Lecture note for Statistics.

1. 1-0
CHAPTER THREE
3. SPECIAL PROBABILITY
DISTRIBUTIONS AND
DENSITIES

2. 1-1
3.0 Introduction
A probability distribution is a statistical
function that describes all the possible values
and probabilities for a random variable within
a given range.
Probability distribution yields the possible
outcomes for any random event.
That is they directly describe the behaviour of
attributes.

3. 1-2
3.1 Discrete Probability Distributions
1. The discrete uniform distribution
If a random variable can take on k different values
with equal probability, we say that it has a discrete
uniform distribution.
Definition 3.1. A random variable x has a discrete
uniform distribution and it is referred to as a discrete
uniform random variable, if and only if its probability
distribution is given by f x =
1
k
, for x = x1, x2, … . xk
and where xi ≠ xj, when i ≠ j.

4. 1-3
Cont..
The mean and the variance of the discrete uniform
distribution are;
 The mean is E x = μ = xi ∗
1
k
k
i<1 and
 The variance is δ2
= xi − μ 2
k
i<1 ∗
1
k
2. The Bernoulli Probability distribution
Historically, the Bernoulli probability distribution
theory was applied only to games of chance. There
fore, an experiment has two possible outcomes.

5. 1-4
Cont...
This type of distribution is concerned with a situation
like success/failure, passing/failing an examination,
gaining/losing a profit in Business activities, etc.
If an experiment has two possible outcomes,
success and failure, and their probabilities are,
respectively P and (1 – P), then the random variable,
0 or 1, has a Bernoulli distribution.
Definition 3.2. A random variable x has a Bernoulli
distribution and it is referred to as a Bernoulli
random variable, if and only if;

6. 1-5
Cont...
Its probability distribution is given by:
 𝐟 𝐱; 𝐩 = 𝐩𝐱 𝟏 − 𝐩 𝟏;𝐱, for 𝐱 = 𝟎, 𝟏
In general, if we have p – success and the random
variable x = 0, 1 then;
 𝐟 𝟎; 𝐩 = 𝐩𝟎 𝟏 − 𝐩 𝟏;𝟎 = (𝟏 − 𝐏) and
 𝐟 𝟏; 𝐩 = 𝐩𝟏
𝟏 − 𝐩 𝟎;𝟏
= 𝐩

7. 1-6
Cont...
Example: In an industry’s quality control system it was
observed that there is a 95% success in producing
quality product. Then;
1. What is the probability of obtaining defective products?
2. List the probabilities for values x = 0, 1
Solution:
1. f 0; 0.95 = 0.950
1 − 0.95 1;0
= 0.05.
2. f 1; 0.95 = 0.951 1 − 0.95 1;1 = 0.95

8. 1-7
Cont...
Theorem 3.1. The mean and the variance of the
Bernoulli distributions are:
 𝐄 𝐱 = 𝛍 = 𝐩 and 𝛅𝟐 = 𝐩(𝟏 − 𝐩),
Examples: If the probability of defective goods in a
particular factory is 0.1 , find the mean and the
variance, for the distribution of defective goods in a
total of 100. Solution:
μ = p =
1
10
= 0.1 and δ2
= p(1 − 𝑝) =∗
1
10
∗
9
10
= 0.09.

9. 1-8
3. The Binomial Probability
distribution
In a binomial experiment our interest is in the
number of successes occurring in the n trials. If we
let x denote the number of successes in n trials, we
see that x can assume the values of 0, 1, 2, 3 … n.
Definition 3.3. A random variable x has a binomial
distribution and it is referred to as a binomial random
variable, if and only if its probability distribution is
given by:
 𝐛 𝐱; 𝐧, 𝐩 = 𝐧
𝐱
𝐩𝐱 𝟏 − 𝐩 𝐧;𝐱, for x = 0, 1, 2, . . . n

10. 1-9
Cont...
Where:
 n- is be the number of trials;
 x- is the number of observed success;
 p- is the probability in each trial;
 n
x
= nCx =
n!
x! n;x !
Example: Suppose a company produces toothpaste.
Historically, eight-tenths of the toothpaste tubes were
correctly filled (successes).

11. 1-10
Cont...
What is the probability of getting exactly three of six
tubes (half a carton) correctly filled?
Solution: Given; p = 0.8, q = 0.2, r = 3 and n = 6
 Probability of r success in n trial =
n!
r! n;r !
pr
qn;r
 Probability of 3 correctly filled=
6!
3! 6;3 !
(0.8)3(0.2)6;3
 b 3 =
6∗5∗4∗3!
3∗2∗1∗3!
0.512 0.008 =
120
6
0.512 0.008
 b(3) = 20 0.512 0.008 = 0.08192.

12. 1-11
Properties of Binomial Experiment
o The data collected are the results of counts.
o The experiment consists of a series of n-identical
trials.
o In each trial there are only two possible outcomes.
o The probability of a success on one trial does not
change from one trial to another and the probability
of a failure also does not change from trial to trial.
o Statistically, the trials are independent.

13. 1-12
Cont...
Theorem 3.2. The mean and the variance of the
binomial distributions are:
 𝐄 𝐱 = 𝛍 = 𝐧𝐩 and 𝛅𝟐
= 𝐧𝐩𝐪, When q = (1 − p).
Examples: If the probability of defective goods in a
particular factory is 0.1, find the mean and the variance,
for the distribution of defective goods in a total of 200.
Solution:
μ = np = 200 ∗
1
10
= 20 and δ2
= npq = 200 ∗
1
10
∗
9
10
=
18.

14. 1-13
Cont...
4. The hyper geometric distribution
Under this distribution choosing is with out
replacement, n of the N elements contain in the set,
there are M
x
ways of choosing x of the M
successes and N;M
n;x
ways of choosing (n– x) of the
(N– M) failures and, hence, M
x
N;M
n;x
ways of
choosing x successes and (n– x) failures.

15. 1-14
Cont...
Since there are N
n
ways of choosing n of the N
elements in the set, assuming that they are all
equally likely, there fore, the probability of “ x
successes in n trials” is
M
x
N−M
n−x
N
n
.
Definition 3.4. A random variable x has a hyper
geometric distribution and it is referred to as a hyper
geometric random variable if and only if; its
probability distribution is given by:

16. 1-15
Cont...
h x; n, N, M =
MCx.N;MCn−x
NCn
, for, x = 0, 1, 2, … n, where,
 x ≤ M and n − x ≤ N − M
 N- is the size of the population
 M- is number of success in the population
 n- the size of sample (number of trials)
 x- number of success in the sample (n)
 C- is the symbol for a combination

17. 1-16
Cont...
The hyper geometric distribution should be
applied:
 If the sample is selected from a finite population.
 If the size of the sample n is more than or equal to
5% of the population N.
 It is especially appropriate when the size of the
population is small.
Example. 1: A factory employs 50 people in the
assembly department 40 of the employees belong to
a union.

18. 1-17
Cont...
5 employees are selected at random to form a
committee to meet with management regarding shift
starting times. What is the probability that 4 of the 5
selected for the committee belong to a union?
Solution: Given;
 N = 50 the number of employees
 M = 40 the number of union employees
 X = 4 the number of union employees selected
 n = 5 the number of employees selected

19. 1-18
Cont...
h 4; 5, 50, 40 =
40C4.50;40C5−4
50C5
=
40!
4!36!
10!
1!9!
50!
5!45!
=
91390 10
2118760
= 0.431.
Theorem 3.3. The mean and the variance of the
hyper geometrical distribution are:
𝐄(𝐱) = 𝛍 =
𝐧𝐌
𝐍
and 𝛅𝟐 =
𝐧𝐌 𝐍;𝐌 𝐍;𝐧
𝐍𝟐 𝐍;𝟏

20. 1-19
Cont..
 Example. 2: Suppose the Economics Department of Oromia
State University wants to recruit four people for a teaching
position and for this 16 applicants have applied. Among the
16 applicants for a teaching position, 10 have university
degree. If 4 of the applicants are randomly chosen for
interviews, what are the mean and the variance?
 Solution: Given; N = 16, M = 10 and n = 4.
 𝛍 =
𝟒∗𝟏𝟎
𝟏𝟔
=
𝟒𝟎
𝟏𝟔
= 𝟐. 𝟓 and 𝛅𝟐
=
𝟒∗𝟏𝟎 𝟏𝟔;𝟏𝟎 𝟏𝟔;𝟒
𝟏𝟔𝟐 𝟏𝟔;𝟏
=
𝟒𝟎∗𝟔∗𝟏𝟐
𝟏𝟔∗𝟏𝟔∗𝟏𝟓
=
𝟑
𝟒
= 𝟎. 𝟕𝟓.

21. 1-20
Cont...
Based on the example above, if two of the four will
be taken for the job, what will be the probability?
Solution:
h 2; 4, 16, 10 =
10
2
16−10
4−2
16
4
=
10
2
6
2
16
4
=
10!
2!8!
6!
2!4!
16!
4!12!
=
135
364
=
0.371.

22. 1-21
Cont...
5. The Poisson distribution
As the number of trials (n) become large the
calculation of binomial probabilities with binomial
distribution formula will involve a prohibitive amount
of work.
Therefore the limiting form of the binomial
distribution where the probability of success p is
very small and n is large is called the Poisson
distribution i. e. p → 0, n → ∞ .

23. 1-22
Cont...
Definition 3.5. A random variable X has a poison
distribution and it is referred to as a poison random
variable if and only if; its probability distribution is given
by:
 𝐏 𝐱; 𝐦 = 𝐏 𝐗 = 𝐱 =
𝐦𝐱𝐞−𝐦
𝐱!
, for x = 0, 1, 2, …,
Where;
 P X = x − Probability of x occurrence in an interval
 m = np − The expected value or the average number of
occurrences in an interval
 e = constant equals to 2.71828 …

24. 1-23
Cont...
Example.1: A certain restaurant has a reputation for
good food. The restaurant management boasts that
on a Saturday night, groups of customers arrive at a
rate of 15 groups every half an hour, on average.
a) What is the probability that 5 minutes will pass with
no groups of customers arriving?
b) What is the probability that 8 groups of customers
will arrive in 10 minutes?

25. 1-24
Cont..
Solution:
a) Given, m= 15 group in 30 minutes
15 = 30min
? = 5min
 On average 2.5 groups in 5 minutes
P x =
mxe;m
x!
P 0; 2.5 =
2,50e;2.5
0!
=
1 ∗ e;2.5
1
= 0.0821

26. 1-25
Cont..
b) Given, m= 15 group in 30 minutes
15 = 30min
? = 10min
 On average 5 groups in 10 minutes
P x =
mxe;m
x!
P 0; 5 =
58
e;5
8!
=
2632.0105465178
40,320
= 0.0653

27. 1-26
Cont...
Example. 2: Calculate the probability that 2 of 1000
persons celebrating a ceremony of “ Timket” on a
very hot day will suffer from heat exhaustion, if the
probability is 0.005, that any one of the 1000
persons celebrating the ceremony will suffer from
heat exhaustion.
Solution: m = np = 1000 ∗
5
1000
= 5. x = 2
P 2; 5 =
52e−5
2!
=
25∗0.0067
2∗1
=
0.1675
2
= 0.08375

28. 1-27
Cont...
Poison distribution may be obtained as a limiting
case of Binomial probability distribution under the
following conditions:
1. n , the number of trials is indefinitely large
i. e. n → ∞ .
2. P, the constant probability of success for each trial
is indefinitely small i. e. p → 0 .
3. np = m, (say which is finite.)

29. 1-28
Cont..
Poisson distribution will provide a good approximation to
binomial probabilities when n ≥ 30 and p ≤ 0.05 and
np < 5.
Properties of Poisson Distribution
1. The probability of an occurrence of the event is the
same for any two intervals of equal length.
2. The occurrence or non-occurrence of the event in any
interval is independent of the occurrence or non-
occurrence in any other interval.

30. 1-29
Cont...
Theorem 3.4. The mean and the variance of the
poison distribution are;
 𝐄(𝐱) = 𝛍 = 𝐦 and 𝛅𝟐
= 𝐦.
Example: In poison distribution, if p(x) for x = 0 is
10% , find the mean and the variance of the
distribution.
Solution:
P x = 0 =
m0e−m
0!
= 10% = e;m → lne;m = ln10% →
− m = ln0.1 ≈ 2.3026.

31. 1-30
Cont...
If the standard deviation of a Poisson distribution is
2, find the probability that x = 0, and μ, and δ2.
Solution:
P x = 0 =
20e− 2
2
0!
= e;2 =
1
e2 = 0.135
μ = δ2 = 2
2
= 2

32. 1-31
3.2 Continuous Probability
Distributions
1. The Uniform Density (Rectangular Distribution)
Definition 3. 6. A random variable has a uniform
distribution and it is referred to as a continuous
uniform random variable if and only if its probability
density is given by:
𝐔 𝐱; 𝐚, 𝐛 =
𝟏 𝐛 − 𝐚 , 𝐚 ≤ 𝐱 ≤ 𝐛
𝟎, 𝐞𝐥𝐬𝐞𝐰𝐡𝐞𝐫𝐞
The parameter a and b of this probability density are
real constants, with a < b.

33. 1-32
Cont...
Theorem 3 .4. The mean and the variance of the
uniform distribution are:
 The mean: 𝐄 𝐱 = 𝛍 =
𝒂:𝒃
𝟐
 The variance: 𝐕𝐚𝐫 𝐱 = 𝛅𝟐
=
𝐛;𝐚 𝟐
𝟏𝟐
Example: Slater customers are charged for the
amount of salad they take. Sampling suggests that
the amount of salad taken is uniformly distributed
between 5 ounces and 15 ounces.

34. 1-33
Cont...
The probability density function is:
U x; a, b =
1 10 , 5 ≤ x ≤ 15
0, elsewhere
Where: x = salad plate filling weight
 What is the probability that a customer will take
between 12 and 15 ounces of salad?
Solution: U 12, 15; 5, 15 =
1
15;5
dx
15
12
=
1
10
𝑥
15
12
=
1
10
15 − 12 =
1
10
3 =
3
10

35. 1-34
Cont..
 What is the mean and the variance of the
distribution?
Solution:
Expected value of x is;
E x = μ =
a + b
2
=
5 + 15
2
=
20
2
= 10
The Variance of x is;
Var x =
b − a 2
12
=
15 − 5 2
12
=
10 2
12
=
100
12
= 8.33

36. 1-35
2. Gamma & Beta Distribution
2.1 Gamma Distribution
The random variable X with probability density
function:
𝐟 𝐱 =
𝛌𝐫
г 𝐫
𝐱𝐫;𝟏
𝐞;𝛌𝐱
, 𝐟𝐨𝐫, 𝐱 > 𝟎 is a
gamma random variable with parameters λ > 0 and
r > 0.
Gamma distribution has:
Mean = 𝐄 𝐱 =
𝐫
𝛌
and Variance = 𝛅𝟐 =
𝐫
𝛌𝟐

37. 1-36
2.2 Beta Distribution
The random variable X with probability density
function;
𝐟 𝐱 =
г 𝛂:𝐛
г 𝛂 .г 𝐛
𝐱𝛂;𝟏
𝟏 − 𝐱 𝐛;𝟏
, 𝐟𝐨𝐫 𝟎 < 𝐱 < 𝟏is a beta
random variable.
Where, X α, b = xα;1
1 − x b;1
dx
1
0
A Beta distribution has;
 Mean = 𝐄 𝐱 =
𝛂
𝛂:𝐛
and Variance 𝛅𝟐
=
𝛂𝐛
𝛂:𝐛 𝟐 𝛂:𝐛:𝟏

38. 1-37
Cont...
All modern programming languages have a package
for calculating Beta CDFs. You will not be expected
to compute the CDF by hand.

39. 1-38
3. The exponential density
distribution
The exponential distribution applies not only to the
occurrence of the first success in Poisson process,
but also it applies to the waiting times between
successes.
Definition 3.7. A random variable x has an
exponential distribution and it is referred to as an
exponential random variable if and only if its
probability density is given by:

40. 1-39
Cont...
𝐠 𝐱; 𝛉 =
𝟏
𝛉
𝐞
;𝐱
𝛉 , 𝐱 ≥ 𝟎, 𝛉 > 𝟎
𝟎, 𝐞𝐥𝐬𝐞𝐰𝐡𝐞𝐫𝐞
 If θ =
1
m
, then the above equation will be;
g x; m =
me;xm, 0 ≤ x < ∞,
0, elsewhere
Theorem 3.5. The mean and variance of the
exponential distributions are given by:
 𝐄 𝐱 = 𝛍 = 𝛉 =
𝟏
𝐦
and 𝐕𝐚𝐫 𝐱 = 𝛅𝟐 = 𝛉𝟐 =
𝟏
𝐦𝟐

41. 1-40
Cont...
Cumulative exponential distribution function can
be given as:
𝐏 𝐱 ≤ 𝐱𝟎 = 𝟏 − 𝐞
;𝐱
𝛉
When x0 is some specific value of x.
Example: A certain kind of appliance requires repairs
on the average once every 2 year. Assuming that the
times between repairs are exponentially distributed,
what is the probability that such an appliance will
work at least 3 years with out requiring repairs?

42. 1-41
Cont...
Solution: Given; μ = 2 = θ, x - at least 3 years.
g x; 2 =
1
2
e;
1
2
x
dx
∞
3
= e;
1
2
x ∞
3
=
−e;
1
2 ∞
− −e;
1
2 3
= 0 + e;
3
2 = 0.2231
4. The Normal Distribution
The most widely used model for the distribution of a
random variable is a normal distribution.

43. 1-42
Cont...
f(x)
x


44. 1-43
Cont...
Definition 3.8. A random variable x has a normal
distribution and it is referred to as a normal random
variable if and only if its probability density is given
by:
𝐧(𝐱; 𝛍, 𝛅𝟐) =
𝟏
𝛅 𝟐𝛑
𝐞
;
𝟏
𝟐
∗
𝐱;𝛍
𝛅
𝟐
 Where; −∞ < x < ∞, δ2 > 0, and −∞ < μ < ∞
 μ = mean, 𝛅 = standard deviation
 𝛑 = 3.14159 e = 2.71828

45. 1-44
4.1 Characteristics of the normal
probability distribution
 The normal curve is a bell-shaped curve and symmetric.
 Two parameters, mean and standard deviation, determine
the location and shape of the distribution.
Standard Deviation s
Mean 
x

46. 1-45
Cont..
 The mean of the normal distribution can be any
numerical value: negative, zero, or positive.
x
25
0
-10

47. 1-46
Cont..
 The highest point on the normal curve is at the
mean, which is also the median and mode.
 The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
x

48. 1-47
Cont...
 The normal probability distribution is asymptotic,
gets closer and closer to the x- axis but never
touches it.
x
s = 15
s = 25

49. 1-48
Cont...
 Probabilities for the normal random variable are
given by areas under the curve.
 The total area under the curve is 1 (0.5 to the left of
the mean and 0.5 to the right of the mean).
x
.5
.5

50. 1-49
 Some areas under normal curve are
given as follow;
A. U. N. C.
99.72%
95.44%
68.26%
 – 3s
 – 2s
 – 1s   + 1s
 + 2s
 + 3s
x

51. 1-50
4.2 Standard Normal Probability
Distribution
A random variable that has a normal distribution with
a mean of zero and a standard deviation of one is
called a standard normal probability distribution.
s = 1
z
0

52. 1-51
Cont..
The standard normal probability distribution is given
as:
𝐧(𝐱; 𝛍, 𝛅𝟐
) =
𝟏
𝛅 𝟐𝛑
𝐞;
𝟏
𝟐𝐱𝟐
The letter z is commonly used to designate this
standard normal distribution of random variable.
To convert a normally distributed random variable to
the standard normal distribution we use;
𝐳 =
𝐱 − 𝛍
𝛅

53. 1-52
Cont...
The standard normal value, 𝐳 =
𝐱;𝛍
𝛅
, then the
standard normal variable, z has mean = 0 and
standard deviation = 1
We can think of z as a measure of the number of
standard deviations x is from its mean.
The probability density function of standard normal
variable z is given by:
 ∅(𝐙) =
𝟏
𝟐𝛑
𝐞;
𝐙𝟐
𝟐 , Where; −∞ < 𝐙 < ∞

54. 1-53
Cont...
Theorem 3.6. The mean and the variance of normal
distribution is;
 The mean is; 𝐄 𝐱 = 𝛍
 The variance is; 𝐕𝐚𝐫 𝐱 = 𝛅𝟐
4.3 How to read the standard normal probability
distribution
This method calls for standardizing the distribution
or changing from the x scale to z- scale.

55. 1-54
Cont...
To find the area between a value of (x) and the
mean (μ), we first compute the difference between
the value (x) and the mean (μ), then we express
that difference in units of standard deviation. In other
words we compute the value by using: 𝐳 =
𝐱;𝛍
𝛅
Finally, we find the desired area under the curve, or
the probability, by referring to a table whose entry
corresponds to the calculated value of Z. Thus we
can convert any normal distribution to the standard
normal distribution.

56. 1-55
Cont...
Therefore the valve of Z follows a normal probability
distribution with a mean of zero and standard
deviation of one unit.
Examples 1: The monthly incomes of employees in
a textile factory are normally distributed with a
means of birr 2000 and a standard deviation of birr
200.
a) What is the Z value for an in come of birr 2200?
b) What is the Z value for an income of 1700?

57. 1-56
Cont...
Solution:
Given x = 2200, μ = 200 and δ = 200
a) z =
x;μ
δ
=
2200;200
200
=
200
200
= 1
b) z =
x;μ
δ
=
1700;200
200
=
;300
200
= −1.5
Hence a Z- value of 1 indicates that the value of
2200 is one standard deviation above the mean of
2000, while a Z- value of 1700 is 1.5 standard
deviation below the mean of 2000.

58. 1-57
Cont...
Examples 2: Pep Zone sells auto parts and supplies
including a popular multi-grade motor oil. When the
stock of this oil drops to 20 gallons, a replenishment
order is placed. The store manager is concerned
that sales are being lost due to stock outs while
waiting for an order. It has been determined that
lead-time demand is normally distributed with a
mean of 15 gallons and a standard deviation of 6
gallons.

59. 1-58
Cont...
 The manager would like to know the probability of a
stock out, P(x > 20).
Solution:
Given x = 20, μ = 15 and δ = 6
 Z =
x;μ
δ
=
20;15
6
=
5
6
= 0.83
The Standard Normal table shows an area of .2967
for the region between the z = 0 and z = .83 lines
below. The shaded tail area is .5 - .2967 = .2033.
The probability of a stock-out is .2033.

60. 1-59
Cont...
 Therefore, P(x > 20) is 0.2033.
0 .83
Area = .2967
Area = .5
Area = .5 - .2967
= .2033
z
Area = .5

61. 1-60
5. The Normal Approximation to the
Poisson and Binomial Distribution
5.1. The Normal Approximation to the Poison
distribution
If X is a poisson random variable, with E x =
m = mean and Var x = δ2 = m for m → ∞ thus
standard poison variable becomes;
𝐒𝐏𝐕 =
𝐱 − 𝐄(𝐱)
𝛅
=
𝐱 − 𝐦
𝐦
The approximation is good for 𝐦 ≥ 𝟓. When we use
normal approximation to Poisson distribution the
continuity correction must be applied.

62. 1-61
5.2. The Normal Approximation to
the binomial distribution.
Normal distribution is a limiting case of binomial
probability distribution under the condition that
𝐧 → ∞, and either 𝐏 or(𝟏 − 𝐏)is very small. Then
since 𝐄(𝐱) = 𝐧𝐩 and 𝐯𝐚𝐫(𝐱) = 𝐧𝐩𝐪 under binomial
probability distribution. Hence standard Binomial
variable can be specified as:
𝐙 =
𝐱 − 𝐄(𝐱)
𝛅
=
𝐱 − 𝐧𝐩
𝐧𝐩𝐪

63. 1-62
Cont...
Normal approximation to binomial distribution can be
applied, when number of trials (𝐧) is very large and
𝐩 or (𝟏 − 𝐏) is very small.
The normal probability distribution is a good
approximate to the binomial probability distribution
when, 𝐧𝐩 ≥ 𝟓 and 𝐧(𝟏 − 𝐩) ≥ 𝟓 are both at least 5.
However, before we apply the normal
approximation, we must make sure that our
distribution of interest is in fact a binomial
distribution.

64. 1-63
Cont...
So, when we approximate a discrete probability
distribution by a continuous one we need to make
an adjustment called the continuity correlation. In
this adjustment, we add or subtract 𝟎. 𝟓 to the value
that the binomial distribution is approximated by the
normal as follows.
 For the probability at least X occur, use the area
above (X – 0.5).
 For the probability that more than X occur, use the
area above (X + 0.5).

65. 1-64
Cont...
 For the probability that X or fewer occur, use the
area below (X + 0.5).
 For the probability that fewer than X occur, use the
area below (X – 0.5).
Example 1: A restaurant found that 70% of their new
customers return for another meal. For a week in
which 80 new customers were denied occurs at that
restaurant’s. What is the probability that 60 or more
will return for another meal?

66. 1-65
Cont...
 Solution: Given, n = 80, P = 0.70 x ≥ 60, μ = np =
80 ∗ 0.70 = 56
 δ = np(1 − p) = 80 ∗
70
100
∗
30
100
= 16.8 = 4.10.
 Z =
x;μ
δ
=
59.5;56
4.10
= 0.85, P Z = 0.85 = 0.3023
 P 60 ≤ x < 80 = P Z ≥ 0.85 = 0.5– 0.3023 =
0.1977

67. 1-66
Cont...
Example 2: Assume a binomial probability
distribution with n = 50 and P = 0.25, then, compute;
1. The mean and the standard deviation of the
random variable
2. The probably that x is 15 or more
3. The probability that x is 10 or less
 Solution; 1. μ = np = 50 ∗ 0.25 = 12.5
2. δ = np(1 − p) = 50 ∗
25
100
∗
75
100
= 9.375 = 3.0619

68. 1-67
Cont...
 Z =
x;μ
δ
=
14.5;12.5
3.0619
= 0.65, P Z = 0.65 = 0.2422
 P x ≥ 15 = P Z ≥ 0.65 = 0.5– 0.2422 = 0.2578
 3. Z =
x;μ
δ
=
10.5;12.5
3.0619
= −0.65, P Z = 0.65 = 0.2422
 P x ≤ 10 = P 0 ≤ x ≤ 10 = P Z ≤ 0.65 = 0.5
− 0.2422 = 0.2578

69. 1-68
Cont...

70. 1-69
Cont...

71. 1-70
Cont...