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Chapter 9
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CHAPTER 9: Applications of Numbers
Weariness.—Nothing is so insufferable to man as to be completely at rest, without passions, without
business, without diversion, without study. He then feels his nothingness, his forlornness, his
insufficiency, his dependence, his weakness, his emptiness. There will immediately arise from the
depth of his heart weariness, gloom, sadness, fretfulness, vexation, despair.28
Averages
The ordinary average is more officially called the (arithmetic) mean. We will use this term later
when it helps us distinguish the mean from other definitions of "average". The average of a collection of
numbers is found by adding them and dividing by how many numbers there are in the collection. The
justification behind this procedure is something like this: your average score is the score you could have
gotten on all three tests and achieved the same point total.
Example 1: This course includes 3 tests. If you get scores of 86, 78, and 90, your test average
86+78+90
254
would be
=
= 84.7.
3
3
A more general type of average is commonly called a weighted average. (The term "mean" can
also be used here.) In this case, one or more of the numbers in the collection are more important or
significant in some way, and so receive more "weight" in the calculations.
Example 2: Suppose you take a course in which there are 2 tests during the semester and a
comprehensive final exam which is worth as much as 2 tests. If your test scores were 84 and 72,
and 85 on the final exam, what is your weighted average?
Solution: In this case, the final exam score needs to be multiplied by 2 because it is worth the same
as two tests, and you would divide by 4, not 3. The weighted average is
84 + 72 + (2•85)
4
=
326
4
=
81.5.
The same result could be found by multiplying the test scores by the appropriate percentages; the
two tests are worth 25% each, and the final exam is worth 50%:
.25 (84) + .25 (72) + .50 (85) = 21 + 18 + 42.5 = 81.5
Example 3: Suppose you take a class in which your semester average is computed by weighting the
various components of your work. The weights are expressed in terms of percentages on the course
outline. Your average on the three exams is worth 66%, the paper is 14%, and your homework is
worth 20%. Here's the formula to compute your semester (weighted) average:
.66 (test average) +.14 (paper) + .20 (homework)
So suppose you get 84.7 average on the three tests (see Example 1), get a 90 on the paper, and do 100% of
the homework. What is your semester average?
28Pascal,
Pensees, p, 50.
66
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Chapter 9
Solution:
0.66 (84.7)+ 0.14 (90)+ 0.20 (100)
= 55.9 + 12.6 +20
= 88.5
A type of average which all college students encounter is the grade point average (GPA). Consider
a simple system for computing grade point averages in which an A is worth 4 points, a B is worth 3 points,
a C is worth 2 points, a D is worth 1 point, and an F is worth 0 points. Biola's system is more complicated
because of pluses and minuses, but the simpler system is adequate to illustrate the principles. The GPA is
an average in which grades are weighted according to the number of credits the class was worth.
Example 4: Esther took two 4-unit classes, and got two A's. She took three 3-unit classes, and got
an A and two B's. She also took a 1-unit class, and got a C. Find her semester GPA.
Solution: Esther's total number of grade points is found this way:
grade
A
A
A
B
B
C
grade points
4
4
4
3
3
2
TOTALS
credits
4
4
3
3
3
1
18
course total
16
16
12
9
9
2
64
64
Esther's semester GPA is 18 = 3.56.
Combining your past cumulative GPA with your current semester GPA to get your new cumulative
GPA can also be done as a weighted average. (The following procedure involves approximation. The
process can also be done exactly by going back to the original information of grades and units for each class,
or by doing some other calculations. The Registrar's Office would do exact calculations. We will be content
with an approximation that illustrates a way of applying weighted averaging.) The weighting will be
provided by the number of units for which the average is computed.
Example 5: Vern's GPA last semester was 2.95 for 17 units. For his previous 65 units, Vern's
cumulative GPA was 3.32. Find his current cumulative GPA.
Solution: GPA =
(3.32) (65) + (2.95) (17)
65 + 17
=
265.95
82
= 3.24
Average Speed
A somewhat different average is average speed. In its most basic form, an average speed is not
found by working with a collection of speeds at all. The average speed is found by taking the total distance
traveled, and dividing by the total time traveled. Doing this is justified by noting that the average speed is
the speed a person could travel for the entire trip and still go the same distance.
Example 6: Suppose a woman drives a car 2 hours at 60 mph and 2 hours at 40 mph. What is her
average speed?
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Chapter 9
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The total distance she drove is (2•60) + (2•40) = 120 + 80 = 200 miles. The total time
200
was 2 + 2 = 4 hours. So her average speed was
= 50 mph . Notice this is simply the average
4
of her two speeds. Does it always work like this? If not, why not?
Solution:
Example 7: Suppose a man drives a car for 300 miles at 50 miles per hour and for another 100
miles at 40 miles per hour. What is his average speed?
Solution: Since average speed is total distance divided by total time, we should divide 400 miles
300
by the total time of the trip. The first 300 miles would take
= 6 hours; the second 100 miles
50
400
100
would take
= 2.5 hours. So the total time is 6 + 2.5 = 8.5 hours, and the average speed is 8.5
40
= 47.1 miles per hour.
Notice that this is not the same as simply taking the average of 40 and 50, the two speeds, because
they were not maintained for the same portions or percentages of the time for the trip. This observation
leads to a second way of computing the average speed in this case. We could weight the two speeds
according to the time each was maintained. In this case, the work looks like this:
average speed
=
(6 hr) 50 mph+ (2.5 hr) 40 mph
6 hr + 2.5 hr
400
= 8.5
= 47.1 mph
This weighting has the same effect as computing the percentage of the time of the trip taken by each
segment of the trip, and multiplying by those numbers:
6
2.5
= 70.6%
= 29.4%
8.5
8.5
average speed = .706 (50) + .294 (40) = 35.3 + 11.7 = 47.0 mph
(This answer is slightly different because of rounding during the calculations.)
This weighting is justified algebraically in the following way: let s1 be the first speed maintained for
time t1, and s2 be the second speed maintained for time t2. Then
Average speed
=
total distance
total time
=
=
where
s1t1 + s 2t 2
t1 + t 2
t
t2
s1 • 1 + s 2•
,
t1 + t 2
t1 + t2
t1
is the fraction or percentage of the total time traveled at speed s1 and
t1 + t2
t2
is the percentage of the total time traveled at speed s2.
t1 + t2
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Chapter 9
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Basic Counting Principle
If one thing can be done in m ways, and another thing can be done independently in n ways,
then the two things can be done together in m • n ways.
Example 8: As a part of Biola's General Education requirement, you need a Philosophy class (6
options), a Behavioral Science class (3 options), and a Fine Arts class (5 options). How many ways
are possible to meet these three requirements?
Solution: 6 • 3 • 5 = 90.
Example 9: At Prince of Burgers, you can order your hamburger with or without mustard, with or
without mayonnaise, with or without ketchup, with or without onion, with or without lettuce, with
or without pickle and with or without tomato, and the hamburger patty can be ordered rare,
medium, or well-done. How many different ways can you order a hamburger at Prince of Burgers?
Solution: 2•2•2•2•2•2•2•3 = 384
A much more profound and significant example of the application of the Basic Counting Principle
involves genetics, the study of the inherited characteristics of living organisms. In the 1800's, an
Augustinian monk named Mendel discovered some basic principles which govern the patterns of
inheritance. These principles provided the ability to predict inherited characteristics. But the mechanism
which explained inheritance remained a mystery. This mystery was solved in this century with the
discovery of DNA, a chemical which in its structure contains the hereditary information for each organism.
The relevant structure consists of a sequence of chemicals called nucleotides, with each position in the
sequence being occupied by one of only 4 nucleotides. The fact that there are only 4 nucleotides involved
may seem very simple, and in a sense it is. But every different order of the nucleotides makes a different
organism. The great variety of organisms we observe among the members of a given species (like the
members of this class!) are a result of the length of the sequence.
For instance, a virus is a rather simple organism. A typical DNA sequence for a virus would
contain 5,000 (or more) nucleotides. Each of these nucleotides can be any one of the 4 possibilities. How
many different genetics patterns are possible for this virus? According to the Basic Counting Principle,
45000 . This is an astronomically large number.
Human beings are much more complicated organisms than viruses. Our DNA sequences are much
longer, more like 5 billion nucleotides. The variety possible for human beings becomes 45,000,000,000 . No
wonder no two of us are alike! God has made each of us unique in so many ways.
CHAPTER 9: Applications of Numbers
Homework
1. Suppose you take a course in which there are 2 tests during the semester and a comprehensive final
exam which is worth as much as 2 tests. If your scores were 90 78, and 83 on the final exam, what is you
weighted average?
2. Suppose you take a course in which there are 3 tests during the semester and a comprehensive final
exam which is worth as much as 2 tests. If your scores were 70, 77, 80, and 85 on the final exam, what is
your weighted average?
3. Suppose a course requires 3 short papers worth 10% each, a research paper worth 30% and two tests
worth 20% each. Suppose your scores are 85, 75, and 90 on the three short papers, 84 on the research paper,
and 68 and 83 on the tests. Find your weighted average for the semester.
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Chapter 9
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4. Suppose a course requires 3 tests worth 22% each, a short paper worth 14%, and homework worth 20%.
Suppose your scores are 65, 78, and 70 on the three tests, 90 on the paper, and 81 on the homework. Find
your weighted average for the semester.
5. Suppose a woman drives a car for 150 miles at 50 miles per hour and for another 60 miles at 40 miles
per hour. What is her average speed?
6. Suppose a man takes a bus for 500 miles at 50 miles per hour and takes a taxicab for 20 miles at 25 miles
per hour. What is his average speed?
7. An airplane travels round trip from Los Angeles to Denver, roughly 1000 miles each way. If the speed of
the airplane to Denver is 600 miles per hour, and the speed is 400 miles per hour on the return trip, what is
the airplane's average speed on the round trip?
8. Sack lunches are being prepared in the Cafe'. Everyone gets a sandwich (turkey, ham, or peanut butter
and jelly), a drink (lemonade or fruit punch), a piece of fruit (apple, orange, or banana), and dessert (cookie
or brownie). How many different lunches are possible?
9. In a given area code and a given three-digit prefix, how many possible telephone numbers are there?
That is, how many ways can you choose the remaining four digits?
10. Last semester, Tim got a B in a 4-unit class, two A’s and a B in three 3-unit classes, and a B in a 2-unit
class. Find his semester GPA.
11. Last semester, Mark got a D in a 4-unit class, an A, a B, and a D in three 3-unit classes, and an A in a 2unit class. Find his semester GPA.
12. Last semester, Teri had a GPA of 3.23 for 16 units. For her previous 45 units, she had a cumulative GPA
of 3.46. Find her current cumulative GPA.
13. Last semester, Amy had a GPA of 3.72 for 17 units. For her previous 83 units, she had a cumulative
GPA of 3.85. Find her current cumulative GPA.
14. Last semester, Martha had a GPA of 2.73 for 13 units. For her previous 105 units, she had a cumulative
GPA of 2.49. Find her current cumulative GPA.
Selected Answers:
1. 83.5
3. 80.4
5. 46.7 mph
7. 480 mph
9. 10,000
11. 2.40
13. 3.83