Theoretical Models
1
1.1
Discrete Probability Distributions
Descriptive statistics
We discussed frequency tables and characteristics before.
1.1.1
Example
2
3
Outcomes Abs. Freq. Rel.Freq. Cum.Rel.Freq
7
6
1
10
0; 10
0; 10
7
6
7
6
2
15
0;
15
0;
25
7
6
7
6
3
15
0; 15
0; 40
6
7
6
7
4
20
0;
20
0;
60
6
7
6
7
5
30
0;
30
0;
90
6
7
5
4
6
10
0; 10
1; 00
total
100
1
We see that the possible outcomes are 1; 2; :::; 6. The relative frequencies
are nonnegative and sum to 1.
1.1.2
Graphs
needle graph = the graph of the relative frequencies
EDF = the graph of the cumulative relative frequencies
EDF (x) = proportion of outcomes
1.1.3
x.
Characteristics
We calculated mean,variance and standarddeviation:
mean
x =
1
10 + 2
15 + 3
15 + 4
100
20 + 5
10
15
10
+2
+ ::: + 6
100
100
100
= sum(outcomes x relative frequencies)
= 1
1
30 + 6
10
second moment
x2
12
10 + 22
15 + ::: + 62 10
=
100
= sum(outcomes2 x rel.frequencies)
variance
s2 = x 2
1.2
(x)2
Probability theory: discrete variables
We use X; Y; Z; ::: to denote random variables
1.2.1
Image of X:
* image(X) = the set of possible values of X = fx1 ; x2 ; x3 ; :::g
* with each outcome xi we associate a number pi
pi = P (X = xi ) = the probability that X is equal to xi
* we assume that (cf. relative frequencies):
pi
0, for each i
p1 + p2 + p3 + ::: = 1
* Theoretical distribution function: F (x) = T DF (x) = FX (x) = P (X
x)
Example
2
6
6
6
6
6
6
6
6
6
6
4
X = x P (X = x) P (X x).
1
0; 10
0; 10
2
0; 15
0; 25
3
0; 15
0; 40
4
0; 20
0; 60
5
0; 30
0; 90
6
0; 10
1; 00
total
1; 00
2
3
7
7
7
7
7
7
7
7
7
7
5
1.2.2
Mean and variance
* mean
= E(X) = sum(outcomes x probabilities)
= x1 p1 + x2 p2 + :::
* second moment
E(X 2 ) = sum(outcomes2 x probabilities)
= x21 p1 + x22 p2 + :::
2
* variance:
1.2.3
= V ar(X) = E(X 2 )
(E(X))2
Examples
Example 1 The variable X has the following pdf:
P (X = 1) = 1=3; P (X = 2) = 1=6;
P (X = 3) = 1=9; P (X = 4) = 2=9;
P (X = 5) = p
(a) Because the sum of the probabilities has to be 1, we have
1 1 1 2
+ + + +p=1
3 6 9 9
and we …nd that p = 1=6.
(b) The mean is given by
1
1
1
2
1
=1 +2 +3 +4 +5
3
6
9
9
6
18 + 18 + 18 + 48 + 45
=
54
147
=
t 2; 72
54
E(X) =
(c) The second moment is given by
1
1
1
2
1
E(X 2 ) = 12 + 22 + 32 + 42 + 52
3
6
9
9
6
525
=
t 9; 72
54
3
(d) The variance is given by
V ar(X) =
=
2
=
525
54
(
147 2
)
54
6741
t 2; 3117
2916
The standard deviation is given by
p
= 2; 3117 t 1; 52
(e) Draw a needle graph and draw the theoretical distribution function.
(f) We …nd some probabilities as follows:
P (X 2; 5) = P (X = 1) + P (X = 2) = 1=3 + 1=6 = 1=2;
P (2 < X < 6) = P (X = 3) + P (X = 4) + P (X = 5) = 1=9 + 2=9 + 1=6 =
1=2;
P (X is even) = P (X = 2) + P (X = 4) = 1=6 + 2=9 = 7=18;
P (2 X 6) = P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)
= 1 P (X = 1) = 1 1=3 = 2=3;
P (X > 3) = 1 P (X 3) = 1 1=3 1=6 1=9 = 7=18;
P (2 < X 6) = P (X = 3) + P (X = 4) + P (X = 5) = 1=2;
P (X 3) = 1 P (X 2) = 1 1=2 = 1=2;
P (2 X < 6) = P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) = 2=3;
P (X = 2; 5) = 0;
P (X > 8) = 0.
Example 2 Suppose X has the following pdf:
2
X = x P (X = x). P (X x).
6
1
0; 10
0; 10
6
6
2
0; 15
0; 25
6
6
3
0; 15
0; 40
6
6
4
0;
20
0; 60
6
6
5
0; 30
0; 90
6
4
6
0; 10
1; 00
total
1; 00
Find
= E(X) and
2
= V ar(X)
4
3
7
7
7
7
7
7
7
7
7
7
5
1.2.4
Discrete vectors
When we deal with two variables X and Y we have
Im(X; Y ) = f(xi ; yj ) : xi 2 Im(X); yj 2 Im(Y )g
With each of the possible outcomes we associate a number pi;j :
pi;j = P (X = xi ; Y = yj )
where the numbers pi;j have to satisfy
0
sum of all pi;j equals 1.
pi;j
We call these numbers the joint probability distribution of the vector (X; Y ).
Example 1 Consider the following table:
Y
X
0
2
4
Total
0
0; 05
0; 20
0
0; 25
1
2
Total
0; 15
0
0; 20
0; 25 0; 10 0; 55
0; 10 0; 15 0; 25
0; 50 0; 25
1
Here we have
Im(X; Y ) = f(0; 0); 0; 2); 0; 4); (1; 0); (1; 2); (1; 4); (2; 0); (2; 2); (2; 4)g.
The numbers in the table are nonnegative and the sum equals 1. The
numbers in the table are called the joint pdf of (X; Y ).
The numbers in the margins are called the marginal pdf of X resp. Y .
We can represent the joint pdf also by the following table
x
0
0
0
1
1
1
2
2
2
y
0
2
4
0
2
4
0
2
4
P (X = x; Y = y)
0; 05
0; 20
0
0; 15
0; 25
0; 10
0
0; 10
0; 15
1
5
As before we can calculate several types of characteristics for X and Y .
We …nd for example:
* E(X) = 0 0; 25 + 1 0; 50 + 2 0; 25 = 1
* E(X 2 ) = 02 0; 25 + 12 0; 50 + 22 0; 25 = 1; 50
* V ar(X) = 1; 50 12 = 0; 50 and (X) t 0; 7071
* E(Y ) = 0 0; 20 + 2 0; 55 + 4 0; 25 = 2; 10
* E(Y 2 ) = 02 0; 20 + 22 0; 55 + 42 0; 25 = 6; 2
* V ar(Y ) = 6; 2 (2; 10)2 = 1; 79 and (Y ) t 1; 3379
* E(XY ) = 0 0 0; 05 + 0 2 0; 20 + 0 4 0
+1 0 0; 15 + 1 2 0; 25 + 1 4 0; 10
+2 0 0 + 2 2 0; 10 + 2 4 0; 15
= 2; 5
* Cov(X; Y ) = E(XY ) E(X)E(Y ) = 2; 5 (1 2; 10) = 0; 40
* (X; Y ) = Cov(X; Y )= (X) (Y ) t 0; 106.
All properties of variance and covariance and correlation coe¢ cient remain the same as in the descriptive statistics.
Consider for example the sum X + Y . Using the table, we …nd all sums:
x y
0 0
0 2
0 4
1 0
1 2
1 4
2 0
2 2
2 4
P (X = x; Y = y)
x+y
0; 05
0
0; 20
2
0
4
0; 15
1
0; 25
3
0; 10
5
0
2
0; 10
4
0; 15
6
1
It follows that:
P (X + Y = 0) = 0; 05; P (X + Y = 1) = 0; 15, P (X + Y = 2) = 0; 20;
P (X + Y = 3) = 0; 25; P (X + Y = 4) = 0; 10; P (X + Y = 5) = 0; 10;
P (X + Y = 6) = 0; 15.
As a consequence, we …nd
* E(X + Y ) = 0 0; 05 + 1 0; 15 + ::: + 6 0; 15 = 3; 10
Note that E(X + Y ) = E(X) + E(Y ).
* E(X + Y )2 = 02 0; 05 + 12 0; 15 + ::: + 62 0; 15 = 12; 7
6
* V ar(X + Y ) = 3; 09
* Alternatively, we …nd that
V ar(X + Y ) = V ar(X) + V ar(Y ) + 2 Cov(X; Y )
= 0; 5 + 1; 79 + 2 0; 4
= 3; 09.
Example 2 In the following table, X = the age of children and Y = the
number of books they have read in the last week.
X!
Y #......
0
1
2
3
4
T otal
8
0; 06
0; 02
0; 01
0; 00
0; 00
0; 09
9
0; 08
0; 06
0; 06
0; 04
0; 02
0; 26
10
0; 02
0; 08
0; 10
0; 06
0; 01
0; 27
11
0; 07
0; 07
0; 03
0; 02
0; 03
0; 22
12 T otal
0; 01 0; 24
0; 00 0; 23
0; 02 0; 22
0; 08 0; 20
0; 05 0; 11
0; 16 1; 00
Find
* E(X); E(X 2 ); V ar(X) and (X);
* E(Y ); E(Y 2 ); V ar(Y ) and (Y );
* E(XY ); Cov(X; Y ) and (X; Y ).
1.2.5
Rules of Probability
0
P (A)
1
P (A [ B) = P (A) + P (B)
P (Ac ) = 1
P (A \ B)
P (A)
P (A j B) = P (A \ B)=P (B)
P (AnB) = P (A)
P (A \ B)
Problems
(1) A and B are events with P (A) = 3=8, P (B) = 1=2 and P (A \ B) =
1=4.
Find P (Ac ), P (A [ B), P (B c ), P (Ac \ B c ), P (Ac [ B c ), P (A \ B c ).
7
Find P (A j B), P (A j B c ), P (A \ B j A), P (A j Ac ).
(2) Let A and B be events with P (A [ B) = 3=4, P (Ac ) = 2=3 and
P (A \ B) = 1=4.
Find P (A), P (B), P (A \ B c ), P (AnB), P (Ac nB).
1.3
Discrete models
1.3.1
Uniform discrete
In this case X has a …nite number of outcomes, and each outcome has the
same probability.
Example 1. Toss a coin and let X = 0 if head and X = 1 if tail. If the
coin is not false, each outcome has probability 1=2.
Example 2. Toss a die. Then Im(X) = f1; 2; 3; 4; 5; 6g each with equal
probability 1=6.
1.3.2
Techniques of counting
Rule 1.
If procedure 1 can be performed in n(1) ways, and if procedure 2 can be
performed in n(2) ways, and if procedure 3 can be performed in n(3) ways,
and if ...., then the number of ways the procedures can be performed in this
order, is the product n(1) n(2) n(3) :::.
Examples
(1) Car number plates in country A consist of 6 numbers out of the
numbers f0; 1; :::; 9g. We can make
10
10
10
10
10
10 = 1:000:000
di¤erent number plates. This is not enough to give all cars in Belgium a
unique number.
(2) Using the alphabet fA; B; :::; Zg we …nd the number of words consisting of at most 5 symbols.
We …nd
- the number of words with 1 symbol is 26;
- the number of words with 2 symbols is 26 26;
...
- the number of words with 5 symbols is 26 26 26 26 26.
8
The total number of words of 5 symbols or less is given by
26 + 26
26 + ::: + 26
26
26
26
26 = 12:356:630
(3) In my bookshelve, 5 books can be re-arranged in
5
4
3
2
1 = 5! = 120
ways.
(4) In a class there are 10 boys and 15 girls. We have to choose a team
consisting of
- one president
- one female secretary
- one male secretary
- one female bookkeeper
- one male logistic manager
We can make such a team in several ways
Case 1. First we choose a male boss (10), and then choose a female
secretary (15), a male secretary (9), a female bookkeeper (14) and a male
logistic manager (8). We …nd a total number of
10
15
9
14
8 = 930
possible teams.
Case 2. Starting with a female boss, we …nd a total number of
15
14
10
13
9 = 2820
possible teams.
As a total, we …nd 2820 + 930 = 3750 di¤erent teams.
Rule 2
The number of ways to choose k objects from a set of n objects is given
by
n!
n
n
=
=
k
n k
k!(n k)!
These numbers are called binomial coe¢ cients. The number gives the number
of combinations of k objects out of n objects.
Examples
(1) Suppose we have 4 people in a class fA; B; C; Dg.
9
Case 1. Choose a boss and a secretary
The number of possibilities is 4 3 = 12 and they are given as follows:
AB; BA; AC; CA; AD; DA; BC; CB; BD; DB; CD; DC
Case 2. Choose a team of 2 people
Since the team AB is the same as team BA, we …nd 4x3=2 = 6 possible
teams.
(2) A student is to answer 8 out of 10 questions in an exam.
(i) How many choices has he?
10
Answer:
= 45
8
(ii) How many choices has he if he has to answer the …rst 3 questions?
Answer. He can freely choose 5 questions out of the remaining 7 questions
7
and has
= 21 choices.
5
(iii) How many choices has he if he has to answer at least 4 of the …rst 5
questions?
Answer.
Case 1. He chooses 4 questions out of the …rst 5 questions (5 ways) and
then chooses 4 questions out of the remaining 5 questions (5 ways). This
gives a total of 25 possibilities.
Case 2. He chooses the …rst 5 questions (1 way) and then chooses 3 questions out of the remaining 5 questions (10 ways). This gives a total of 10
possible exams.
The total of the 2 cases gives 25 + 10 = 35 di¤erent exams.
1.3.3
Probability example 1
A container contains 10 products of which 4 are defect (and 6 are correct).
We randomly take 3 out of these 10 objects and count the number X of
defects.
10
- To choose 3 objects out of 10 can be done in
= 120 ways.
3
4
6
- Choosing 0 defect and 3 correct objects can be done in
=
0
3
20 ways;
4
6
- Choosing 1 defect and 2 correct objects can be done in
=
1
2
60 ways;
10
- Choosing 2 defect and 1 correct objects can be done in
4
2
6
1
=
4
3
6
0
=
36 ways;
- Choosing 3 defect and 0 correct objects can be done in
4 ways.
Now we …nd
P (X
P (X
P (X
P (X
We …nd that
1.3.4
= 1; 2 and
2
=
=
=
=
0) = 20=120
1) = 60=120
2) = 36=120
3) = 4=120.
= 0; 56.
Probability examples 2
1) We choose 5 people from a classroom that constist of 8 girls and 4 boys.
Find the pdf of
- X = the number of boys in the selection;
- Y = the number of girls in the selection.
Also …nd E(X), E(Y ), V ar(X) , V ar(Y ) and E(X + Y ), V ar(X + Y ).
2) Two cards are drawn from an ordinary pack of 52 cards.
- Let X = the number of red cards. Find the pdf of X together with
mean and standard deviation.
- what is the probability that
- both cards are hearts;
- both are spades;
- one card is red;
- one card is heart and one is a spade.
1.3.5
Bernoulli distribution
A Bernoulli experiment is an experiment with precisely two outcomes. Usually the outcomes are denoted by SUCCES and FAILURE. And we take
X = 1 when there is a succes and X = 0 when there is a failure.
Let p = P (X = 1) denote the succes-probability and 1 p = q = P (X =
0) the probability of a failure.
11
We …nd
= E(X) = 1 p + 0 (1 p) = p
E(X ) = 12 p + 02 (1 p) = p
2
= E(X 2 ) E 2 (X) = p p2 = p(1
2
p)
Bernoulli experiments are important because they form the basis of many
other probabilistic models.
1.3.6
Binomial distribution
We consider n repeated and independent Bernoulli trials and we are interested in the number X of successes (and not in the order in which they
occur). If n = 3, we …nd the following possibilities:
2
3
exp
X
probability
6 SSS
7
3
ppp
6
7
6 SSM
7
2
pp(1
p)
6
7
6 SM S 2
7
pp(1 p)
6
7
6 M SS 2
7
pp(1
p)
6
7
6 SM M 1
7
p(1
p)(1
p)
6
7
6 M SM 1
7
p(1 p)(1 p)
6
7
4 MMS 1
5
p(1 p)(1 p
M M M 0 (1 p)(1 p)(1 p)
We …nd that
P (X
P (X
P (X
P (X
=
=
=
=
0) = (1 p)3
1) = 3p(1 p)2
2) = 3p2 (1 p)
3) = p3
In a more compact notation, we obtain that
P (X = k) =
n k
p (1
k
p)n k , 0
k
n.
This pdf is called the binomial distribution.Notation X s BIN (n; p).
12
Properties of this distribution are:
2
= E(X) = np
= np(1 p)
We can calculate P (X = k) and/or P (X
calculator or EXCEL or
For the calculator TI 84 we have
P (X = k) = binompdf(n; p; k)
P (X k) = binomcdf(n; p; k).
k) by using tables or a
Examples
(1) A fair coin is tossed 6 times and we count the number X of heads.
In this case X s BIN (n = 6; p = 1=2).
The probability that exactly two heads occur is P (X = 2) =binompdf(6; 1=2; 2) =
0; 23.
The probability of getting at most 3 heads is P (X 3) =binomcdf6; 1=2; 3) =
0; 66
The probability of getting at least 4 heads is
P (X
4) = 1
P (X
3) = 1
0; 66 = 0; 34.
(2) Team A has probability p = 2=3 of winning any set in a game. If
A plays 4 sets, then the number of games that team A wins is given by
X s BIN (n = 4; p = 2=3).
We …nd P (X = 2) = 0; 30; P (X 1) = 1 P (X 0) = 0; 99.
(3) The number of books read by 12 year old children is a random variable
X with
x
0
1
2
3
4
P (X = x) 1=10 2=10 4=10 2=10 1=10
(i) Determine E(X) and 2 (X).
(ii) We do a small survey and ask 10 children about their reading behaviour.
- What is the probability that we …nd exacly 5 children that read 2 books.
- What is the probability that we …nd at most 5 children that read 2
books.
13
- What is the probabiltiy that we …nd at least 5 children that read 2
books.
- What is the probabiltiy that we …nd exactly 5 children that read 2 or
more books.
- What is the probabiltiy that we …nd at most 5 children that read 2 or
more books.
- What is the probabiltiy that we …nd at least 5 children that read 2 or
more books.
(4) The probability of a man hitting a target is 1/3.
(i) If he …res 5 times, what is the probability of hitting the target at least
2 times?
(ii) If he …res 5 times, what is the probability of hitting the target at most
2 times?
(iii) If he …res 5 times, what is the probability of hitting the target exactly
2 times?
(iv) If he …res 5 times, what is the probability of hitting the target at
least 1 time?
(v) How many times must he …re so that the probability of hitting the
target at least once is more that 95%?
(5) Of the light bulbs produced by a factory, 2% are defective. In a
shipment of 5000 light bulbs, …nd the expected number of defective bulbs
and the standard deviation.
(6) During a multiple choice exam a student has to answer 20 questions.
Each question has 4 answers of which exactly one is correct. The result of
the exam is given by X = the number of correct answers.
(i) A student has not studied and for each question he randomly chooses
an answer. What is P (X 10).
(ii) For each good answer the student gets 2 points and for each wrong
answer the student gets 1. Let Y denote the …nal result. Find E(Y ) and
V ar(Y ).
(iii) A student has studied part of the materials and answers a question
correctly with probability p = 2=3. What is P (X 10) now? What is E(Y )
and V ar(Y ) now?
(7) The management of the posto¢ ces claims that 95% of the mail inside
Belgium is delivered within 2 days of the time of the mailing. Six letters are
14
randomly sent to di¤ferent locations.
(i) What is the probability that all six arrive within two days?
(ii) What is the probability that exactly 5 arrive within 2 days?
(iii) What is the mean number of letters that arrive within 2 days?
(8) The Police Association has found that 60% of the drivers use their
seat belt. A sample of 10 drivers is selected.
(i) What is the probability that exactly 7 were wearing seat belts?
(ii) What is the probability that 7 or fewer of the drivers are wearing seat
belts?
1.3.7
Negative binomial distribution
In the binomial model we count the number of successes in a …xed number
of experiments. In the negative binomial we are interested in the number of
experiments until we have a …xed number of successes.
Assume that the probability of a success is p.
Let X(1) denote the number of experiments until the …rst success.
Celarly we have
P (X(1) = 0) = 0
P (X(1) = 1) = p (…rst experiment is a success)
P (X(1) = 2) = (1 p) p (…rst a failure and then a success)
P (X(1) = 3) = (1 p)2 p (…rst two failures and then a success)
...
P (X(1) = k) = (1 p)k 1 p, for k = 1; 2; 3; :::
One can prove that
P (X(1) > k) = (1
and that
E(X(1)) =
p)k , k = 0; 1; 2; :::
1 p
1
and V ar(X(1)) =
.
p
p2
Let p = 0; 1 and let X(5) denote the number of experiments until we
have the 5th success.
15
To …nd P (X(5) = 20), we proceed as follows. The event X(5) = 20
corresponds to the following picture:
1 2 3 ... ... ... ... ... ...
20
5th S
In the picture we have 20 experiments. The only thing that we know is
that the 20th experiment is a success. In the preceeding 19 experiments we
should have exactly the 4 remaining successes. The probability P (X(5) = 20)
is given by
P (X(5) = 20) = binompdf (19; p; 4) p
= 0; 0798 0; 1 = 0; 00798.
In a similar way we …nd that
P (X(5) = 40) = binompdf (39; p; 4) p
= 0; 2059 0; 1 = 0; 02059.
To …nd the probability P (X(5) > 35) we proceed as follows: The event
X(5) > 35 corresponds to the following picture:
1 2 3 .. ... ... ... ... ... 35 ... ... ... ... X(5)
S
In the picture we put X(5) experiments where X(5) > 35. The only
thing that we know is that in the experiments up to 35 we have a maximum
of 4 sucesses. Conversely, if in the …rst 35 experiments we have at most 4
successes, then the 5th success should happen in an experiment later than
35. In this case we …nd that
P (X(5) > 35) = binomcdf (35; p; 4)
= 0; 7308.
In a similar way, we …nd that
P (X(5) > 50) = binomcdf (50; p; 4)
= 0; 43.
16
One can prove that
E(X(5)) = 5
1
and V ar(X(5)) = 5
p
1
p
p2
.
Excercises
(1) The Police Association has found that 60% of the drivers use their
seat belt.
(i) The police checks car by car. What is the probability that the 4th
driver is the …rst one without a seat belt?
Answer: X(1) = the number of experiments until the …rst success where
success = without belt.
P (X(1) = 4) = (0; 6)3 0; 4 = 0; 0864.
(ii) What is the probability that the 10th driver is the 3rd one without
seat belts.
Answer: X(3) = the number of experiments until the 3rd success.
P (X(3) = 10) = binompdf (9; 0; 4; 2) 0; 4 = 0; 0645.
(2) In 15% of the cases, a bag of chips contains a point. If you collect 10
points, you take part in a big lottery. What is the probability that we have
to buy more than 90 packets of chips to have collected 10 points?
1.3.8
Poisson distribution
The Poisson model is a model in which Im(X) = f0; 1; 2; :::g and with pdf
given by
1 k
e , k = 0; 1; 2; :::
P (X = k) =
k!
We use the notation X s P OISSON ( ) and say that X has a Poisson
distribution with parameter > 0.
One can show that for a Poisson distribution we have
E(X) =
V ar(X) =
Usually X counts a number of successes or events in a certain time-unit
and is the expected number of successes per time unit. If we change the
time-unit, we still have a Poisson distribution with a di¤erent parameter.
We can calculate probabilities concerning POISSON using tables, a calculator or EXCEL or...
17
For the calculator TI 84 we have
P (X = k) = poissonpdf( ; k);
P (X k) = poissoncdf( ; k):
Examples
(1) The number of lost bags per ‡ight follows a Poisson distribution with
= 0; 30.
(i) The probability of losing not any bag is
P (X = 0) t 0; 7408
(ii) The probability of losing exactly one bag is
P (X = 1) t 0; 2222
(iii) The expected number of lost bags is given by E(X) = = 0; 30.
(iv) The number of lost bags per 10 ‡ights follows a Poisson distribution
with parameter 10 0; 3 = 3.
(2) Suppose that the weekly number of absent employees in a companie
is given by X s P OISSON (7).
(i) What is the probability that in a given week we have X 5?
(ii) What is the probability that in a given week we have X > 5?
(iii) What is the probability that in a given week we have X = 7?
(iv) What is the probability that in a given week we have X 7?
(v) What is the probability that in a given week we have 5 X 9?
(vi)What is the expected number of absent people during a given month
(4 weeks)?
(vii) What is the probability that in a given month we have 30 or more
absent people?
Remark
Suppose that we have to deal with a problem concerning Y s BIN (n =
5000; p = 0; 01). This is a problem with n very large and p rather small. In
this case we …nd for example that
P (Y
5000
(0; 01)2 (0; 99)4998
2
5000 4999
(0; 01)2 (0; 99)4998
=
2!
1
4999
=
(5000 0; 01)2
(0; 99)4998
2!
5000
= 2) =
18
One can prove that this probability can be approximated as follows: we
have
1 2
e
P (Y = 2) t
2!
where = np = 5000 0; 01 = 50.
Similar calculations can be made for P (Y = 3) or P (Y = 4) or P (Y = k).
Moreover, we have the following property
If Y s BIN (n; p) with n large and p small
Then X t Y where Y s P OISSON ( = np).
Examples
(1) The number of lost bags per ‡ight follows a Poisson distribution with
= 0; 30.
(i) The probability of losing not any bag is
P (X = 0) =
1
(0; 30)0 e
0!
0;30
t 0; 7408
(ii) The probability of losing exactly one bag is
P (X = 1) =
1
(0; 30)1 e
1!
0;30
t 0; 2222
(iii) The expected number of lost bags is given by E(X) = = 0; 30.
(2) Suppose that the weekly number of absent employees in a companie
is given by X s P OISSON (7).
(i) What is the probability that in a given week we have X 5?
(ii) What is the probability that in a given week we have X > 5?
(iii) What is the probability that in a given week we have X = 7?
(iv) What is the probability that in a given week we have X 7?
(v) What is the probability that in a given week we have 5 X 9?
(vi)What is the expected number of absent people during a given month
(4 weeks)?
(vii) What is the probability that in a given month we have 30 or more
absent people?
19
1.3.9
"Negative Poisson" distribution
Example 1 Suppose that the number of customers in a shop has a Poisson
distribution with a mean value of 9 customers per hour.
If we take as a time-unit "one hour", we have X(1) s P oisson(9). Per
two hours we have X(2) s P oisson(18). Per half and hour we have X(0; 5) s
P oisson(4; 5).
Let T (1) denote the time (in hours) until the …rst customer enters the
shop;
let T (2) denote the time (in h) until the second customer enters the shop;
...
let T (k) denote the time (in h) until the kth customer enters the shop.
Recall that we take "one hour" as a time-unit.
We can calculate (for example) P (T (1) > 1=6) as follows. The event
(T (1) > 1=6) means that in the …rst 1=6 of an hour there has been no
customer! This is the same as the event (X(1=6) = 0). Since X(1=6) s
P oisson(9=6), we …nd that
P (T (1) > 1=6) = P (X(1=6) = 0)
= poissonpdf (9=6; 0) = 0; 22.
To …nd (for example) P (T (12) > 2) we proceed as follows. The event
(T (12) > 2) means that in the …rst 2 hours we had at most 11 customers.
This is the same as the event (X(2) 11). Since X(2) s P oisson(18), we
…nd that
P (T (12) > 2) = P (X(2) 11)
= poissoncdf (18; 11) = 0; 05.
Example 2 The police randomly checks drivers for alcohol. The number
of drivers under in‡uence has a Poisson distribution with a mean value of 7:1
drivers per hour.
a) Let Y (1) = # of drivers under in‡uence when the check for 1 hour.
- Find P (Y (1) = 8), P (Y (1) 8), P (Y (1) 8).
b) Let Y (3; 2) = # of drivers under in‡uence when they check for 3; 2
hours.
- Find P (Y (3; 2) = 23), P (Y (3; 2) 23), P (Y (3; 2) 23).
c) Let T (k) = the time until the police …nds the k th driver under in‡uence.
20
- Find P (T (1) > 1=10), P (T (1) > 2=10), P (T (7) > 1; 3), P (T (13) > 2).
Remark
The distributions of T (1), T (2), T (3),... are called ERLANG distributions
or GAMMA distributions.
More information can be found in the following links:
http://en.wikipedia.org/wiki/Erlang_distribution
http://en.wikipedia.org/wiki/Gamma_distribution
21
2
2.1
Continuous probability distributions
Descriptive statistics
In descriptive statistics, for "continuous data", we used a frequency table with
classes of outcomes and we used the histogram and the empirical distribution
function. The EDF was a ’nice’ function’ non-decreasing function starting
at 0 and ending at 1.
The histogram was a set of blocks and the total area covered by the blocks
was equal to 1.
2.2
Probability theory: continuous models
- The image of X is assumed to be an (a …nite or in…nite) interval of the
form [a; b].
- A probability density function is a function f (x) such that
Z
f (x)
1
0;
f (t)dt = area between x
axis and f (:) is 1.
1
- The theoretical distribution function F (x) is the cumulative area between the horizontal axis and f (:) up to the value x:
Z x
F (x) =
f (t)dt.
1
Clearly f (x) = F 0 (x) for (almost) all values of x.
2.2.1
Uniform(a; b)
We say that X has a U N IF ORM (a; b) distribution if the density of X is
given by:
1
,a x
b a
= 0, otherwize.
f (x) =
Example 1.
22
b
In the case of UNIF(5, 25), we …nd that f (x) = 1=20 = 0; 05 for 5
x
25 and 0 elswewhere. Note that intervals (inside [a; b]) of the same
lenght,have the same probability. We …nd for example
P (5
X
10) = P (7
X
12) = P (15
X
20) = 5=20.
To calculate the TDF F (x), we calculate areas under the density function.
We …nd
F (x) = 0, x 5
x 5
,5 x
=
20
= 1, x 25.
25
Thinking about the interpretation of the mean
= E(X), we …nd that
E(X) = (b + a)=2 = 15.
It can be proved that for the general case, the second moment and the
variance are given by
b 3 a3
1
= (b2 + ab + a2 )
3(b a)
3
1
V ar(X) =
(b a)2
12
E(X 2 ) =
Example 2. X has a standard uniform distribution if X s U N IF (0; 1).
In this case we …nd
= E(X) = 1=2
E(X ) = 1=3, V ar(X) =
2
2
= 1=12
Clearly X s U N IF (0; 1) implies that aX + b s U N IF (a; b).
Example 3.
(i) Suppose that a trip with the train from A to B takes a random time
X s U N IF (25; 35) (in minutes). In this case we …nd that the mean time af
the trip is given by = E(X) = 30 minutes. It takes longer than 30 minutes
with probability P (X 30) = 1=2. The trip takes less than 26 minutes with
probability P (X 26) = 1=10.
(ii) When 12 year old girls run, they run a 100 meters in a time (seconds)
given by X s U N IF (a; b). If we know that = 20 and = 5, …nd the
corresponding values of a and b.
23
More information about the uniform distribution can be found in the
following link:
http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)
24
TRIANG(1,4, 5)
0
2.2.2
1
2
3
4
5
6
7
Triangular
It is clear that many real situations don’t correspond to a uniform distribution. The train, for example, is in my experience more often late than that it
is early; the number of accidents is not spread out uniformly throughout the
year. As an alternative, one can often work with a triangular distribution.
In general, the density function looks as in …gure 1.
To …nd the precise formula of the density, …rst notice that the area below
the curve should be equal to 1. Hence we …nd
1
1 = area = (c
2
and so h = 2=(c
a)
h
a). Now we …nd the line that connects (a; 0) with (b; h):
y
so that
y=
0=
h
b
a
(x
h
b
0
(x
a
a), a
25
a)
x
b.
The line that connects (b; h) and (c; 0) is given by
y
so that
y=
h=
h
c
b
0
c
(c
h
(x
b
x), b
b)
x
c.
To conclude, we …nd
f (x) =
h
a), a
x
b;
(c x), b
c b
= 0, elsewhere.
x
c;
=
b
a
(x
h
For the TDF, we …nd (cf classroom)
F (x) = 0, x a;
h
=
(x a)2 , a x b;
2(b a)
h
= 1
(c x)2 , b x c;
2(c b)
= 1, x c.
One can prove that
= E(X) =
a+b+c
3
Mode
= Mo = b
a2 + b 2 + c 2
V ar(X) =
ab
ac
bc
18
For more information,see for example
http://en.wikipedia.org/wiki/Triangular_distribution.
Exercises
(1) At the age of 10, children can throw a ball a distance (in meters) given
by X s T RIAN G(0; 3; 8).
(i) What is = E(X) and 2 (x)?
(ii) What is P (X 3)?, P (X 6)? P (2 X 5)?
26
(iii) Determine P (X
) and P (X > ).
(iv) Determine the median. This is the number x such that P (X x) =
P (X x) = 0; 5.
(2) In an election the proportion of votes that a party received was given
by X s T RIAN G(0; 0; 2; 0; 25).
(i) Calculate mean and variance of X.
(ii) What is the mode?, the median?
(iii) What is the probability that X is between and the median?
27
0,045
0,04
0,035
0,03
0,025
0,02
0,015
0,01
0,005
0
-4
2.2.3
-3
-2
-1
0
1
2
3
4
Normal distribution
A famous continuous distribution is the general normal X s N ( ;
bution and the standardnormal distibution Z s N (0; 1).
The form of the density of Z is given in the …gure 2.
2
) distri-
One can prove that = E(X) and that 2 = V ar(X). Moreover, normal
distributions are symmetric around the mean .
The density function has the following (rather complicated form):
X s N( ;
2
(x
1
): fX (x) = p exp(
2
2
)2
2
),
1 < x < 1;
and
x2
1
p
Z s N (0; 1) : fZ (x) =
exp(
),
2
2
One can prove that
X s N( ;
2
) if and only if Z =
28
X
1 < x < 1.
s N (0; 1).
Unfortunately, there is no closed form for the formula for the distribution
function F (x). In order to make calculations related to normal distributions
one usually standardizes and then use tables, excel or calculator to …nd the
desired probabilities.
With the TI 84, we use the following instuctions:
P (Z
P (a
a) = normalcdf ( 10; a)
Z b) = normalcdf (a; b).
To solve (for example) the equation P (Z
x) = 0; 90, we use invNorm:
invN orm(0; 9) = 1; 28
Examples and exercises (1) The weekly income is given by X s N ( =
1000; = 100).
(i) We …nd P (X 790) = P (Z (790 1000)=100) = P (Z
2; 10) =
0; 0179.
(ii) We …nd P (790 X 1000) = P ( 2; 10 Z 0) = 0; 4821.
(iii) We …nd P (790 X 1200) = P ( 2; 10 Z 2) = 0; 9224.
(iv) We …nd P (X 1245) = P (Z 2; 45) = 0; 0071.
(v) To …nd the value of x such that P (X
x) = 0; 95, we …nd 0; 95 =
P (Z (x 1000)=100) and then (x 1000)=100 = 1; 645 so that x = 1164; 5.
(2) For the standard normal distribution, we …nd some popular values:
P (Z 1; 284) = P (Z
1; 285) = 0; 90
P (Z 1; 645) = P (Z
1; 645) = 0; 95
P (Z 1; 96) = P (Z
1; 96) = 0; 975
P (Z 2; 33) = P (Z
2; 33) = 0; 99
P (Z 2; 57) = P (Z
2; 57) = 0; 995
(3) A normal population X has a mean of 50 and a variance of 16.
(i) Compute P (44 X 55)
(ii) Compute P (X 55)
(iii) Determine x such that P (X x) = 0; 95.
(iv) Determine x such that P (X x) = 0; 80
(v) Determine x such that P (X x) = 0; 25 (this is the …rst quartile)
(vi) Determine x such that P (X x) = 0; 75 (this is the 3rd quartile)
(4) If X s N (9; 9), …nd P (5 < X > 11).
29
(5) A …rm makes light bulbs that have a lifetime (in hours) given by
X s N ( = 800; 2 = 1600)
(i) What is the probability that a bulb works longer than 778 hours?
(ii) What is the probability that a light bulb works between 778 and 835
hours?
(iii) Given that a lightbulb has worked for 800 hours, what is the probability that it will work at least 850 hours?
(this is P (X 850 j X 800))
(6) Co¤ee is sold in packets with weight (in grammes) X s N ( =
250; = 5).
(i) What is the probabiltity that a customer buys a packet that contains
less than 250 gr?
(ii) What is the probability that a customer buys a packet that contains
less than 245 gr?
(iii) A customer buys n = 10 packets. What is the probability of …nding
at most 2 packets that contain
- less than 250 gr?
- less than 245 gr?
(7) In measuring the content (in cl) of bottles of parfume, the measurement errors are given by X s N ( = 0; = 0; 1).
(i) Find P (jXj 0; 1).
(ii) Find P (X 0; 1)
(8) Suppose X s N ( 4; 2 = 4). Compute
(i) P (X
2)
(ii) P ( 5 X
2)
(iii)P (X
6)
(iv) P (jX + 3j 1)
(v) Find a such that P ( a X + 4 a) 95%.
An important property An important property of the family of normal distributions is that it is closed under taking linear combinations. More
precisely we have the following property.
Property. Suppose that X s N ( 1 ; 21 ) and Y s N ( 2 ; 22 ) and suppose
that X and Y are independent. Then for all numbers a; b where exclude the
case where a = b = 0, we have
aX + bY s N (a
1
+ b 2 ; a2
30
2
1
+ b2
2
2 ).
Of course a similar property holds for 3 or more independent normal
distributed variables. We have, for example the following property.
Property. Suppose that X1 ; X2 ; :::; Xn are independent variables all having a normal distribution N ( ; 2 ).Then
Sn = X1 + X2 + ::: + Xn s N (n ; n 2 )
and
2
Sn
s N( ; )
n
n
Even the previous property holds starting from any distribution with
…nite variance, upon the condition that n is su¢ ciently large, cf. below.
X=
Examples and exercises (1) People in B have a weight that is distributed
according to a normal distribution N ( = 75; 2 = 625). Taking a sample of
10 prsons, they have independent weights X1 ; X2 ; :::; X10 . Their total weight
is given by S10 = X1 + X2 + ::: + X10 and S10 s N (750; 6250).
The probability that they total a weight of 800 or more is given by
p
P (S10 > 800) = P (Z > (800 750)= 6250) = P (Z > 0; 63) = 0; 26.
(2) Light bulbs have a life time (in hours) given by X s N (2000; = 250).
We want to have nonstop light during a period of 10 year of 365 days. How
many light bulbs do we need to have nonstop light with a probability of 95%
or more.
To answer this question we need bulbs with lifetimes X1 ; X2 ; :::; Xn and
total amount of hours of light given by Sn = X1 + X2 + ::: + Xn .We want to
…nd n such that P (Sn > 10 365 24) 0; 95.
Now observe that Sn s N (n 2000; n 2502 ). Standardizing gives
0; 95
P (Sn > 87600)
= P (Z > (87600 2000
Now recalling that P (Z >
p
n)=(250 n))
1; 645) = 0; 95, we …nd that
87600 2000
p
250
n
n
=
1; 645
so that
87600
2000
n=
31
1; 645
250
p
n
or
2000
Now set
p
n
1; 645
250
p
n
87600 = 0
a
87600 = 0
n = a to …nd that
2000
a2
1; 645
250
Solving this quadratic equation we …nd 2 roots:
a1 =
6; 5161
a2 = 6; 72176
p
Clearly the …rst root doen’t solve our problem (recall that a = n > 0)
and we …nd n (6; 72176)2 = 45; 18. We conclude that we need at least 46
light bulbs.
(3) Car batteries produced by company A have a lifetime in years given
by XA s N ( A = 3; 5; A = 0; 4). A similar battery produced by company
B has a lifetime XB s N ( B = 3; 3; B = 0; 3).
(i) What is the distribution of XA XB ?
(ii) What is P (XA > XB )?
(iii) We take a sample of 25 batteries of each company and calculate X A
and X B .
Find P (X A 3; 5); P (X A > 3); P (X B 3; 5); P (X B > 3); P ( X A X B >
0; 4).
2.3
The normal approximation of the binomial
Recall that for X s BIN (n; p) we have = np and 2 = np(1 p). In calculating probabilies related to a binomial variable, one can often approximate
X by a normally distributed variable Y s N ( ; 2 ). Comparing moments,we
should use Y s N (np; np(1 p)).
2.3.1
Example
Take X s BIN (n = 10; p = 0; 25) and the corresponding Y s N ( =
2; 5; 2 = 1; 975). In the table below we list the TDF for both cases. Then
we calcalate the absolute value of the di¤erences.
32
k P (X k) P (Y
k) Di¤erence:
0
0; 056
0; 034
0; 022
1
0; 244
0; 137
0; 107
2
0; 526
0; 358
0; 168
3
0; 776
0; 642
0; 133
4
0; 922
0; 863
0; 059
5
0; 980
0; 966
0; 014
6
0; 996
0; 995
0; 014
7
1; 000
0; 999
0; 000
8
1; 000
1; 000
0; 000
9
1; 000
1; 000
0; 000
10
1; 000
1; 000
0; 000
The quality of the approximation is at some places good and at some
places bad. One way to obtain better approximations is by using a continuitiy
correction. Note that for X we have the following equalities:
P (X
4) = P (X
4; 2) = P (X
4; 5) = P (X
4; 99).
For the continuous variable Y in each of these cases we …nd a di¤erent
value!
P (Y
P (Y
P (Y
4) = 0; 863
4; 5) = 0; 928
4; 9) = 0; 960
One strategy is to use the normal approximation Y using the values 0; 5,
1; 5, ...4; 5, ...:
P (X
k) = P (X
k + 0; 5) t P (Y
We …nd the following new table:
33
k + 0; 5).
k P (X k) P (Y
k) Di¤erence:
0
0; 056
0; 072
0; 016
1
0; 244
0; 233
0; 011
2
0; 526
0; 500
0; 026
3
0; 776
0; 767
0; 008
4
0; 922
0; 928
0; 006
5
0; 980
0; 986
0; 005
6
0; 996
0; 998
0; 002
7
1; 000
1:000
0; 000
8
1; 000
1; 000
0; 000
9
1; 000
1; 000
0; 000
10
1; 000
1; 000
0; 000
Notice the large improvement of the approximations! We can further
improve the approximation, by using the approximation only for larger values
of n.
We summarize in the following result.
2.3.2
Law of The Moivre and Pascal
Suppose that X s BIN (n; p) and that n is large (say n 35). Also suppose
that p is not extremely small nor extremely large (say np 5 and n(1 p)
5) Let Y s N (np; np(1 p)). Then
P (X
k) = P (X k + 0; 5) t P (Y
k + 0; 5),
P (X = k) = P (k 0; 5 X k + 0; 5) t P (k 0; 5
Y
k + 0; 5).
Remark
If n is large and np is small, one can approximate X s BIN (n; p) by
Y s P OISSON ( = np).
2.3.3
Examples and exercises
(1) In an insurance company one has observed that stolen goods are not
recovered in 80% of the reported thefts. During a period in which n = 200
thefts occurred, we count the number X of unrecovered goods.
Clearly X s BIN (n = 200; p = 0; 80). To calculate probabilities concerning X we shall use
Y s N ( = 200
0; 80;
2
= 200
34
0; 80
0; 20) = N (160; 32).
(i) We calculate P (X
170). We …nd
P (X
t
=
=
=
(ii) We calculate P (X
P (X
170) = P (X 170; 5)
P (Y
170; 5)
170; 5 160
p
P (Z
)
32
P (Z 1; 86)
0; 968.
150). We …nd
150) = P (X 149; 5)
t P (Y
149; 5)
= P (Z
1; 86)
= 0; 968.
(2) In a production facility units are produced and 10% of the units are
defective. Suppose that a (produced) lot is accepted if there are 2 or less
defective items in a sample of n = 50. What is the probability that a lot is
accepted?
To answer the queston note that X = number of defective items s
BIN (n = 50; p = 0; 10). The probability that X
2 can be calculated
2
as follows: take Y s N ( = 50 0; 10; = 50 0; 10 0; 90) = N (5; 4; 5)
to …nd that
P (X
2) = P (X 2; 5)
t P (Y
2; 5)
= P (Z
1; 178)
= 0; 119
We conclude that in about 12% of the cases a lot will be accepted. This is
not too much!
(3) A pair of fair dice (one red and one white) is thrown 120 times. What
is the probability of throwing at least 25 times a "seven"?
- Throwing a pair of dice gives outcomes f(1; 1); (1; 2); :::; (1; 6); (2; 1); (2; 2); :::; (6; 6)g.
Since the dice are fair, each outcome has equal probability 1=36. To have a
"seven" means that we should have one of the following outcomes:
f(1; 6); (2; 5); (3; 4); (4; 3); (5; 2); (6; 1)g
35
so we have a "seven" with probability 1=6.
- Now we consider throwing 120 times and each time we are interested
in "seven" or "no seven". Clearly the number of times we have "seven" is
given by X s BIN (n = 120; p = 1=6). Now we approximate with Y s
N (120 1=6; 120 1=6 5=6) and we …nd
P (X
25) = P (X 24; 5)
t P (Y
24; 5)
= P (Z
= P (Z
= 0; 13
(24; 5
1; 10)
p
20)= 100=6)
(4) A multiple choice test has 80 questions,each with four possible answers
of which only one answer is correct. Suppose that a student has no knowledge
at all and answers the questions by pure guesswork. Let X denote the number
of correct answers.
(i) Find P (X 40)
(ii) Find P (25 X 30).
(iii) Answer (i) and (ii) assuming that each question is answered correct
with probability
(a) p = 1=3, (b) p = 0; 5.
2.4
The Normal approximation of the Poisson distribution
Also for the Poisson distribution, we have a normal approximation. Recall
that if X s P OISSON ( ), then E(X) = V ar(X) = and nowweshould
use Y s N ( ; ). We have the following property.
Property
Suppose that X s P OISSON ( ) and that is large (say
35). Let
Y s N ( ; ). Then
P (X
k) = P (X k + 0; 5) t P (Y
k + 0; 5),
P (X = k) = P (k 0; 5 X k + 0; 5) t P (k 0; 5
Y
k + 0; 5).
This property is called the central limit theorem for the Poisson distribution.
36
Examples and exercises
(1) In a shop the weekly demand for new TV’s is given by X s P OISSON (5).
(i) The mean number of sold TV-sets is given by E(X) = 5.
(ii) We…nd P (X 5) = 0; 616
(iii) We …nd P (X = 5) = 0; 175
(iv) We …nd P (X 4) = 1 0; 265 = 0; 735
(v) The yearly demand is P OISSON (52 5) = P OISSON (260). The
probability of selling 280 TV’s or more is given by
P (X
t
=
=
=
280) = P (X 279; 5)
P (Y
279; 5)
279; 5 260
p
P (Z
)
260
P (Z 1; 21)
0; 113.
(2) In a book the number of printing errors per page is P OISSON (2)distributed. If the book contains 250 pages, then the total number of errors
is given by P OISSON (500). We …nd for example
P (480
2.5
X 510) = P (479; 5 X 510; 5)
t P (479; 5 Y
510; 5)
479; 5 500
510; 5 500
p
= P( p
Z
)
500
500
= 0; 68 0; 18 = 0; 50.
The central limit theorem
Starting from independent variables X1 ; X2 ; :::; Xn all having a normal distribution N ( ; 2 ).Then earlier we have seen that
Sn = X1 + X2 + ::: + Xn s N (n ; n 2 )
and
2
Sn
s N( ; )
n
n
The previous property holds approximately when we start from any arbitrary distribution with …nite variance, upon the condition that n is su¢ ciently large.
X=
37
The Central Limit Theorem
Suppose that X1 ; X2 ; :::; Xn are independent and all having a distribution
W ( ; 2 ).Then as n ! 1, we have
Sn = X1 + X2 + ::: + Xn t N (n ; n 2 )
and
2
Sn
t N( ; )
n
n
In practise we use the central limit theorem (CLT) for n 35. If X is a
discrete variable, we use a continuity correctiinwhen we apply the CLT.
Examples and exercises
(1) In a shop the weekly demand for new TV-sets is given by X and has
the following pdf:
P (X = 0) = 0; 20
P (X = 1) = 0; 25
P (X = 2) = 0; 40
P (X = 3) = 0; 15
We easily …nd
= 0; 25 + 0; 80 + 0; 45 = 1; 50
E(X 2 ) = 0; 25 + 1; 60 + 1; 35 = 3; 2
2
= 3; 2 (1; 5)2 = 0; 95 and = 0; 9747.
We study the sales S during one yearand …nd that Y = X1 +X2 +:::+X52
where Xi denotes the number of sold TV’s in week i.
Assuming that the sales are independent from week to week, the CLT
shows that
X=
S t Y s N (52
1; 50; 52
0; 95) = N (78; 49; 4)
(i) Using a continuity correction we …nd for example
P (S
t
=
=
=
70) = P (S 70; 5)
P (Y
70; 5)
70; 5 78
p
)
P (Z
49; 4)
P (Z
1; 067)
0; 143.
38
(ii) In a similar way we …nd
P (S >
t
=
=
84) = P (S 84; 5)
P (Y
84; 5)
P (Z 0; 925)
0; 177.
(iii) Now we determine the minimum number of weeks so that the sales
are 200 or more with probability 95% or more. To this end we write
0; 95
P (S
t P (Y
= P (Z
We …nd that
299; 5
p
n
and then
1; 5
n
1; 64
and then
p
300) = P (S 299; 5)
299; 5)
299; 5 1; 5 n
p
)
n 0; 95
1; 5 n
=
0; 95
p
n
1; 64
0; 95
299; 5 = 0
n t 14; 67
n t 215; 3
We conlcude that it will take at least 216 weeks.
(2) The weights (in gram) of cans of beans is distributed according to a
variable X with mean = 250 and standard deviation = 10. In order to
ensure the quality, from time to time we take 36 cans and calculate the mean
X.
(i) In how many cases the mean is less than the norm 250?
The CLT gives
100
X t N (250;
).
36
and hance we …nd
P (X
250) = P (Z
0) = 0; 50.
It follows that in 50% of the cases, the cans have a weight less than 250.
39
(ii) In how many cases the mean is less than 248?
Now we …nd
P (X
248) = P (Z
= P (Z
= 0; 115.
1; 2)
248 250
p
)
100=36
(iii) Now we determine such that P (X 250)
We …nd
250
p
P (X 250) = P (Z
)
100=36
It follows that
and then
250
p
=
100=36
= 250 + 1; 64
p
0; 05.
0; 05
1; 64
100=36 t 252; 73
(3) The delay (inminutes) of trains going from Brussels to Ostend and
vice versa is a variable X s U N IF (0; 5). A person lives in Ostend and works
in Brussels 200 days per year. Suppose that the delays are independent from
day to day. Let S denote the total yealry delay time and X the daily average.
(i) Find E(S), V ar(S), E(X) and V ar(X).
(ii) Find P (S 520)
(iii) Find P (500 S 550)
(iv) Find P ( X 2; 5
0; 10)
(4) If you pay 50 euro, you can play the following game. You toss a fair
coin 40 times. Each time a "head" appears, you win a = 2 euro. The total
amount of money that you get is given by T .
(i) What is the probability that you get more than you paid? P (T > 50)?
(ii) What is the probability that you don’t loose money? P (T 50)?
(iii)Are you willing to play the game?
(iv) Repeat these questions with other values of a (take a = 3; a = 2; 5,
a = 1; 9):
40