Class PPT About Quantitative Business Analysis

Quantitative for Business Analysis
(Semester 3 - Entrepreneurship)
M Miqdad Robbani, M.Sc.

Chapter 2
To accompany
Quantitative Analysis for Management, Eleventh
Edition, by Render, Stair, and Hanna
Power Point slides created by Brian Peterson
Probability Concepts and
Applications

Copyright ©2012 Pearson Education, Inc. publishing as
Prentice Hall
2-3
Learning Objectives
1. Understand the basic foundations of probability
analysis.
2. Describe statistically dependent and independent
events.
3. Use Bayes’ theorem to establish posterior probabilities.
4. Describe and provide examples of both discrete and
continuous random variables.
5. Explain the difference between discrete and continuous
probability distributions.
6. Calculate expected values and variances and use the
normal table.
After completing this chapter, students will be able to:

Prentice Hall
2-4
Chapter Outline
2.1 Introduction
2.2 Fundamental Concepts
2.3 Mutually Exclusive and Collectively
Exhaustive Events
2.4 Statistically Independent Events
2.5 Statistically Dependent Events
2.6 Revising Probabilities with Bayes’
Theorem
2.7 Further Probability Revisions

Prentice Hall
2-5
Chapter Outline
2.8 Random Variables
2.9 Probability Distributions
2.10 The Binomial Distribution
2.11 The Normal Distribution
2.12 The F Distribution
2.13 The Exponential Distribution
2.14 The Poisson Distribution

Prentice Hall
2-6
Introduction
• Life is uncertain; we are not sure what the
future will bring.
• Probability is a numerical statement about the
likelihood that an event will occur.

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2-7
Fundamental Concepts
1. The probability, P, of any event or state of
nature occurring is greater than or equal to 0
and less than or equal to 1. That is:
0  P (event)  1
2. The sum of the simple probabilities for all
possible outcomes of an activity must equal 1.

Prentice Hall
2-8
Chapters in This Book
That Use Probability
CHAPTER TITLE
3 Decision Analysis
4 Regression Models
5 Forecasting
6 Inventory Control Models
12 Project Management
13 Waiting Lines and Queuing Theory Models
14 Simulation Modeling
15 Markov Analysis
16 Statistical Quality Control
Module 3 Decision Theory and the Normal Distribution
Module 4 Game Theory
Table 2.1

Prentice Hall
2-9
Diversey Paint Example
• Demand for white latex paint at Diversey Paint and Supply
has always been either 0, 1, 2, 3, or 4 gallons per day.
• Over the past 200 days, the owner has observed the
following frequencies of demand:
QUANTITY
DEMANDED NUMBER OF DAYS PROBABILITY
0 40 0.20 (= 40/200)
1 80 0.40 (= 80/200)
2 50 0.25 (= 50/200)
3 20 0.10 (= 20/200)
4 10 0.05 (= 10/200)
Total 200 Total 1.00 (= 200/200)

Prentice Hall
2-10
Diversey Paint Example
• Demand for white latex paint at Diversey Paint and Supply
has always been either 0, 1, 2, 3, or 4 gallons per day
• Over the past 200 days, the owner has observed the
following frequencies of demand
QUANTITY
DEMANDED NUMBER OF DAYS PROBABILITY
0 40 0.20 (= 40/200)
1 80 0.40 (= 80/200)
2 50 0.25 (= 50/200)
3 20 0.10 (= 20/200)
4 10 0.05 (= 10/200)
Total 200 Total 1.00 (= 200/200)
Notice the individual
probabilities are all between 0
and 1
0 ≤ P (event) ≤ 1
And the total of all event
probabilities equals 1
∑ P (event) = 1.00

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2-11
Determining objective probability :
• Relative frequency
• Typically based on historical data
Types of Probability
P (event) =
Number of occurrences of the event
Total number of trials or outcomes
 Classical or logical method
 Logically determine probabilities without
trials
P (head) =
1
2
Number of ways of getting a head
Number of possible outcomes (head or tail)

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2-12
Types of Probability
Subjective probability is based on the
experience and judgment of the person making
the estimate.
• Opinion polls
• Judgment of experts
• Delphi method

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2-13
Mutually Exclusive Events
Events are said to be mutually exclusive if only one
of the events can occur on any one trial.
 Tossing a coin will result
in either a head or a tail.
 Rolling a die will result in
only one of six possible
outcomes.

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2-14
Collectively Exhaustive Events
Events are said to be collectively exhaustive if the list
of outcomes includes every possible outcome.
• Both heads and
tails as possible
outcomes of
coin flips.
• All six possible
outcomes
of the roll
of a die.
OUTCOME
OF ROLL
PROBABILITY
1 1
/6
2 1
/6
3 1
/6
4 1
/6
5 1
/6
6 1
/6
Total 1

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2-15
Drawing a Card
Draw one card from a deck of 52 playing cards
P (drawing a 7) = 4
/52 = 1
/13
P (drawing a heart) = 13
/52 = 1
/4
 These two events are not mutually exclusive
since a 7 of hearts can be drawn
 These two events are not collectively
exhaustive since there are other cards in the
deck besides 7s and hearts

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2-16
Table of Differences
DRAWS MUTUALLY
EXCLUSIVE
COLLECTIVELY
EXHAUSTIVE
1. Draws a spade and a club Yes No
2. Draw a face card and a
number card
Yes Yes
3. Draw an ace and a 3 Yes No
4. Draw a club and a nonclub Yes Yes
5. Draw a 5 and a diamond No No
6. Draw a red card and a
diamond
No No

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2-17
Adding Mutually Exclusive Events
We often want to know whether one or a
second event will occur.
 When two events are mutually
exclusive, the law of addition is:
P (event A or event B) = P (event A) + P (event B)
P (spade or club) = P (spade) + P (club)
= 13
/52 + 13
/52
= 26
/52 = 1
/2 = 0.50

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2-18
Adding Not Mutually Exclusive Events
P (event A or event B) = P (event A) + P (event B)
– P (event A and event B both occurring)
P (A or B) = P (A) + P (B) – P (A and B)
P(five or diamond) = P(five) + P(diamond) – P(five and diamond)
= 4
/52 + 13
/52 – 1
/52
= 16
/52 = 4
/13
The equation must be modified to account
for double counting.
 The probability is reduced by
subtracting the chance of both events
occurring together.

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2-19
Venn Diagrams
P (A) P (B)
Events that are mutually
exclusive.
P (A or B) = P (A) + P (B)
Figure 2.1
Events that are not
mutually exclusive.
P (A or B) = P (A) + P (B)
– P (A and B)
Figure 2.2
P (A) P (B)
P (A and B)

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2-20
Statistically Independent Events
Events may be either independent or dependent.
• For independent events, the occurrence of one event
has no effect on the probability of occurrence of the
second event.

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2-21
Which Sets of Events Are Independent?
1. (a) Your education
(b) Your income level
2. (a) Draw a jack of hearts from a full 52-card
deck
(b) Draw a jack of clubs from a full 52-card
deck
3. (a) Chicago Cubs win the National League pennant
(b) Chicago Cubs win the World Series
4. (a) Snow in Santiago, Chile
(b) Rain in Tel Aviv, Israel
Dependent events
Dependent
events
Independent
events
Independent events

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2-22
Three Types of Probabilities
• Marginal (or simple) probability is just the probability of a
single event occurring.
P (A)
 Joint probability is the probability of two or more events occurring and is
equal to the product of their marginal probabilities for independent
events.
P (AB) = P (A) x P (B)
 Conditional probability is the probability of event B given that event A
has occurred.
P (B | A) = P (B)
 Or the probability of event A given that event B has occurred
P (A | B) = P (A)

Prentice Hall
2-23
Joint Probability Example
The probability of tossing a 6 on the first
roll of the die and a 2 on the second roll:
P (6 on first and 2 on second)
= P (tossing a 6) x P (tossing a 2)
= 1
/6 x 1
/6 = 1
/36 = 0.028

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2-24
Independent Events
1. The probability of a black ball drawn on first draw is:
P (B) = 0.30 (a marginal probability)
2. The probability of two green balls drawn is:
P (GG) = P (G) x P (G) = 0.7 x 0.7 = 0.49
(a joint probability for two independent events)
A bucket contains 3 black balls and 7 green balls.
 Draw a ball from the bucket, replace it, and
draw a second ball.

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2-25
Independent Events
3. The probability of a black ball drawn on the second draw
if the first draw is green is:
P (B | G) = P (B) = 0.30
(a conditional probability but equal to the marginal
because the two draws are independent events)
4. The probability of a green ball drawn on the second draw
if the first draw is green is:
P (G | G) = P (G) = 0.70
(a conditional probability as in event 3)
A bucket contains 3 black balls and 7 green balls.
 Draw a ball from the bucket, replace it, and
draw a second ball.

Prentice Hall
2-26
Statistically Dependent Events
The marginal probability of an event occurring is computed
in the same way:
P (A)
The formula for the joint probability of two events is:
P (AB) = P (B | A) P (A)
P (A | B) =
P (AB)
P (B)
Calculating conditional probabilities is slightly more complicated. The
probability of event A given that event B has occurred is:

Prentice Hall
2-27
When Events Are Dependent
Assume that we have an urn containing 10 balls of
the following descriptions:
 4 are white (W) and lettered (L)
 2 are white (W) and numbered (N)
 3 are yellow (Y) and lettered (L)
 1 is yellow (Y) and numbered (N)
P (WL) = 4
/10 = 0.4 P (YL) = 3
/10 = 0.3
P (WN) = 2
/10 = 0.2 P (YN) = 1
/10 = 0.1
P (W) = 6
/10 = 0.6 P (L) = 7
/10 = 0.7
P (Y) = 4
/10 = 0.4P (N) = 3
/10 = 0.3

Copyright ©2012 Pearson Education, Inc.
publishing as Prentice Hall
2-28
4 balls
White (W)
and
Lettered (L)
2 balls
White (W)
and
Numbered (N)
3 balls
Yellow (Y)
and
Lettered (L)
1 ball Yellow (Y)
and Numbered (N)
Probability (WL) =
4
10
Probability (YN) =
1
10
Probability (YL) =
3
10
Probability (WN) =
2
10
The urn
contains 10
balls:
Figure 2.3

Prentice Hall
2-29
The conditional probability that the ball drawn
is lettered, given that it is yellow, is:
P (L | Y) = = = 0.75
P (YL)
P (Y)
0.3
0.4
We can verify P (YL) using the joint probability formula
P (YL) = P (L | Y) x P (Y) = (0.75)(0.4) = 0.3

Prentice Hall
2-30
Joint Probabilities
for Dependent Events
P (MT) = P (T | M) x P (M) = (0.70)(0.40) = 0.28
If the stock market reaches 12,500 point by January,
there is a 70% probability that Tubeless Electronics
will go up.
 You believe that there is only a 40% chance the
stock market will reach 12,500.
 Let M represent the event of the stock market
reaching 12,500 and let T be the event that
Tubeless goes up in value.

Prentice Hall
2-31
Posterior
Probabilities
Bayes’
Process
Revising Probabilities with
Bayes’ Theorem
Bayes’ theorem is used to incorporate additional information and help
create posterior probabilities.
Prior
Probabilities
New
Information
Figure 2.4

Prentice Hall
2-32
Posterior Probabilities
A cup contains two dice identical in appearance but one is fair
(unbiased), the other is loaded (biased).
• The probability of rolling a 3 on the fair die is 1
/6 or 0.166.
• The probability of tossing the same number on the loaded die is
0.60.
• We select one by chance,
toss it, and get a 3.
• What is the probability that
the die rolled was fair?
• What is the probability that
the loaded die was rolled?

Copyright ©2012 Pearson Education, Inc.
publishing as Prentice Hall
2-33
We know the probability of the die being fair or loaded
is:
P (fair) = 0.50 P (loaded) = 0.50
And that
P (3 | fair) = 0.166 P (3 | loaded) = 0.60
We compute the probabilities of P (3 and fair)
and P (3 and loaded):
P (3 and fair) = P (3 | fair) x P (fair)
= (0.166)(0.50) = 0.083
P (3 and loaded) = P (3 | loaded) x P (loaded)
= (0.60)(0.50) = 0.300

Prentice Hall
2-34
We know the probability of the die being fair or loaded
is
P (fair) = 0.50 P (loaded) = 0.50
And that
P (3 | fair) = 0.166 P (3 | loaded) = 0.60
We compute the probabilities of P (3 and fair)
and P (3 and loaded)
P (3 and fair) = P (3 | fair) x P (fair)
= (0.166)(0.50) = 0.083
P (3 and loaded) = P (3 | loaded) x P (loaded)
= (0.60)(0.50) = 0.300
The sum of these probabilities
gives us the unconditional
probability of tossing a 3:
P (3) = 0.083 + 0.300 = 0.383

Prentice Hall
2-35
P (loaded | 3) = = = 0.78
P (loaded and 3)
P (3)
0.300
0.383
The probability that the die was loaded is:
P (fair | 3) = = = 0.22
P (fair and 3)
P (3)
0.083
0.383
If a 3 does occur, the probability that the die rolled
was the fair one is:
 These are the revised or posterior probabilities for the
next roll of the die.
 We use these to revise our prior probability estimates.

Prentice Hall
2-36
Bayes’ Calculations
Given event B has occurred:
STATE OF
NATURE
P (B | STATE
OF NATURE)
PRIOR
PROBABILITY
JOINT
PROBABILITY
POSTERIOR
PROBABILITY
A P(B | A) x P(A) = P(B and A) P(B and A)/P(B) = P(A|B)
A’ P(B | A’) x P(A’) = P(B and A’) P(B and A’)/P(B) = P(A’|B)
P(B)
Table 2.2
Given a 3 was rolled:
STATE OF
NATURE
P (B | STATE
OF NATURE)
PRIOR
PROBABILITY
JOINT
PROBABILITY
POSTERIOR
PROBABILITY
Fair die 0.166 x 0.5 = 0.083 0.083 / 0.383 = 0.22
Loaded die 0.600 x 0.5 = 0.300 0.300 / 0.383 = 0.78
P(3) = 0.383
Table 2.3

Prentice Hall
2-37
General Form of Bayes’ Theorem
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
|
(
A
P
A
B
P
A
P
A
B
P
A
P
A
B
P
B
A
P




We can compute revised probabilities more
directly by using:
where
the complement of the event ;
for example, if is the event “fair die”,
then is “loaded die”.
A
A
A
=
¢
A

Prentice Hall
2-38
General Form of Bayes’ Theorem
This is basically what we did in the previous example:
Replace with “fair die”
Replace with “loaded die
Replace with “3 rolled”
We get
A
A
B
)
|
( rolled
3
die
fair
P
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
loaded
loaded
3
fair
fair
3
fair
fair
3
P
P
P
P
P
P


22
0
383
0
083
0
50
0
60
0
50
0
166
0
50
0
166
0
.
.
.
)
.
)(
.
(
)
.
)(
.
(
)
.
)(
.
(




Prentice Hall
2-39
Further Probability Revisions
We can obtain additional information by performing
the experiment a second time
 If you can afford it, perform experiments
several times.
We roll the die again and again get a 3.
50
0
loaded
and
50
0
fair .
)
(
.
)
( 
 P
P
36
0
6
0
6
0
loaded
3
3
027
0
166
0
166
0
fair
3
3
.
)
.
)(
.
(
)
|
,
(
.
)
.
)(
.
(
)
|
,
(




P
P

Prentice Hall
2-40
several times
We roll the die again and again get a 3
50
0
loaded
and
50
0
fair .
)
(
.
)
( 
 P
P
36
0
6
0
6
0
loaded
3
3
027
0
166
0
166
0
fair
3
3
.
)
.
)(
.
(
)
|
,
(
.
)
.
)(
.
(
)
|
,
(




P
P
)
(
)
|
,
(
)
( fair
fair
3
3
fair
and
3,3 P
P
P 

013
0
5
0
027
0 .
)
.
)(
.
( 

)
(
)
|
,
(
)
( loaded
loaded
3
3
loaded
and
3,3 P
P
P 

18
0
5
0
36
0 .
)
.
)(
.
( 


Prentice Hall
2-41
several times
We roll the die again and again get a 3
50
.
0
)
loaded
(
and
50
.
0
)
fair
( 
 P
P
36
.
0
)
6
.
0
)(
6
.
0
(
)
loaded
|
3
,
3
(
027
.
0
)
166
.
0
)(
166
.
0
(
)
fair
|
3
,
3
(




P
P
)
(
)
|
,
(
)
( fair
fair
3
3
fair
and
3,3 P
P
P 

013
0
5
0
027
0 .
)
.
)(
.
( 

)
(
)
|
,
(
)
( loaded
loaded
3
3
loaded
and
3,3 P
P
P 

18
0
5
0
36
0 .
)
.
)(
.
( 

067
0
193
0
013
0
3
3
fair
and
3,3
3
3
fair .
.
.
)
,
(
)
(
)
,
|
( 


P
P
P
933
0
193
0
18
0
3
3
loaded
and
3,3
3
3
loaded .
.
.
)
,
(
)
(
)
,
|
( 


P
P
P

Prentice Hall
2-42
After the first roll of the die:
probability the die is fair = 0.22
probability the die is loaded = 0.78
After the second roll of the die:
probability the die is fair = 0.067
probability the die is loaded = 0.933

Prentice Hall
2-43
Random Variables
Discrete random variables can assume only a finite
or limited set of values.
Continuous random variables can assume any one
of an infinite set of values.
A random variable assigns a real number to
every possible outcome or event in an
experiment.
X = number of refrigerators sold during the day

Prentice Hall
2-44
Random Variables – Numbers
EXPERIMENT OUTCOME
RANDOM
VARIABLES
RANGE OF
RANDOM
VARIABLES
Stock 50
Christmas trees
Number of Christmas
trees sold
X 0, 1, 2,…, 50
Inspect 600
items
Number of acceptable
items
Y 0, 1, 2,…, 600
Send out 5,000
sales letters
Number of people
responding to the
letters
Z 0, 1, 2,…, 5,000
Build an
apartment
building
Percent of building
completed after 4
months
R 0 ≤ R ≤ 100
Test the lifetime
of a lightbulb
(minutes)
Length of time the
bulb lasts up to 80,000
minutes
S 0 ≤ S ≤ 80,000
Table 2.4

Prentice Hall
2-45
Random Variables – Not Numbers
EXPERIMENT OUTCOME RANDOM VARIABLES
RANGE OF
RANDOM
VARIABLES
Students
respond to a
questionnaire
Strongly agree (SA)
Agree (A)
Neutral (N)
Disagree (D)
Strongly disagree (SD)
5 if SA
4 if A..
X = 3 if N..
2 if D..
1 if SD
1, 2, 3, 4, 5
One machine
is inspected
Defective
Not defective
Y = 0 if defective
1 if not defective
0, 1
Consumers
respond to
how they like a
product
Good
Average
Poor
3 if good….
Z = 2 if average
1 if poor…..
1, 2, 3
Table 2.5

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2-46
Probability Distribution of a
Discrete Random Variable
The students in Pat Shannon’s statistics
class have just completed a quiz of five
algebra problems. The distribution of
correct scores is given in the following
table:
For discrete random variables a probability
is assigned to each event.

Prentice Hall
2-47
Probability Distribution of a
Discrete Random Variable
RANDOM VARIABLE
(X – Score)
NUMBER
RESPONDING
PROBABILITY
P (X)
5 10 0.1 = 10/100
4 20 0.2 = 20/100
3 30 0.3 = 30/100
2 30 0.3 = 30/100
1 10 0.1 = 10/100
Total 100 1.0 = 100/100
The Probability Distribution follows all three rules:
1. Events are mutually exclusive and collectively exhaustive.
2. Individual probability values are between 0 and 1.
3. Total of all probability values equals 1.
Table 2.6

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2-48
Probability Distribution for
Dr. Shannon’s Class
P
(X)
0.4 –
0.3 –
0.2 –
0.1 –
0 –
| | | | | |
1 2 3 4 5
X
Figure 2.5

Prentice Hall
2-49
Probability Distribution for
Dr. Shannon’s Class
P
(X)
0.4 –
0.3 –
0.2 –
0.1 –
0 –
| | | | | |
1 2 3 4 5
X
Figure 2.5
The central tendency of
the distribution is the
mean, or expected value.
The amount of variability
is the variance.

Prentice Hall
2-50
Expected Value of a Discrete Probability Distribution
   



n
i
i
i X
P
X
X
E
1
  )
(
...
)
( 2
2
1
1 n
n X
P
X
X
P
X
X
P
X 



The expected value is a measure of the central tendency of the
distribution and is a weighted average of the values of the random
variable.
where
i
X
)
( i
X
P


n
i 1
)
(X
E
= random variable’s possible values
= probability of each of the random variable’s
possible values
= summation sign indicating we are adding all n
possible values
= expected value or mean of the random sample

Prentice Hall
2-51
Expected Value of a Discrete Probability Distribution
   



n
i
i
i X
P
X
X
E
1
9
.
2
1
.
6
.
9
.
8
.
5
.
)
1
.
0
(
1
)
3
.
0
(
2
)
3
.
0
(
3
)
2
.
0
(
4
)
1
.
0
(
5











For Dr. Shannon’s class:

Prentice Hall
2-52
Variance of a
Discrete Probability Distribution
For a discrete probability distribution the
variance can be computed by
)
(
)]
(
[





n
i
i
i X
P
X
E
X
1
2
2
Variance
σ
where
i
X
)
(X
E
)
( i
X
P
= random variable’s possible values
= expected value of the random variable
= difference between each value of the random
variable and the expected mean
= probability of each possible value of the
random variable
)]
(
[ X
E
Xi 

Prentice Hall
2-53
Variance of a
)
(
)]
(
[
variance
5
1
2




i
i
i X
P
X
E
X




 )
.
(
)
.
(
)
.
(
)
.
(
variance 2
0
9
2
4
1
0
9
2
5 2
2



 )
.
(
)
.
(
)
.
(
)
.
( 3
0
9
2
2
3
0
9
2
3 2
2
)
.
(
)
.
( 1
0
9
2
1 2

29
1
361
0
243
0
003
0
242
0
441
0
.
.
.
.
.
.







Prentice Hall
2-54
Variance of a
A related measure of dispersion is the
standard deviation.
2
σ
Variance
σ 

where
σ
= square root
= standard deviation

Prentice Hall
2-55
Variance of a
A related measure of dispersion is the
standard deviation.
2
σ
Variance
σ 

where
σ
= square root
= standard deviation
Variance
σ 
14
1
29
1 .
. 


Prentice Hall
2-56
Using Excel
Program 2.1A
Formulas in an Excel Spreadsheet for the Dr.
Shannon Example

Prentice Hall
2-57
Using Excel
Program 2.1B
Excel Output for the Dr. Shannon Example

Prentice Hall
2-58
Probability Distribution of a Continuous
Random Variable
Since random variables can take on an infinite
number of values, the fundamental rules for
continuous random variables must be modified.
 The sum of the probability values must still
equal 1.
 The probability of each individual value of the
random variable occurring must equal 0 or
the sum would be infinitely large.
The probability distribution is defined by a
continuous mathematical function called the
probability density function or just the probability
function.
 This is represented by f (X).

Prentice Hall
2-59
Probability Distribution of a Continuous
Random Variable
Probability
| | | | | | |
5.06 5.10 5.14 5.18 5.22 5.26 5.30
Weight (grams)
Figure 2.6

Prentice Hall
2-60
The Binomial Distribution
• Many business experiments can be characterized
by the Bernoulli process.
• The Bernoulli process is described by the
binomial probability distribution.
1. Each trial has only two possible outcomes.
2. The probability of each outcome stays the same from
one trial to the next.
3. The trials are statistically independent.
4. The number of trials is a positive integer.

Prentice Hall
2-61
The binomial distribution is used to find the probability
of a specific number of successes in n trials.
We need to know:
n = number of trials
p = the probability of success on any
single trial
We let
r = number of successes
q = 1 – p = the probability of a failure

Prentice Hall
2-62
The binomial formula is:
r
n
r
q
p
r
n
r
n
n
r 


)!
(
!
!
trials
in
successes
of
y
Probabilit
The symbol ! means factorial, and
n! = n(n – 1)(n – 2)…(1)
For example
4! = (4)(3)(2)(1) = 24
By definition
1! = 1 and 0! = 1

Prentice Hall
2-63
NUMBER OF
HEADS (r) Probability = (0.5)r
(0.5)5 – r
5!
r!(5 – r)!
0 0.03125 = (0.5)0
(0.5)5 – 0
1 0.15625 = (0.5)1
(0.5)5 – 1
2 0.31250 = (0.5)2
(0.5)5 – 2
3 0.31250 = (0.5)3
(0.5)5 – 3
4 0.15625 = (0.5)4
(0.5)5 – 4
5 0.03125 = (0.5)5
(0.5)5 – 5
5!
0!(5 – 0)!
5!
1!(5 – 1)!
5!
2!(5 – 2)!
5!
3!(5 – 3)!
5!
4!(5 – 4)!
5!
5!(5 – 5)!
Table 2.7
Binomial Distribution for n = 5 and p = 0.50.

Prentice Hall
2-64
Solving Problems with the Binomial Formula
We want to find the probability of 4 heads in 5 tosses.
n = 5, r = 4, p = 0.5, and q = 1 – 0.5 = 0.5
Thus
4
5
4
5
.
0
5
.
0
)!
4
5
(
!
4
!
5
)
trials
5
in
successes
4
( 



P
15625
0
5
0
0625
0
1
1
2
3
4
1
2
3
4
5
.
)
.
)(
.
(
)
!
)(
)(
)(
(
)
)(
)(
)(
(



Prentice Hall
2-65
Solving Problems with the Binomial Formula
Probability
P
(r)
| | | | | | |
1 2 3 4 5 6
Values of r (number of successes)
0.4 –
0.3 –
0.2 –
0.1 –
0 –
Figure 2.7
Binomial Probability Distribution for n = 5 and p = 0.50.

Prentice Hall
2-66
Solving Problems with Binomial Tables
MSA Electronics is experimenting with the
manufacture of a new transistor.
 Every hour a random sample of 5 transistors is
taken.
 The probability of one transistor being defective
is 0.15.
What is the probability of finding 3, 4, or 5 defective?
n = 5, p = 0.15, and r = 3, 4, or 5
So
We could use the formula to solve this problem, but using the table is
easier.

Prentice Hall
2-67
P
n r 0.05 0.10 0.15
5 0 0.7738 0.5905 0.4437
1 0.2036 0.3281 0.3915
2 0.0214 0.0729 0.1382
3 0.0011 0.0081 0.0244
4 0.0000 0.0005 0.0022
5 0.0000 0.0000 0.0001
Table 2.8 (partial)
We find the three probabilities in the table
for n = 5, p = 0.15, and r = 3, 4, and 5 and
add them together.

Prentice Hall
2-68
Table 2.8 (partial)
We find the three probabilities in the table
for n = 5, p = 0.15, and r = 3, 4, and 5 and
add them together
P
n r 0.05 0.10 0.15
5 0 0.7738 0.5905 0.4437
1 0.2036 0.3281 0.3915
2 0.0214 0.0729 0.1382
3 0.0011 0.0081 0.0244
4 0.0000 0.0005 0.0022
5 0.0000 0.0000 0.0001
)
(
)
(
)
(
)
( 5
4
3
defects
more
or
3 P
P
P
P 


0267
0
0001
0
0022
0
0244
0 .
.
.
. 




Prentice Hall
2-69
It is easy to find the expected value (or mean) and
variance of a binomial distribution.
Expected value (mean) = np
Variance = np(1 – p)
For the MSA example:
6375
0
85
0
15
0
5
1
Variance
75
0
15
0
5
value
Expected
.
)
.
)(
.
(
)
(
.
)
.
(







p
np
np

Prentice Hall
2-70
Using Excel
Function in an Excel 2010 Spreadsheet for Binomial
Probabilities
Program 2.2A

Prentice Hall
2-71
Using Excel
Excel Output for the Binomial Example
Program 2.2B

Prentice Hall
2-72
The Normal Distribution
The normal distribution is the one of the most
popular and useful continuous probability
distributions.
• The formula for the probability density function is
rather complex:
2
2
2
2
1 



)
(
)
(



x
e
X
f
 The normal distribution is specified
completely when we know the mean, µ,
and the standard deviation,  .

Prentice Hall
2-73
• The normal distribution is symmetrical, with the
midpoint representing the mean.
• Shifting the mean does not change the shape of
the distribution.
• Values on the X axis are measured in the number
of standard deviations away from the mean.
• As the standard deviation becomes larger, the
curve flattens.
• As the standard deviation becomes smaller, the
curve becomes steeper.

Prentice Hall
2-75
µ
Figure 2.9
Same µ, smaller 
Same µ, larger 

Prentice Hall
2-76
Using the Standard Normal Table
Step 1
Convert the normal distribution into a standard
normal distribution.
 A standard normal distribution has a mean of 0
and a standard deviation of 1
 The new standard random variable is Z




X
Z
where
X = value of the random variable we want to measure
µ = mean of the distribution
 = standard deviation of the distribution
Z = number of standard deviations from X to the mean, µ

Prentice Hall
2-77
For example, µ = 100,  = 15, and we want to
find the probability that X is less than 130.
15
100
130 





X
Z
dev
std
2
15
30


| | | | | | |
55 70 85 100 115 130 145
| | | | | | |
–3 –2 –1 0 1 2 3
X = IQ




X
Z
µ = 100
 = 15
P(X < 130)
Figure 2.10

Prentice Hall
2-78
Step 2
Look up the probability from a table of normal curve
areas.
 Use Appendix A or Table 2.9 (portion below).
 The column on the left has Z values.
 The row at the top has second decimal places
for the Z values.
AREA UNDER THE NORMAL CURVE
Z 0.00 0.01 0.02 0.03
1.8 0.96407 0.96485 0.96562 0.96638
1.9 0.97128 0.97193 0.97257 0.97320
2.0 0.97725 0.97784 0.97831 0.97882
2.1 0.98214 0.98257 0.98300 0.98341
2.2 0.98610 0.98645 0.98679 0.98713
Table 2.9 (partial)
P(X < 130)
= P(Z < 2.00)
= 0.97725

Prentice Hall
2-79
Haynes Construction Company
Haynes builds three- and four-unit apartment
buildings (called triplexes and quadraplexes,
respectively).
 Total construction time follows a normal
distribution.
 For triplexes, µ = 100 days and  = 20
days.
 Contract calls for completion in 125 days,
and late completion will incur a severe
penalty fee.
 What is the probability of completing in 125
days?

Prentice Hall
2-80
From Appendix A, for Z = 1.25
the area is 0.89435.
 The probability is about
0.89 that Haynes will not
violate the contract.
20
100
125 





X
Z
25
1
20
25
.


µ = 100 days X = 125 days
 = 20 days
Figure 2.11

Prentice Hall
2-81
Suppose that completion of a triplex in 75 days or
less will earn a bonus of $5,000.
What is the probability that Haynes will get the
bonus?

Prentice Hall
2-82
But Appendix A has only
positive Z values, and the
probability we are looking for
is in the negative tail.
20
100
75 





X
Z
25
1
20
25
.




Figure 2.12
µ = 100 days
X = 75 days
P(X < 75 days)
Area of
Interest

Prentice Hall
2-83
Because the curve is
symmetrical, we can look at
the probability in the positive
tail for the same distance
away from the mean.
20
100
75 





X
Z
25
1
20
25
.




P(X > 125 days)
Area of
Interest

Prentice Hall
2-84
 We know the probability
completing in
125 days is 0.89435.
 So the probability
completing in more
than 125 days is
1 – 0.89435 = 0.10565.

Prentice Hall
2-85
µ = 100 days
X = 75 days
The probability of completing in less than 75
days is 0.10565.
Going back to the
left tail of the
distribution:
The probability
completing in more than
125 days is
1 – 0.89435 = 0.10565.

Prentice Hall
2-86
What is the probability of completing a triplex
within 110 and 125 days?
We know the probability of completing in 125
days, P(X < 125) = 0.89435.
We have to complete the probability of
completing in 110 days and find the area
between those two events.

Prentice Hall
2-87
From Appendix A, for Z = 0.5 the
area is 0.69146.
P(110 < X < 125) = 0.89435 –
0.69146 = 0.20289.
20
100
110 





X
Z
5
0
20
10
.


Figure 2.13
µ = 100
days
125
days
 = 20 days
110
days

Prentice Hall
2-88
Using Excel
Program 2.3A
Function in an Excel 2010 Spreadsheet for the
Normal Distribution Example

Prentice Hall
2-89
Using Excel
Program 2.3B
Excel Output for the Normal Distribution Example

Prentice Hall
2-90
The Empirical Rule
For a normally distributed random variable with
mean µ and standard deviation  , then
1. About 68% of values will be within ±1 of the mean.
2. About 95.4% of values will be within ±2 of the
mean.
3. About 99.7% of values will be within ±3 of the
mean.

Prentice Hall
2-91
The Empirical Rule
Figure 2.14
68%
16% 16%
–
1
+1
a µ b
95.4%
2.3% 2.3%
–2 +2
a µ b
99.7%
0.15% 0.15%
–3 +3
a µ b

Prentice Hall
2-92
The F Distribution
 It is a continuous probability distribution.
 The F statistic is the ratio of two sample variances.
 F distributions have two sets of degrees of freedom.
 Degrees of freedom are based on sample size and used to calculate the
numerator and denominator of the ratio.
 The probabilities of large values of F are very small.
df1 = degrees of freedom for the numerator
df2 = degrees of freedom for the denominator

Prentice Hall
2-93
F
The F Distribution
Figure 2.15

Prentice Hall
2-94
The F Distribution
df1 = 5
df2 = 6
 = 0.05
Consider the example:
From Appendix D, we get
F, df1, df2
= F0.05, 5, 6 = 4.39
This means
P(F > 4.39) = 0.05
The probability is only 0.05 F will exceed 4.39.

Prentice Hall
2-95
The F Distribution
Figure 2.16
F = 4.39
0.05
F value for 0.05 probability
with 5 and 6 degrees of
freedom

Prentice Hall
2-96
Using Excel
Program 2.4A

Prentice Hall
2-97
Using Excel
Program 2.4B
Excel Output for the F Distribution

Prentice Hall
2-98
The Exponential Distribution
• The exponential distribution (also called the
negative exponential distribution) is a continuous
distribution often used in queuing models to
describe the time required to service a customer.
Its probability function is given by:
x
e
X
f 
 

)
(
where
X = random variable (service times)
µ = average number of units the service facility can
handle in a specific period of time
e = 2.718 (the base of natural logarithms)

Prentice Hall
2-99
The Exponential Distribution
time
service
Average
1
value
Expected 


2
1
Variance

 f(X)
X
Figure 2.17

Prentice Hall
2-100
Arnold’s Muffler Shop
• Arnold’s Muffler Shop installs new mufflers on
automobiles and small trucks.
• The mechanic can install 3 new mufflers per hour.
• Service time is exponentially distributed.
What is the probability that the time to install a new
muffler would be ½ hour or less?

Prentice Hall
2-101
Here:
X = Exponentially distributed service time
µ = average number of units the served per time period =
3 per hour
t = ½ hour = 0.5hour
P(X≤0.5) = 1 – e-3(0.5)
= 1 – e -1.5
= 1 = 0.2231 = 0.7769

Prentice Hall
2-102
P(X≤0.5) = 1 – e-3(0.5)
= 1 – e -1.5
= 1 = 0.2231 = 0.7769
Note also that if:
Then it must be the case that:
P(X>0.5) = 1 - 0.7769 = 0.2231

Prentice Hall
2-103
Probability That the Mechanic Will Install a Muffler in
0.5 Hour
Figure 2.18

Prentice Hall
2-104
Using Excel
Function in an Excel Spreadsheet for the
Exponential Distribution
Program 2.5A

Prentice Hall
2-105
Using Excel
Program 2.5B
Excel Output for the Exponential Distribution

Prentice Hall
2-106
The Poisson Distribution
• The Poisson distribution is a discrete distribution that
is often used in queuing models to describe arrival
rates over time. Its probability function is given by:
!
)
(
X
e
X
P
x 
 

where
P(X) = probability of exactly X arrivals or occurrences
 = average number of arrivals per unit of time
(the mean arrival rate)
e = 2.718, the base of natural logarithms
X = specific value (0, 1, 2, 3, …) of the random
variable

Prentice Hall
2-107
The Poisson Distribution
The mean and variance of the distribution are both .
Expected value = 
Variance = 

Prentice Hall
2-108
Poisson Distribution
We can use Appendix C to find Poisson probabilities.
Suppose that λ = 2. Some probability calculations are:
2706
.
0
2
)
1353
.
0
(
4
!
2
2
)
2
(
2706
.
0
1
)
1353
.
0
(
2
!
1
2
)
1
(
1353
.
0
1
)
1353
.
0
(
1
!
0
2
)
0
(
!
)
(
2
2
2
1
2
0














e
P
e
P
e
P
X
e
X
P
x 


Prentice Hall
2-109
Figure 2.19
Sample Poisson Distributions with λ = 2 and λ = 4

Prentice Hall
2-110
Using Excel
Program 2.6A
Functions in an Excel 2010 Spreadsheet for the

Prentice Hall
2-111
Using Excel
Program 2.6B
Excel Output for the Poisson Distribution

Prentice Hall
2-112
Exponential and Poisson Together
• If the number of occurrences per time period follows a Poisson
distribution, then the time between occurrences follows an
exponential distribution:
• Suppose the number of phone calls at a service center followed a Poisson
distribution with a mean of 10 calls per hour.
• Then the time between each phone call would be exponentially distributed
with a mean time between calls of 6 minutes (1/10 hour).

Prentice Hall
2-113
Copyright
All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted, in
any form or by any means, electronic, mechanical,
photocopying, recording, or otherwise, without the prior
written permission of the publisher. Printed in the United
States of America.

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