mean of which
is of course also 7. Therefore, the median of the above list is:
xmed = 7
Now the median is larger than the average the distribution. The larger
numbers in the list outnumber the smaller ones. The frequency
distribution of the entries in Fig. 7-10 illustrates that clearly. This is
called a left-tail distribution.
7.2 Key Figures: Centre
P a g e | 201
Frequency Distribution
Frequencies
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
Numbers
Figure 7-10: Left-tail distribution
In general, the average is greater than the median in right-tail
distributions, the average is less than the median in left-tail distributions,
and the average equals the median in symmetric distributions (see
Fig. 7-11).
Average = Median
Average
Average
50%
50%
Median
Median
Left tail: Average < Median
50%
Symmetric: Average = Median
Right tail: Average > Median
Figure 7-11: The tails of frequency distributions
Taking the average is a powerful way of summarising data, but it also
hides the diversity of the data. However, average and median together
provide some additional information about the shape of the frequency
distribution.
202 | P a g e
7. Basics of Statistics
7.2.3 Mode
Knowing the most frequent entry in the list can also provide additional
information about the statistical properties of the attribute values. For
instance, if we want a “representative value” for the micro loan portfolio,
it may make more sense to choose the most frequent value of all micro
loans, rather than the middle point of the micro loan group. (Remember
that we discussed the problem of selecting the representative value when
we calculated the statistical mean and encountered a bias due to the
selection.)
Hence, it is quite common in statistics to look at
•
the minimum value,
•
the maximum value, and
•
the most frequent value in the list.
MINIMUM VALUE
The minimum value is simply the entry xm such that all other entries in
the list are greater than or equal to the minimum: x j ≥ xm for all j.
Accordingly, we will define the maximum value in the list as follows:
MAXIMUM VALUE
The maximum value is simply the entry xM such that all other entries in
the list are less than or equal to the maximum: x j ≤ xM for all j.
As an example, let us look again at the list of numbers in Tab. 7.4 (see
Section 7.1.2; Page 163).
7.2 Key Figures: Centre
25
25
49
66
57
15
17
22
13
1
16
19
21
23
24
30
25
43
29
28
26
28
26
27
28
5
26
27
39
33
P a g e | 203
20
52
19
29
27
30
26
29
28
29
24
30
32
35
38
37
26
27
27
26
21
21
30
40
63
25
45
12
48
41
After a quick inspection you will probably find:
•
The minimum value is xm = 1
•
The maximum value is xM = 66
It should also be noted that analysing the whole list, although it is
relatively short, is not a trivial task any more. For lists with many more
entries it is useful to get the information using Excel.
EXCEL:
One may use the two functions MIN and MAX, which
return the minimum and the maximum values,
respectively, of an array.
In addition, watch the status bar at the bottom of your
Excel sheet when you mark an array. The minimum and
maximum values are among the information displayed.
MODE
The mode xmod of a statistical list is defined as the most frequent attribute
value in the list.
In order to find the most frequent value it is useful to look at the
frequency distribution. Without this information it would be quite tedious
to find this value.
The mode in the list of numbers in the above table is:
xmod = 26
204 | P a g e
EXCEL:
7. Basics of Statistics
The statistical function MODE returns the most frequently
occurring entry (attribute value) in an array or range of
data.
If more than one value are candidates, i.e. if they appear
equally often with maximal frequency, Excel returns only
one value. It is the value which appears first of all equally
qualified values in the list.
EXAMPLES
1.
The mode of the list L1 = {1, 2, 4,5, 4, 2,3, 2,1,5} is:
xmod = 2
2.
The list L2 = {2, 7,5, 6,5, 4, 5, 6, 7, 6, 4, 2, 7,3} has more than one
most frequent element. Each of the numbers 5, 6 and 7 appear
three times. Hence they all are qualified. Excel, though, returns 7
because, of the three candidates, the 7 appears first in the list.
ATTENTION – TYPICAL MISTAKES
Calculating the average
average ≠
minimum + maximum
2
average ≠
minimum + maximum
number of entries
7.2 Key Figures: Centre
EXERCISE 7.2:
1. a)
P a g e | 205
KEY FIGURES: CENTRE
Calculate the average of the numbers in the list:
1
-2
4
5
6
-3
0
3
7
-4
3
5
-6
-3
4
5
-3
4
8
5
b) Sort the numbers in ascending order and determine the
median.
2. Given is the list of grades according to the German system (from
1 to 5) in the statistics exam:
Student i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Grade
a)
1 3 4 3 2 1 2 4 2 5 3 4 3 2 1 5 2 3 4 3
Calculate the average grade.
b) Compile a frequency table.
c)
Verify the average by applying the two alternative formulas
using the absolute and relative frequencies.
206 | P a g e
7. Basics of Statistics
3. The following list of 70 grades was already used in exercise 7.1,
problem 2:
2
2
4
3
2
1
1
4
2
5
3
4
3
2
1
5
1
1
2
1
4
4
3
3
3
2
3
4
5
3
4
5
2
1
5
2
3
2
1
3
3
4
3
3
3
2
4
3
2
2
5
4
1
5
2
2
1
1
1
4
2
3
4
5
4
2
3
3
5
2
Use Excel to determine
a)
the average,
b) the median,
c)
the mode of the list.
4. Use the frequency list which you compiled in exercise 7.1,
problem 2, and determine the average without using Excel.
5. Given is a list of 50 randomly generated numbers between 1 and
50:
a)
25
29
16
30
26
5
20
30
24
37
35
17
24
18
33
26
45
16
30
26
21
22
48
28
50
23
19
27
32
7
12
13
7
19
27
39
29
3
45
37
34
1
24
8
28
33
17
29
45
36
Group the data in five classes, each with width 10, i.e. 1-10,
11-20, etc., and use a tally list to compile a frequency table.
7.2 Key Figures: Centre
P a g e | 207
b) Calculate the average using the absolute and relative
frequencies, respectively, and the class middle points (5, 15,
etc.).
c)
Calculate the exact arithmetic mean of the numbers with
Excel, and compare it with the results of 5 b).
6. Given is a list of 14 integers:
-3
a)
4
2
-2
5
-1
1
6
4
1
6
0
1
4
Calculate the average, median, and mode.
b) Determine whether the frequency distribution is basically
left-tail or right-tail?
7. Use the “Base-Problem”:
a)
Calculate the mean, median and mode with the appropriate
Excel functions. What conclusions can be drawn from the
values of average and median?
b) Use the frequency tables (with class width 10) compiled in
exercise 7.1, problem 5 b) and calculate the average again.
Compare the result with that of 7 a).
c)
Calculate the average using the smaller class width of 5 (see
exercise 7.1, problem 5 d)) and compare the result with that
of 7 b).
208 | P a g e
7. Basics of Statistics
ANSWERS 7.2:
1. a)
b)
KEY FIGURES: CENTRE
x = 1.95
xmed = 3.5
2. Given is the list of grades according to the German system (from
1 to 5) in the statistics exam:
a) g =
57
= 2.85
20
b) Frequency table.
Grades Freq. rel. Freq.
1
3
15%
2
5
25%
3
6
30%
4
4
20%
5
2
10%
20
100%
c)
g = 2.85
3. a)
2.842857
b) 3
c)
2
7.2 Key Figures: Centre
4. g =
1
70
P a g e | 209
⋅199 = 2.84286
5. a)
Right Middle Freq. rel. Freq.
10
5
6
12%
20
15
10
20%
30
25
20
40%
40
35
9
18%
50
45
5
10%
50
100%
Sum
b)
x = 24.4
c)
The exact mean calculated using Excel is 25.5
6. a)
x = 2 ; xmed = 1.5 ; xmod = 4
b) Right-tail
7. a)
x = 45.036 ; xmed = 45 ; xmod = 40 → symmetrical
b)
x = 44.98
c)
x = n1 ∑ mk ⋅ f k =
12
k =1
22555
= 45.11
500
210 | P a g e
7. Basics of Statistics
7.2.4 Progress Test “Key Figures: Centre”
You should assign yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Don’t use the reader to look
for the solution. The aim of the test is to give you feedback on how much
you know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you may continue and start the next chapter. Otherwise
(if you got the wrong answer or no answer at all) you should repeat the
corresponding units in order to close the gap.
1. Calculate the arithmetic mean, the median, and the mode of the
following list of numbers:
L = {0, 7, 3, −1, −1, 4, −2, 0, 4, 2.5, 4, 3.5,1, 4,1}
2. The average height of ten people in a room is 175 cm. An 11th
person who is 180 cm tall enters the room. Find the average
height of all 11 people.
3. The average height of 21 people in a room is 175 cm. A 22nd
person enters the room. How tall would he have to be to raise the
average height of all 22 people by 2 cm?
4. The average of the 40 numbers in exercise 7.1, problem 4 is
26.05. The table of grouped frequencies was:
7.2 Key Figures: Centre
P a g e | 211
Group
Right
Middle
Freq.
17-19
20-22
23-25
26-28
29-31
32-34
35-37
Sum
19
22
25
28
31
34
37
18
21
24
27
30
33
36
3
8
10
7
5
4
3
40
Rel.
Freq.
7.5%
20.0%
25.0%
17.5%
12.5%
10.0%
7.5%
Sum of
rel. Fr.
7.5%
27.5%
52.5%
70.0%
82.5%
92.5%
100.0%
On the basis of the grouped data with the middle point as
reference value, calculate the average.
5. Given is the sum curve of a distribution of numbers with mean =
25.325.
Sum Curve
Sum of relative frequencies
120.0%
100.0%
80.0%
60.0%
40.0%
20.0%
0.0%
17-19
20-22
23-25
26-28
29-31
32-34
35-37
Groups
Mark the median in the graph and decide whether the distribution
is left-tail or right-tail.
212 | P a g e
7. Basics of Statistics
7.3 Key Figures: Spread
7. Basics of Statistics
7.0 Sum
Symbol
7.1 Presentation
Techniques
7.2 Key
Figures: Centre
7.3 Key
Figures: Spread
Tables
Mean
Variance
Diagrams
Median
Standard
Deviation
Sum
Curves
Mode
Normal
Curve
Prerequisites:
In this section it is again important to be familiar
with the sum symbol introduced and discussed in
Section 7.0.1. We are still dealing with large
amounts of data.
data This requires a notation which
allows us to argue without writing long expressions.
Again, we would like to remind the reader that Excel
provides a wide range of functions which can be
used to calculate the figures needed. In most cases
we will refer to these
t
functions explicitly.
7.3 Key Figures: Spread
P a g e | 213
Learning Targets: Descriptive statistics is the art of summarising large
sets of data.
Having summarised the data with respect to centre
information we now turn our attention to the
question of how widely or narrowly they are spread
around this centre. It makes a big difference whether
the data are mostly concentrated around the average,
or are equally distributed over the whole range of
data. Sets of data with the same statistical mean can
have very different spreads.
While the average was a key figure which many
people are very familiar with, the spread figures
need some more explanation based on statistical
argumentation.
Variance and standard deviation are the most
commonly used spread figures. Together with the
information about the concentration of the data they
provide quite an accurate picture of the distribution
of data without our having to draw a diagram.
The average of the following list L1 of 15 numbers is.
L1 : x j
2
3
3
3
4
1
5
1
3
2
3
2
3
6
4
We get the same average x = 3 for the second list L2 .
L2 : x j
2
7
4
-2
4
1
3
3
-2
4
5
6
-1
3
8
However, if we look at the frequency distribution of the two lists, we will
notice a distinctive difference between them. The attribute values of list
L1 are concentrated narrowly around the average of x = 3 while the
values of list L2 are widely spread.
214 | P a g e
7. Basics of Statistics
Distribution in List 1
7
Frequencies
6
5
4
3
2
1
0
1
2
3
4
5
6
Numbers
Figure 7-12: Narrowly spread attribute values
Distribution in List 2
3.5
Frequencies
3
2.5
2
1.5
1
0.5
0
-2
-1
0
1
2
3
4
5
6
Numbers
Figure 7-13: Widely spread attribute values
7
8
7.3 Key Figures: Spread
P a g e | 215
In order to quantify the difference between the two statistical lists we
obviously have to characterise the distances between the values. Since
distances need to be based on a metric, the following calculations of
spread figures can only be applied if the attribute values are cardinal, i.e.
measured by numbers. In this case we can determine variance and
standard deviation.
7.3.1 Variance
When we talk about a distance we immediately think of the range or
interval between two points, one of them being the reference point to
which we measure the distance. A “natural” reference point could be the
origin (zero point) of the array of real numbers. However, it does not
make sense to refer all kinds of numbers, which might range up to
100,000 (e.g., number of outstanding loans), to a point which is not
associated with the statistical list itself. A much better choice is therefore
the arithmetic mean.
A first idea to define “spread” might be to calculate the distances from
each attribute value of the list to the mean point and calculate the
arithmetic mean of those distances.
Carrying out this operation for the above list L1 of 15 cardinal numbers
with the arithmetic mean x = 3 we get:
L1 : x j
2
3
3
3
4
1
5
1
3
2
3
2
3
6
4
(x j − x )
-1
0
0
0
1
-2
2
-2
0
-1
0
-1
0
3
1
Table 7.10:
Number list
The average distance is the sum of all distances divided by the number of
entries. The sum of the above distances is equal to zero. Therefore the
mean is also zero. Interpreting this result would mean: There is no
measurable spread of the numbers around the mean value. “No spread”
would imply that all numbers are equal to the mean. This is obviously not
216 | P a g e
7. Basics of Statistics
the case; therefore the implication is false, which means that the method
of measuring spread is not appropriate.
What caused the failure?
Apparently, we produce “negative distances” when subtracting the mean
from numbers which are smaller. But negative distances do not make
sense when discussing spread, and – this caused the zero mean – the
negative figures cancel out the positive ones. The question is therefore
how to handle the fact that negatives cancel out the positives? A measure
for the spread of the numbers is naturally the distance between them. In
order to overcome the sign, one can calculate the squares of the
distances.
L1 : x j
(x j − x )
( x j − x )2
2
3 3 3 4
1
5
1
3
2
3
2
3 6 4 sum
-1 0 0 0 1 -2 2 -2
0 -1 0 -1 0 3 1
0
1
0
26
0 0 0 1
4
4
4
1
0
1
Table 7.11:
Calculating the variance
0 9 1
The sum of the squares does not make sense either, because it depends
basically on the number of entries in the list. Therefore, we calculate the
mean of the squares. Following the concept of the arithmetic mean, we
have to divide the sum by the length of the list, i.e. the number of entries
in the list. The resultant figure is called variance.
Thus, the variance of list L1 is:
26
15
= 1.7333
7.3 Key Figures: Spread
P a g e | 217
VARIANCE
The variance of a statistical list L of n numbers x j is defined as the mean
of the attribute values’ squared distances from the mean:
n
∑ ( x j − x )2
VAR =
j =1
=
n
1 n
∑ ( x j − x )2
n j =1
EXAMPLE
Let us calculate the variance of the numbers in list L2 with the same
arithmetic mean of x = 3 as list L1 :
L2 : x j
2
7
4 -2 4 1
3 3 -2 4 5 6 -1
3 8
∑
(x j − x )
-1
4
1 -5 1 -2
0 0 -5 1 2 3 -4
0 5
0
( x j − x )2
1
16 1 25 1 4
0 0 25 1 4 9 16 0 25
128
= 8.5333
Dividing the sum of the squares by 15 gives: VAR = 128
15
If we now compare the variances of the two lists L1 and L2 we realise
that there is a significant difference in the margin of the result. The
variance of list L1 with numbers narrowly spread around its mean is
Var ( L1) = 1.7333 which is relatively small compared to the variance of
list L2 with Var ( L2 ) = 8.5333 .
EXCEL:
If you apply the function VAR from the statistical toolbox
the results differ surprisingly from the calculated values
above. However, if you look at the definition of the
218 | P a g e
7. Basics of Statistics
function VAR you will find: “Estimates variance based on
a sample (ignores logical values and text in the sample)”.
This explains the difference. Statisticians distinguish
between the variance for the whole population – which is
calculated by the above formulas – and a slightly different
formula, which should be used for samples. The latter is
called the unbiased variance. The two formulas differ in
the denominator by 1.
The unbiased variance is:
n
∑ ( x j − x )2
VARu =
j =1
n −1
=
1 n
( x j − x )2
∑
n − 1 j =1
In Excel the function VAR returns the unbiased variance.
For our purpose – we are almost exclusively using
statistics for whole sets of data, i.e. for populations rather
than samples – we need to apply the function VARP,
where the P stands for “population”. The definition of this
function differs from the former quite significantly:
“Calculates variance based on the entire population…”
Hence, if we apply the function VARP the calculated
variances are exactly those that we calculated for lists L1
and L2 .
AN ALTERNATIVE FORMULA FOR THE VARIANCE
A few algebraic calculations involving the sum symbol can significantly
simplify the formula for the variance and especially its application.
VAR =
1 n
1 n
2
(
x
−
x
)
=
( x j 2 − 2 x j ⋅ x + x 2 ) ← after squaring
∑
∑
j
n j =1
n j =1
VAR =
1 n 2 1 n
1 n 2
x
−
2
x
⋅
x
+
∑ j n ∑ j n ∑ x ← each term is a sum
n j =1
j =1
j =1
7.3 Key Figures: Spread
VAR =
1 n

n 
1 n 2
2 1


x
−
2
x
⋅
x
+
x
⋅

∑ j
∑1
j
n∑

n j =1
n

 j =1 
 j =1 
P a g e | 219
← The terms 2x and
x 2 are independent of the sum index j and can therefore be taken out of
the sum and placed as factors before it.
VAR =
1 n 2
1 n
2 1
x
−
2
x
⋅
x
+
x
⋅
⋅
n
←
by
replacing:
∑ j
∑ x j = x (in the
n j =1
n
n j =1
n
round brackets) and
∑1 = n [in the square brackets]
j =1
VAR =
1 n 2
1 n 2
2
2
x
−
x
+
x
⋅
=
x j − x 2 ← after composing equal
2
1
∑
∑
j
n j =1
n j =1
terms
Finally we get:
VAR =
1 n
1 n 2
2
(
x
−
x
)
=
xj − x2
∑
∑
j
n j =1
n j =1
ATTENTION
The above algebraic exploitation is a nice example of the application of
the sum symbol and the necessity of brackets (see: the beginning of this
chapter; Page 147) in order to keep the uniqueness of the expression:
3
The sum symbol replaces a pair of brackets:
∑ x j = x1 + x2 + x3 ;
j =1
therefore:
3
c ⋅ ∑ x j = x1 + x2 + x3 = c ⋅ ( x1 + x2 + x3 )
j =1
If the operand of the sum is a sum itself, brackets must be used, in order
to avoid any ambiguity:
220 | P a g e
7. Basics of Statistics
3
∑ ( x j + a ) = ( x1 + a ) + ( x2 + a ) + ( x3 + a )
j =1
Hence, without brackets the term b is not the operand of the sum symbol
and therefore a single summand:
3
∑ x j + a = ( x1 + x2 + x3 ) + a
j =1
The brackets on the right-hand side may be deleted. They are used only
for demonstration.
VARIANCE RELOADED
An alternative way of calculating the variance is: VAR =
1 n 2
xj − x2
∑
n j =1
Notice that the first term on the right-hand side is the mean of the
squared values x j 2 and that the last term x 2 is not part of the summation
denoted by the sum symbol. Therefore, it is helpful to remember that:
“THE VARIANCE IS THE MEAN OF THE SQUARES MINUS THE SQUARE OF
THE MEAN.”
Calculating the variance “by hand”, i.e. if Excel is not available, is much
more efficient if you apply the above formula.
7.3 Key Figures: Spread
P a g e | 221
EXAMPLE
Calculate the variance of the numbers in list L2 with the arithmetic
mean of x = 3 using the above formula:
L2 : x j
2
x j2
7
4
-2
4
4 49 16
4
16 1 9 9
The mean of the squares is
1 3 3 -2
263
15
4
4
5
6
-1
3
8
16 25 36
1
9 64 263
= 17.5333 ; thus:
VAR = 17.5333 − 9 = 8.5333
VARIANCE WITH FREQUENCY TABLES
Frequency tables (see Section 7.1.1; Page 159) are a first step towards
concentrating large sets of data. We took advantage of this form when
calculating the arithmetic mean. Since the variance is simply another
mean, we can in the same manner modify the above formula using the
frequencies:
m
∑ f k ⋅ ( xk − x )2
VAR =
k =1
m
∑ fk
k =1
m
Since
∑ fk = n we obtain:
k =1
VAR =
1 m
f k ⋅ ( xk − x ) 2
∑
n k =1
sum
222 | P a g e
We can also multiply the factor
7. Basics of Statistics
1
n
into the sum and replace the absolute
frequencies using:
fk
= hk
n
by the relative frequencies, and obtain the third alternative formula:
m
VAR = ∑ hk ⋅ ( xk − x ) 2
k =1
The formula “variance = mean of the squares minus square of the mean”
is similarly applicable if the statistical list includes frequencies:
•
Using the absolute frequencies: VAR =
•
Using the relative frequencies:
1 m
∑ f k ⋅ xk 2 − x 2
n k =1
m
VAR = ∑ hk ⋅ xk 2 − x 2
k =1
Remember, applying these formulas reduces the calculation quite
substantially.
EXAMPLE
The statistical list L1 with average x = 3 can be replaced by the
frequency table. Applying the different formulas we always obtain the
same results:
7.3 Key Figures: Spread
entry
frequency
relative freq.
P a g e | 223
1
2
0.133
2
3
0.2
3
6
0.4
4
5
6
2
1
1
0.133 0.067 0.067
( x j − x )2
4
1
0
1
4
9
fk ⋅ ( x j − x )2
8
3
0
2
4
9
26
hk ⋅ ( x j − x ) 2
0.533
0.2
0
0.6
1.733
0.133 0.267
sum
15
1
x j2
1
4
9
16
25
36
91
fk ⋅ x j2
2
12
54
32
25
36
161
hk ⋅ x j 2
0.133
0.8
3.6
2.4
10.73
2.133 1.667
The results are the same in all cases. The originally calculated
variance 1.7333 will also be obtained if the sum in the line with the
absolute frequencies (26) is divided by the sum of frequencies (15):
The same figure is also obtained as the sum in the line with the
relative frequencies.
Alternatively, we can obtain the variance by dividing the sum in the
line with the absolute frequencies (161) by the sum of frequencies
(15) giving 1.7333 after subtracting the square of the mean ( x 2 = 9 ),
which is also obtained when subtracting the squared mean from the
sum in the line with the relative frequencies (10.73).
VARIANCE FOR GROUPED DATA
In Section 7.2.1 (see: Page 190) we discussed how the arithmetic mean
can be calculated if we do not have single attribute values but only
grouped data. The reasons for grouping data have already been discussed:
for instance, our banks group loans based on the principal amount into
classes ranging from “micro-micro” to “large” because this is the only
way to handle the thousands or millions of loans statistically.
When calculating the variance we face the same problem that already
caused a bias of the arithmetic mean. In order to use the above formulas
224 | P a g e
7. Basics of Statistics
with the frequencies, or the relative frequencies, respectively, we need
one single value representing the whole class. The best choice would be
the mean value of all class entries. The middle point of the class is
usually not the mean value! However, in the absence of a better
alternative we may also use the middle point mk of class k.
We can now apply the above formulas for frequency tables by simply
replacing the single attribute values xk by the reference value mk :
•
Grouped data with absolute frequencies:
m
∑ fk ⋅ (mk − x )2
VAR =
k =1
m
∑ fk
=
1 m
∑ fk ⋅ (mk − x )2
n k =1
k =1
•
Grouped data with relative frequencies:
m
VAR = ∑
k =1
m
fk
2
⋅ (mk − x ) = ∑ hk ⋅ (mk − x ) 2
n
k =1
EXAMPLE
We use the data of Tab. 7.7 with the calculated mean of x = 28.3333
compared to the correct average of the complete list calculated with
the Excel function AVERAGE = 29.167.
7.3 Key Figures: Spread
P a g e | 225
k
Pt.
M
mk
Abs.
Freq.
fk
Rel.
Freq.
hk
(mk − x )2
f k⋅ (mk − x )2
hk ⋅ (mk − x )2
1
7.5
5
0.0833
434.03
2170.14
36.17
2
20
16
0.2667
69.44
1111.11
18.52
3
27.5
23
0.3833
0.69
15.97
0.27
4
35
7
0.1167
44.44
311.11
5.19
5
45
5
0.0833
277.78
1388.89
23.15
6
60
4
0.0667
1002.78
4011.11
66.85
60
1
9008.33
150.14
Class
Sum
Table 7.12:
Squares
Calculations using
Abs. Freq.
Rel. Freq.
Variance of grouped data
The calculations in the above are executed using the middle points.
With the sum in the next-to-last column we find the variance:
VAR =
9008.33
= 150.14
60
which is exactly the same value as was obtained by using the formula
with the relative frequencies.
However, both values are based on the “middle point assumption”.
We used the biased assumption for calculating the mean value, which
is used afterwards for the variance. Therefore the variance is also
biased.
The correct variance using the Excel function VARP on the complete
list of the 60 numbers would be: VAR = 146.47
The alternative calculation of the variance (“mean of the squares minus
square of the mean”) may be applied in an equivalent way:
VAR =
m
1 m
2
2
f
⋅
m
−
x
=
∑ k k
∑ hk ⋅ mk 2 − x 2
n k =1
k =1
226 | P a g e
7. Basics of Statistics
The calculations will give exactly the same (biased) results as the original
approach.
ATTENTION
In all calculations the biased mean value was smaller than the correct
average: 28.333 as opposed to 29.167. This does not necessarily imply
that the biased variance is automatically smaller. In fact, it is generally
greater.
7.3.2 Standard Deviation
The variance is a figure which is directly derived from the attribute
values of the statistics, i.e. it has basically the same dimension. However,
it is the mean of a squared figure; therefore its dimension is also squared.
In order to illustrate this, let us look for Examples The statistic about
loans consist of numbers with the dimension “currency” such as USD,
EUR, or any other currency. The mean has the same dimension, i.e. the
mean of all micro loans could be EUR 12433.45.
However, the variance would have the dimension “EUR-square”,
because it is the mean of squared figures. Thus, we have to calculate its
square root in order to obtain a quantity with the same dimension as the
data which were at the origin of the calculations. The result is called
standard deviation; it is the positive square root of the variance.
STANDARD DEVIATION
The positive square root of the variance is called standard deviation:
SD = VAR =
1 n
( x j − x )2
∑
n j =1
In statistics textbooks or in the lectures on statistics you may have
attended in the past, you may have seen the small letter “s”, sometimes
7.3 Key Figures: Spread
P a g e | 227
also the small Greek letter sigma “ σ ” used as symbols for the standard
deviation.
Of course, if the variance is calculated using another method or formula,
the above definition holds for the different formula as well.
Independently of how the variance is calculated, its square root is always
the standard deviation:
SD = VAR
EXAMPLES
1. The standard deviation of list L2
L2 : x j
2 7 4 -2 4 1 3 3 -2
4 5 6 -1 3 8
= 8.5333 is:
with the mean of x = 3 and variance VAR = 128
15
SD = VAR = 8.5333 = 2.9212
2. The calculations for grouped data are totally equivalent. The
variance derived in Tab. 7.12 (see: Page 225) was:
VAR =
EXCEL:
9008.33
= 150.14 → SD = VAR = 150.14 = 12.25
60
In Excel we will again find the function for the standard
deviation in the section “Statistical Functions”. In total
you will find four different types, two of which are of
interest to us: STDEV estimates the standard deviation for
a sample. It is the square root of the VAR, i.e. the
denominator of the mean calculation is n -1.
Since we are mainly operating with entire sets of data (=
populations) we will usually deal with VARP, and
therefore also with SDDEVP, which calculates the
standard deviation for the entire population.
228 | P a g e
7. Basics of Statistics
With the standard deviation we now have a performance figure with the
same dimension as the attributes itself. Thus, if the attributes are the
number of students, the standard deviation can be used as an indicator of
how widely or narrowly the number of students scatters around the mean
value. Thus, mean and standard deviation both have the dimension of the
attribute values. Therefore, they can be added or subtracted.
In Fig. 7-14 we have marked two intervals:
•
The first is called the “one-SD-interval”: AVE ± SD
•
The second is the “two-SD-interval”: AVE ± 2SD
2 SD
1 SD
AVE
1 SD
2 SD
Attribute
Values
Figure 7-14: Spread around mean value
The SD tells us how far away numbers on a list are from their average.
Most entries in the list will be somewhere around SD away from the
average. We can expect to find many of the entries in the interval
AVE ± SD . Certainly many more – if not almost all – will lie in the
interval AVE ± 2SD which is twice as large as the first one.
This kind of interpretation of standard deviation in connection with the
mean value is the most essential conclusion of the calculation.
7.3 Key Figures: Spread
P a g e | 229
EXAMPLE
We once more use the list of 60 entries given in Tab. 7.4 (see: Page
163):
25
15
16
30
26
5
20
30
24
37
21
25
25
17
19
25
28
26
52
26
30
26
21
45
49
22
21
43
26
27
19
29
32
27
30
12
66
13
23
29
27
39
29
28
35
27
40
48
57
1
24
28
28
33
27
29
38
26
63
41
At different steps of analysing the above statistical list we obtained:
•
AVE = 29.1667
•
VAR = 146.472
•
SD = 12.1026
Following the above argumentations, we are interested to know how
many numbers of the list will be in the one-SD-interval, and in the
two-SD-interval, respectively.
•
The one-SD-interval is [17; 41] , i.e. we count the numbers of the
list in the range 17 ≤ i ≤ 41 . The result is that 46 of the 60
numbers or 76.7% are in this range.
•
The two-SD-interval is [5; 53], i.e. we look for numbers in the
interval 5 ≤ i ≤ 53 . We count 56 numbers or 93.3% in that range.
230 | P a g e
7. Basics of Statistics
VARIATION COEFFICIENT
Variance and standard deviation are measurement figures for spread;
however, they cannot be interpreted in their absolute magnitude. The
standard deviation may, for instance, be equal for micro loans and large
loans. If the mean of the micro loans were m = 2500 EUR and the mean
for the large loans were l = 250000 EUR, the spread of the large loans
would be relatively much smaller because the values are spread around a
large mean, whereas the spread around the micro average is relatively
large. In order to avoid such a bias, statisticians have defined a
coefficient which is the standard deviation relative to the mean. It is
called the variation coefficient. Since SD and mean have the same
dimension, the variation coefficient VC is dimension-free.
VARIATION COEFFICIENT
The variation coefficient is the quotient of standard deviation and the
absolute value of the mean:
VC =
SD
x
Assuming the same SD the variation coefficient would be 100 times
larger in the case of micro loans compared to the large loans.
7.3.3 Normal Curve
If one measures the length of all newborn babies in a country, one will
discover that the distribution of the lengths around the mean length of,
say, 52 cm follows a curve which looks pretty much like the curve in
Fig. 7-15.
7.3 Key Figures: Spread
P a g e | 231
Distribution of Lengths
Relative Frequencies
30.0%
25.0%
20.0%
15.0%
10.0%
5.0%
0.0%
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Size of Newborn Babies in cm
Figure 7-15: Distribution of the sizes of newborn babies
It is amazing how many characteristics measured in nature follow this
distribution, as for instance:
•
the sum of n random numbers
•
the weight of all 25-year-old men in a town
•
the test score of the computer test of 100 applicants
232 | P a g e
7. Basics of Statistics
Freqency Distribution
22%
20%
18%
Percentage
16%
14%
12%
10%
8%
6%
4%
2%
0
161-165
166-170
171-173
174-176 177-179
180-182 183-185
186-190
191-195
Sizes (grouped)
Figure 7-16: Histogram and fitted normal curve
If we draw a histogram of the sizes grouped in classes as shown on the
horizontal scale of Fig. 7-16 we notice that the shape of the frequencies
curve closely resembles the bell-shaped normal curve discovered by
Abraham de Moivre from France (around 1720). Carl Friedrich Gauss
(1777-1855), the German mathematician found the function describing
the normal distribution.
It is an exponential function of the two parameters µ = average and σ =
standard deviation. The general function is:
1  x−µ 
σ 
− 
1
y = f(x)=
e 2
σ 2π
2
The normal function involves three of the most famous numbers in the
history of mathematics:
2 , π, and e
7.3 Key Figures: Spread
P a g e | 233
The old 10 Deutschmark bill showed on the front side the portrait of
C. F. Gauss together with the normal curve and the formula of the
function.
Figure 7-17: The former DM 10 bill with normal murve
It is a remarkable fact that many histograms follow the normal curve. For
such histograms, the average and the SD are good summary statistics.
If a histogram follows the normal curve, it looks something like the
sketch in Fig. 7-17.
234 | P a g e
7. Basics of Statistics
Normal Distribution with Standard Units
50
Percent per Standard Unit
40
30
20
10
0
?
?
?
-3
-2
-1
Problem Dependent Dimension
?
?
?
0
1
2
3
Standard Units
Figure 7-18: A normal-shaped histogram
STANDARD NORMAL DISTRIBUTION
Normalised to average µ = 0 and standard deviation σ = 1 (the Greek
letters are used internationally for the average and SD!) the normal curve
(bold line) has the following important properties:
• The graph is symmetric around 0, i.e. mean and median are 0.
• The total area under the curve is 1 (=100%) – as for every distribution
curve. (Note the foregoing discussion about the area of the
histogram!)
With respect to the horizontal scale of standard units (1 unit corresponds
to 1 SD) we find:
• The area under the normal curve between −1 and +1 (= one-SDinterval) is about 68%.
• The area under the normal curve between −2 and +2 (= two-SDinterval) is about 95%.
• The area under the normal curve between −3 and +3 (= three-SDinterval,) is about 99.7%.
7.3 Key Figures: Spread
P a g e | 235
If the histogram for data is similar in shape to the normal curve, we could
convert the horizontal scale to standard units by seeing how many SDs it
is above (+ sign) or below (− sign) average.
For instance, take the data on student size, with an average of 178 cm and
an SD of 6.02 cm. If we assume that the histogram is approximately bellshaped, 178 cm is the most likely size in the class and approximately
68% of the students measure between x − SD ≤ x j ≤ x + SD , i.e. 172 ≤
x j ≤ 184 in height. In fact, none of the students is exactly 178 cm tall,
but 42 = 70% of the students lie in the one-SD-range.
236 | P a g e
7. Basics of Statistics
EXERCISE 7.3:
1. a)
KEY FIGURES: SPREAD
Calculate the variance of the numbers in the list with the
mean x = 1.95 :
1
3
-2
5
4
-6
5
-3
6
4
-3
5
0
-3
3
4
7
8
-4
5
b) Calculate the variance applying the alternative formula.
2. Given is the list of grades according to the German system (from
1 to 5) in the statistics exam. The mean grade was calculated to be
g = 2.85 :
Student i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Grade g j 1 3 4 3 2 1 2 4 2 5 3 4 3 2 1 5 2 3 4 3
a)
Calculate the variance of this list.
b) Use the frequency table compiled in exercise 7.2, problem 2
(see: Page 205) to verify the calculation of a) using the two
alternative formulas (based on the absolute and the relative
frequencies, respectively).
3. The following list of 70 grades was already used in exercise 7.1,
problem 2 (see: Page 175).
2
2
4
3
2
1
1
4
2
5
3
4
3
2
1
1
2
4
3
3
3
5
4
2
5
3
1
3
5
1
1
4
3
2
4
3
5
1
2
2
3
4
3
3
4
2
5
1
2
1
1
2
4
4
3
5
3
2
3
2
4
5
2
1
4
3
5
2
3
2
7.3 Key Figures: Spread
a)
P a g e | 237
Compile a frequency table – with absolute and relative
frequencies – using a tally list.
b) Calculate the variance of the above list using the simplified
formulas.
c)
Use Excel to determine the exact value of the variance and
verify the calculations in b).
4. The following list contains 36 randomly generated numbers
between 2 and 9:
3
9
6
a)
8
7
6
2
5
7
3
7
5
4
8
3
8
8
8
9
2
2
3
6
2
8
3
5
7
6
7
5
2
6
5
7
7
Group the data into 3 classes: 2-4 (middle 3), 5-6 (middle
5.5), and 7-9 (middle 8). Compile a frequency table for these
classes.
b) Calculate the variance for the grouped data using the
simplified formula.
c)
Determine the exact variance using Excel and explain the
difference in the values.
5. Use the “Base-Problem” and calculate the variance using the
Excel function VARP.
6. In exercise 7.1, problem 5 we created two frequency tables for the
“Base-Problem” with different class widths. Calculate for both
tables the variance using the simplified formula. Use Excel and
organise the calculations in appropriate tables.
Compare the values with the exact value of problem 5.
238 | P a g e
7. Basics of Statistics
7. Calculate the standard deviation for the data of the “BaseProblem”. You can either apply the corresponding Excel function,
or use the variance from problem 6.
8. In exercise 7.1, problem 6 we created a histogram of the numbers
of the “Base-Problem”. The shape is very close to the normal
curve.
a)
Determine the one-SD-interval around the mean value.
b)
Decide from the sum curve the percentage of numbers inside
the one-SD-range.
c)
Use Excel to count the exact amount of numbers in the oneSD-range.
d)
Compare these results with the theoretical figures derived
from the normal distributions.
7.3 Key Figures: Spread
ANSWERS 7.3:
1. a)
P a g e | 239
KEY FIGURES: SPREAD
VAR = 16.1475
b) Same as a)
2. a) VAR = 1.4275
b) Same as a)
3. a)
Grade k
fk
hk
fk ⋅ gk 2
hk ⋅ gk 2
1
12
17.14%
12
0.171429
2
18
25.71%
72
1.028571
3
18
25.71%
162
2.314286
4
13
18.57%
208
2.971429
5
9
12.86%
225
3.214286
Sum
70
679
9.7
Absolute frequencies:
VAR = 1.618163
Relative frequencies:
same
b) same
4. a)
k Group rk
mk
fk
hk
hk ⋅ mk
hk ⋅ mk 2
1
2-5
5
3
16
44.4%
1.33
4.00
2
5-6
6
5.5
5
13.9%
0.76
4.20
3
7-9
9
8
15
41.7%
3.33
26.67
5.43
34.87
Sum
36
240 | P a g e
7. Basics of Statistics
b)
VAR = 5.3771
c)
VARP = 4.86034
5. VARP = 83.951
6. Class width 10: VAR = 95.33
Class width 5: VAR = 84.70
7. SD = 9.1625
8. a)
35.8735 ≤ x j ≤ 54.1985
b) Approximately 64% = 320 numbers
c)
With Excel: 305 numbers
d) Theoretical: 340 numbers
7.3 Key Figures: Spread
P a g e | 241
7.3.4 Progress Test “Key Figures: Spread”
You should assign yourself some time for concentrated work on this test.
Try to solve as many problems as you can. Don’t use the reader to look
for the solution. The aim of the test is to give you feedback on how much
you know or have learned up to now.
At the end of the chapter you will find the solutions to the problems.
Each solution is reduced to the final answer; it may even be only a
number, a symbol, a table or a graph. You should check your answers. If
they are correct, you have successfully completed this 2nd volume of the
reader. Otherwise (if you got the wrong answer or no answer at all) you
should repeat the corresponding units in order to close the gap.
1. Calculate the variance and the standard deviation for the
following list of numbers with average = 2:
L = {0, 7, 3, −1, −1, 4, −2, 0, 4, 2.5, 4, 3.5,1, 4,1}
2. For each list below, work out the average, the variance and the
SD.
a)
1, 3, 4, 5, 7
b) 6, 8, 9, 10, 12
3. The average of the 40 numbers in problem 4 of exercise 7.1 is
26.05. After grouping the data we obtained the following
frequency list.
Group
Right
Middle
Freq.
17-19
20-22
23-25
26-28
29-31
32-34
35-37
19
22
25
28
31
34
37
18
21
24
27
30
33
36
3
8
10
7
5
4
3
Rel.
Freq.
7.5%
20.0%
25.0%
17.5%
12.5%
10.0%
7.5%
Sum of
rel. Fr.
7.5%
27.5%
52.5%
70.0%
82.5%
92.5%
100.0%
242 | P a g e
7. Basics of Statistics
Work out the variance and the standard deviation from the
grouped data. Use the alternative (and simpler) formulas.
4. The management of your bank proposes to give all bank
employees a flat raise of EUR 20 a month. What would this do to
the average salary of the bank employees? What would it do to
the SD?
5. Instead of raising all salaries by the same amount the management
of your bank decides to grant a 5% increase in salaries across the
board. What would this do to the average and the SD?
6. The following list of 25 test scores has an average of 50 and an
SD of 10:
39 41 47 58 65 37 37 49 56 59
62 36 48 52 64 29 44 47 49 52
53 54 72 50 50
a)
Use the normal approximation to estimate the number of
scores within a one-SD-range of the average.
b) How many scores really are within the one-SD-range of the
average?
7.4 Answers to Progress Tests
P a g e | 243
7.4 Answers to Progress Tests
7.4.1 Answers to PT “Presentation Techniques”
You should check your answers. If they are correct, you may continue
and start the next chapter. If not (if you got the wrong answer or no
answer at all), repeat the topics of the recommended sections in order to
close the gap.
n
n
j =1
j =1
1. 2∑ x j 2 − 4a ∑ x j + n ⋅ a 2
2. 41% for men; 51% for women.
3. a)
Tally list:
No. Freq. Rel. F.
18
1
5%
19
1
5%
20
0
0%
21
3
15%
22
4
20%
23
2
10%
No. Freq. Rel. F.
24
3
15%
25
1
5%
26
2
10%
27
1
5%
28
1
5%
29
1
5%
20
1
244 | P a g e
7. Basics of Statistics
b) Column diagram:
Frequency Distribution
Relative Frequency
25%
20%
15%
10%
5%
0%
18
19
20
21
22
23
24
25
26
27
Loans per LO
4. Tally list:
Group
Right
Middle
Freq.
17-19
20-22
23-25
26-28
29-31
32-34
35-37
19
22
25
28
31
34
37
18
21
24
27
30
33
36
3
8
10
7
5
4
3
40
Rel.
Freq.
7.5%
20.0%
25.0%
17.5%
12.5%
10.0%
7.5%
100%
Sum of
rel. Fr.
7.5%
27.5%
52.5%
70.0%
82.5%
92.5%
100.0%
28
29
7.4 Answers to Progress Tests
P a g e | 245
Column diagram of grouped data:
Frequency Distribution
Relative Frequencies
30.0%
25.0%
20.0%
15.0%
10.0%
5.0%
0.0%
17-19
20-22
23-25
26-28
29-31
32-34
35-37
Group of Numbers
Sum curve → next problem 5.
5. Using the sum curve from problem 4, estimate the numbers
between 24 and 30. How many numbers really were within this
range?
Sum of Relative Frequencies
Sum Curve
120.0%
100.0%
80.0%
60.0%
40.0%
20.0%
0.0%
17-19
20-22
23-25
26-28
29-31
Group of Numbers
Estimate: 77% - 40% = 37% → 14 numbers
In fact, 12 numbers are really in this range.
32-34
35-37
246 | P a g e
7. Basics of Statistics
7.4.2 Answers to PT “Key Figures: Centre”
You should check your answers. If they are correct, you may continue
and start the next chapter. If not (if you got the wrong answer or no
answer at all) repeat the topics of the recommended sections in order to
close the gap.
1. Mean = 2; Median = 2.5; Mode = 4
2. 176.36 cm
3.
219 cm
4.
26.03
5. Graph with marked median and mean:
Sum of Relative Frequencies
Sum Curve
120%
Mean
100%
80%
60%
Median
40%
20%
0%
17-19
20-22
23-25
25 26
26-28
29-31
Group of Numbers
The distribution is right-tail.
32-34
35-37
Answers to Progress Tests
P a g e | 247
7.4.3 Answers to PT “Key Figures: Spread”
You should check your answers. If they are correct, you have
successfully completed this 2nd volume of the reader. If not (if you got
the wrong answer or no answer at all) repeat the topics of the
recommended sections in order to close the gap.
1. VAR = 5.9; SD = 2.43
2. a)
AVE=4, VAR=4, SD=2
b) AVE=9, VAR=4, SD=2
3. VAR = 24.07; SD = 4.9
4. This would increase the average by EUR 20 but leave the SD
unchanged.
5. This would increase the average and the SD by 5%.
6. a)
One-SD 68% → 17
b) Really: 16
248 | P a g e
7. Basics of Statistics
Index
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Index
A
D
annuities .......................................... 87, 105
arithmetic mean ............................ 187, 188
attribute values ..................................... 188
average .................................................. 187
AVERAGE
function ............................................ 189
average value ........................................ 187
data
grouped .............................................161
decay
radioactive ..........................................25
density distribution ................................166
deposits....................................................82
successive............................................84
diagram ..................................................163
disbursement .........................................100
discount factor .........................................64
distribution
left-tail...............................................199
right-tail ............................................199
symmetric ......................................... 199
distribution curve ...................................169
distribution function ..............................170
domain
exponential function ...........................20
B
balance .................................................... 90
outstanding ....................................... 106
banking year ............................................ 59
base
logarithmic function ........................... 41
basic financial terms ................................ 62
basic rules ................................................ 85
bias ........................................................ 146
E
C
cardinal .................................................. 157
classification
data ................................................... 160
column diagram ..................................... 164
common logarithm ............................ 40, 43
compound factor ..................................... 61
compound interest ............................ 26, 60
compounding
continuously ....................................... 30
period ................................................. 68
periodically ......................................... 72
count index ............................................ 147
COUNTIF
function ............................................ 159
effective interest rate ...................... 70, 113
exponential function ................................20
F
fees ..........................................................70
frequency ...............................................158
absolute ............................................161
relative ...................... 159, 161, 163, 165
table .......................................... 158, 163
FREQUENCY
function .............................................162
frequency sum .......................................167
frequency tables
grouped .............................................160
future value..............................................62
250 | P a g e
G
geometric series ...................... 87, 107, 148
graph
exponential function........................... 19
logarithmic function ........................... 39
grouped data ......................................... 159
Index
mean
normal curve ..................................... 232
mean value ............................................187
measure
spread ...............................................214
median ........................................... 195, 196
normal curve ..................................... 232
middle point ........................... 161, 165, 193
mode ......................................................201
H
N
half-life .................................................... 25
histogram ...................................... 166, 196
normal .............................................. 231
I
in-payment .............................................. 54
instalment.............................. 100, 102, 105
interest .............................................. 26, 57
compound........................................... 61
effective rate ...................................... 60
floating................................................ 71
simple ........................................... 58, 83
interest days ............................................ 83
interest factor .................................... 27, 61
interest period ................................... 58, 68
interest rate ................................. 27, 58, 65
effective ................................ 69, 72, 113
nominal ................................. 68, 70, 113
periodic ............................................... 74
inverse function ....................................... 38
L
loan ........................................................ 100
logarithm ................................................. 40
logarithmic
operations........................................... 42
logarithmic function
properties ........................................... 40
M
maturity ................................................. 107
natural logarithm .....................................43
nominal ..................................................157
normal curve ..........................................230
normal distribution ................................229
number π ................................................28
number e.................................................. 28
number of periods ...................................58
O
one dimensional list ...............................157
one-SD-interval ......................................226
normal curve ..................................... 232
one-to-one function .................................39
ordinal ....................................................157
out-payment ............................................54
outstanding amount ...................... 102, 111
P
payments
constant ..............................................89
polynomial .............................................147
population................ 14, 145, 161, 216, 225
population growth ............................. 15, 24
present value ..................................... 62, 63
principal ...................................................58
principal repayment ....................... 102, 103
probability ..............................................146
properties
exponential function ........................... 22
logarithmic function ............................39
Index
R
radioactive
decay .................................................. 25
random variable .................................... 146
reference point .................................. 55, 85
reference value ...................................... 193
repayment ............................................. 101
repayment plan ..................................... 102
representation
graphical ........................................... 163
S
sample ................................... 145, 216, 225
savings ..................................................... 82
savings rate .............................................. 90
service charge .......................................... 70
spread .................................................... 211
standard deviation................................. 224
statistical features ................................. 189
statistical list .......................................... 188
statistical mean ..................................... 187
statistical sequence ............................... 157
sum curve ...................... 159, 163, 167, 169
sum symbol ........................................... 146
SUMPRODUCT
function ............................................ 192
T
tally list .................................................. 158
three-SD-interval
normal curve..................................... 232
time scale ................................................ 54
time-line .................................................. 55
today-point ........................................ 55, 85
two-SD-interval...................................... 226
normal curve..................................... 232
V
variance ................................................. 215
alternative formula ........................... 218
frequencies ....................................... 219
P a g e | 251
grouped data ..................................... 222
unbiased ............................................216
variation coefficient ...............................228
252 | P a g e
Notes
Notes
P a g e | 253
254 | P a g e
Notes