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STATISTICS FOR MANAGEMENT-I
LEARNING OBJECTIVES:
 To introduce the probability distributions most commonly used in
decision making.
 To show which probability distribution to use and how to find its value.
 To understand the limitations of each of the probability distributions
you use.
 To use the concept of expected value to make decision.
 Characteristics of the Normal probability distribution.
INSTRUCTOR’S NAME:
CHAPTER 5
Probability Distribution:
A list of the outcomes of an experiment with probabilities associated with these
outcomes is called probability distribution .It can be represented in the form of
table.
There are two types of probability distribution.
1-Discrete Probability
Distribution
2-Continuous Probability
Distribution
1-Discrete Probability Distribution:
A discrete probability distribution can take on only limited number of
values, which can be listed.
2-Continuous Probability Distribution:
A continuous probability distribution can take on any value within a given
range. So we cannot list all possible values.
Random Variable:
A variable that takes on different values as a result of the outcomes of a
random experiment.
There are two types of random variable.
1-Discrete Random Variable
2-Continuous Variable Random
1- Discrete Random Variable:
A discrete random variable can take on only a limited number of values, which
can be listed.
2- Continuous Random Variable:
A continuous random variable can take on any value within a given range,
which cannot be listed.
Expected Value:
It is the weighted average of the outcomes of the random experiment.
EX:
(EX Sc 5-1)
Solution:
a)
b)
Graph of probability distribution.
Probability distribution.
x
f
102
105
108
111
114
117
10
20
45
15
20
15
125
c)
Expected value = 109.44
P
10
125
= 0.08
0.16
0.36
0.12
0.16
0.12
Px
102x0.08= 8.16
16.8
38.88
13.32
18.24
14.04
109.44
EX:
(EX 5-8)
Solution:
a)
X
P
8000
9000
10000
11000
12000
13000
PX
0.05
0.15
0.25
0.30
0.20
0.05
8000X0.05= 400
1350
2500
3300
2400
650
10600
b) Expected value = 10600
H.W:
Do EX Sc 5-2 ( pg 229)
Bernoulli process:
A process in which each trail has only two outcomes. Probability of outcomes
remains fixed and trails are independent.
Binomial distribution:
It is a discrete distribution, describing the result of Bernoulli process.
Probability of r success in n trail can be describe as,
P(r success in n trail)
Here
=
𝑛!
𝑟!(𝑛−𝑟)!
pr qn-r
p = Probability of success.
q= Probability of failure.
r= Number of success desire.
n= Number of trails.
= µ = np
= 𝜎 2 = npq
= 𝜎 =√𝑛𝑝𝑞
EX :
(EX 5-20)
Find the mean and standard deviation of the binomial distribution.
Solution:
a)
n= 15 ,p= 0.2
q = 1-p
= 1- 0.2 = 0.8
𝜇 = np
𝜇 = 15x0.2 = 3.0
𝜎 = √𝑛𝑝𝑞
= √15𝑋0.2𝑋0.8 = √2.4 = 1.55
b)
n= 8 ,p= 0.42
q = 1-p
= 1- 0.42 = 0.58
𝜇 = np
𝜇 = 8x0.42 = 3.36
𝜎 = √𝑛𝑝𝑞
= √8𝑋0.42𝑋0.58 = √1.9488 = 1.40
c)
n= 72 ,p= 0.06
q = 1-p
= 1- 0.06= 0.94
𝜇 = np
𝜇 = 72x0.06 = 4.32
𝜎 = √𝑛𝑝𝑞
= √72𝑋0.06𝑋0.94 = √4.0608 = 2.02
H.W:
Do EX 5-20(d, e)
EX:
(EX 5-18)
Find the following probabilities for a binomial distribution with
n=7 and p = 0.2
Solution:
a)
n=7
p= 0.2
q= 1 – p
= 1- 0.2 = 0.8
𝑛!
P(r = 5) =
pr qn-r
𝑟!(𝑛−𝑟)!
7!
=
5!(7−5)!
(0.2)5 (0.8)7-5
= 7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1
(0.0003) (0.64)
(5𝑋4𝑋3𝑋2𝑋1)(2𝑋1)
=
=
b)
42
2
(0.0002)
.0084
2
= 0.0042
n=7
p= 0.2
q= 1 – p
= 1- 0.2 = 0.8
=
𝑛!
𝑟!(𝑛−𝑟)!
pr qn-r
P(r >2) = 1 – P(r ≤2)
= 1 – [P(r=o) + P(r=1) + P(r=2)]
= 1- [
7!
0!(7−0)!
(0.2)0 (0.8)7-0 +
7!
1!(7−1)!
(0.2)1 (0.8)7-1 +
7!
(0.2)2(0.8)7-2]
2!(7−2)!
= 1- [
=1-[
7!
0!7!
(0.2)0 (0.8)7 +
7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1
(1)(7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1)
+
7!
1!6!
(1) (0.2097) +
7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1
(2𝑋1)(5𝑋4𝑋3𝑋2𝑋1)
= 1- [ (0.2097) +
7
1
(0.2)1 (0.8)6 +
(0.2)2(0.8)5]
2!5!
7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1
(1)(6𝑋5𝑋4𝑋3𝑋2𝑋1)
(0.2) (0.2621)
(0.04) (0.3277)]
42
(0.0524) +
= 1- [ 0.2097+ 0.3668 +
7!
0.5502
2
2
(0.0131) ]
]
= 1- [ 0.2097+ 0.3668 + 0.2751]
= 1 – [0.8516] = 0.1484
H.W: Do EX Sc 5-4 (a),pg-247
Poisson distribution:
It is discrete probability distribution. It is used to describe number of events occur
randomly over a specified interval of time. For example number of deaths by
malaria per year and the number of blood cells in a specimen of blood. It has
found wide application in field of Biology, Physics and Operation research etc.The
probability of exactly x occurrences in a Poisson distribution is calculated with
following formula,
P(x) =
Here
λ
𝑒
λ𝑥 ×𝑒 −λ
𝑥!
= the mean number of occurrences per interval of time.
= the base of natural logarithm.
Poisson distribution as an approximation of the binomial distribution:
We use Poisson as approximation of Binomial distribution when n is large(n > 20)
and P is less than or equal to 0.05.In this case we use the following formula,
(𝒏𝒑)𝒙 × 𝒆−𝒏𝒑
P(x) =
𝒙!
λ = np
Here
EX:
Given
λ = 5, for a Poisson distribution, find,
a)
P( x = 0)
b)
P(x=2)
Solution:
P(x) =
λ𝑥 ×𝑒 −λ
𝑥!
𝑒 −λ = 𝑒 −5 = 0.00674
a)
P(x = 0) =
b)
P(x=2)
=
50 x 𝑒 −5
0!
52 x 𝑒−5
2!
=
=
1(0.00674)
1
25(0.00674)
2𝑋1
=
=
0.00674
1
0.1685
2
= 0.00674
= 0.0843
EX:
(EX Sc 5-7)
Solution:
a) P(x) =
λ𝑥 ×𝑒 −λ
𝑥!
𝑒 −λ = 𝑒 −4.2 = 0.01500
P(x ≤2) = P(x=0) + P(x = 1) +P(x = 2)
=
=
=
4.20 x 𝑒 −4.2
0!
1(0.01500)
1
0.01500
1
+
+
+
4.21 x 𝑒 −4.2
1!
4.2(0.01500)
0.063
1
+
1
+
+
EX:
(EX Sc 5-8.b)
Solution:
P(x)
λ
=
λ𝑥 ×𝑒 −λ
𝑥!
= np
= 30x 0.04 = 1.2
2!
17.64(0.01500)
0.2646
2
= 0.01500 + 0.063 + 0.1323
= 0.2103
H.W: Do EX Sc 5-7(c)
4.22 x 𝑒 −4.2
2𝑋1
𝑒 −1.2
= 0.30119
1.23 x 𝑒 −1.2
P(x=3) =
3!
=
(1.728)(0.30119)
3𝑋2𝑋1
=
0.5205
6
= 0.0867
c)
Solution:
P(x) =
λ
λ𝑥 ×𝑒 −λ
𝑥!
= np
= 30x 0.04 = 1.2
= 0.30119
P(x=5) =
1.25 x 𝑒 −1.2
5!
=
(2.4883)(0.30119)
5𝑋4𝑋3𝑋2𝑋1
=
0.7495
120
= 0.0062
Normal distribution:
It is a type of continuous probability distribution. Normal distribution is also called
“MOTHER OF OTHER DISTRIBUTIONS” because many other distributions
are generated from this distribution. This is very important type of continuous
probability distribution.
There are two main reasons why normal distribution has such prominent place in
Statistics.
1-Normal distribution makes the bases for Inferential Statistics (a branch of
Statistics in which we draw conclusions about the population on the basis of
information gained from the sample).
2-Normal distribution comes close to fitting the actual observed frequency
distributions of many phenomena (weight, height etc), outputs from physical
process and other measures of interest to manager in both public and private
sectors.
1-Its curve has single peak, so it is called UNIMODEL.
2-The mean of normally distributed population lies at the centre of its curve.
3-As normal distribution is symmetrical, so mean, median and mode are lie at
the centre of the curve.
4-The two tails of normal distribution extend indefinitely and never touch the
horizontal axis.
5-It has two parameters, mean ( 𝜇 ) and standard deviation ( 𝜎 ).
6-Regardless of the values of 𝜇 and 𝜎 , the total area under normal curve
is 1.
The normal distribution with 𝜇 = 0 and 𝜎 = 1 is called standard normal
probability distribution.
For different values of 𝜇 and 𝜎 we have different normal distributions.
It is not possible to have different tables for every possible normal curve, so we
can use table of standard normal probability distribution to find area or
probability under any normal curve. With the help of this table we can measure
the area or probability that 𝑥 will lie with certain distances from 𝜇 . These
distances are defined in terms of standard deviations (as standard unit) i.e.
standard deviation as unit of measure.
Here is formula for measuring distances 𝑥 from 𝜇 , under normal curve.
Z =
𝒙− 𝝁
𝝈
EX:
Given a normal distribution with
𝜇 = 40 and 𝜎 = 6, find
(a) The area below 32.
(b) The area between 42 and 51.
SOLUTION:
(a)
𝜇 = 40
𝜎=6
Z=
𝑥−𝜇
𝜎
For 𝑥 = 32
Z=
32−40
6
=
−8
6
= -1.33
P (𝑥 < 32) = P (Z < -1.33)
= P (−∞ ≤ Z ≤ 0) - P (-1.33 ≤ Z ≤ 0)
= 0.5 – O.4082
= 0.0918
(b)
𝜇 = 40
𝜎 =6
Z=
𝑥− 𝜇
𝜎
For 𝑥 = 42
Z=
42−40
6
=
2
6
= 0.33
For 𝑥 = 51
Z=
51−40
6
=
11
6
= 1.83
P( 42 ≤ 𝑥 ≤ 51) = P ( 0.33≤ Z ≤ 1.83)
= P (0≤ Z ≤ 1.83) - P (0 ≤ Z ≤ 0.33)
= 0.4664 – O.1293
= 0.3371
EX:
The time taken by person to deliver newspapers in some area is normally
distributed with mean of 12 and standard deviation of 2minutes.
He delivers newspapers every day.
(a) Calculate the number of days during year when he takes longer than
17 minutes.
(b) Calculate the number of days during year when he takes time between
8 and 16 minutes.
Solution:
(a)
𝜇 = 12
𝜎=2
Z=
𝑥−𝜇
𝜎
For 𝑥 = 17
Z=
17−12
2
=
5
2
= 2.5
P (𝑥 > 17) = P (Z > 2.5)
= P (0 ≤ Z ≤ ∞) - P (0 ≤ Z ≤ 2.5)
= 0.5 – O.4938
= 0.0062
Required days = .0062 x 365 = 2
𝜇 = 12
𝜎 =2
(b)
Z=
𝑥− 𝜇
𝜎
For 𝑥 = 8
Z=
8−12
2
=
−4
2
= -2
For 𝑥 = 16
Z=
16−12
2
=
4
2
= 2
P ( 8 ≤ 𝑥 ≤ 16) = P ( -2≤ Z ≤ 2)
= P (-2≤ Z ≤ 0) + P (0 ≤ Z ≤ 2)
= 0.4772 +O.4772
= 0.9544
Required days = .9544 x 365 = 348
EX:
(5-46 pg 271)
(b)What is the probability of finishing the program in fewer than 30 days?
Solution:
(a)
𝜇 = 44
𝜎 = 12
Z=
𝑥−𝜇
𝜎
For 𝑥 = 30
Z=
30−44
12
=
−14
12
= -1.17
P (𝑥 < 30) = P (Z < -1.17)
= P (−∞ ≤ Z ≤ 0) - P (−1.17 ≤ Z ≤ 0)
= 0.5 – O.3790
= 0.1210
H.W:
(a) EX 5-46 (a), pg-271
(b) EX 5-48(a), pg-271
(c) In an intelligence test administered on 100 children. The average I.Q was 42
with standard deviation 24.Find the number of children lying between them
scores of 30 and 54.
Use of expected value in decision making:
EX:(Sc 5-3 pg 235)
How many “every thing but “pizzas should Mario stock each night in order to
minimize expected loss if the number of pizzas ordered has the following
probability distribution?
Number of pizzas demanded
Probability
1
0.40
2
0.30
3
0.20
4
0.10
Solution:
Cost price = $7
Selling price = $12
When supply > demand:
Loss = cost price = $7
When supply < demand:
Loss = selling price – cost price = 12 – 7 = $5
Total loss = 5+5 = $10
(Supply)
Demand
probability
1
2
0.40
1
0
0x0.40
=0
1x7
=7
0.30
2
1x10
=10
10x0.30
=3
0.20
3
2x10
=20
0.10
4
3x10
=30
Expected
loss
3
4
7 x 0.40
= 2.8
2x7
=14
14x0.40
=5.6
3x7
=21
21x.40
= 8.4
0
0X0.30
=0
1x7
=7
7x0.30
=2.1
2x7
= 14
14x.30
= 4.2
20x0.2
=4
1x10
=10
10X0.20
=2
0
0x0.20
=0
1x7
=7
7x.20
= 1.4
30x0.1
=3
2x10
=20
20x0.1
=2
1x10
=10
10x0.1
=1
0
0x.10
=0
10
6.8
Mario should stock two “every thing but” pizzas at night
8.7
14
OBJECTIVE
SECTION
Q-1 WRITE SHORT ANSWERS FOR THE FOLLOWING.
1- Define random variable.
Answer:
A variable that takes on different values as a result of the outcomes
of a random experiment.
2-Define expected Value.
Answer:
It is the weighted average of the outcomes of the random experiment.
3-Define Bernoulli process.
Answer:
A process in which each trial has only two outcomes Probability of all
outcomes remains fixed.
4-Write down the formula for mean and variance of Binomial distribution
Answer:
𝜇 = 𝑛𝑝.
𝜎 2 = npq
5-Write any three properties of normal distribution.
Answer:
i)
Its curve has single peak.
ii)
It is symmetrical distribution.
iii)
Total area under normal curve is 1.
Q-2 Choose the correct one.
1. There are _________types of distribution.
a. Seven.
b. Three.
c. Two.
2. Continuous probability distribution take i=on any value with-in
a given ________ .
a. Range.
b. List.
c. Both of the above.
3. Standard deviation of binomial distribution is ____________.
a. np.
b. npq.
c. √𝑛𝑝𝑞.
4. In binomial distribution q = ____________.
a. 1- p.
b. 1 + P.
c. 2p.
Q-3 Write true or false for the following.
1- 𝜇 = 𝑛𝑝
(
2. We can’t list continuous random variable.
(
)
3. There are four types of probability distributions. (
)
4. A random variable can takes on different values. (
)
)