Chapter 4
Introduction to Probability
 Experiments, Counting Rules,
and Assigning Probabilities
 Events and Their Probability
 Some Basic Relationships
of Probability
 Conditional Probability
 Bayes’ Theorem
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 1
Uncertainties
Managers often base their decisions on an analysis
of uncertainties such as the following:
What are the chances that sales will decrease
if we increase prices?
What is the likelihood a new assembly method
method will increase productivity?
What are the odds that a new investment will
be profitable?
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 2
Probability
Probability is a numerical measure of the likelihood
that an event will occur.
Probability values are always assigned on a scale
from 0 to 1.
A probability near zero indicates an event is quite
unlikely to occur.
A probability near one indicates an event is almost
certain to occur.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 3
Probability as a Numerical Measure
of the Likelihood of Occurrence
Increasing Likelihood of Occurrence
Probability:
0
The event
is very
unlikely
to occur.
.5
The occurrence
of the event is
just as likely as
it is unlikely.
1
The event
is almost
certain
to occur.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 4
Statistical Experiments
In statistics, the notion of an experiment differs
somewhat from that of an experiment in the
physical sciences.
In statistical experiments, probability determines
outcomes.
Even though the experiment is repeated in exactly
the same way, an entirely different outcome may
occur.
For this reason, statistical experiments are sometimes called random experiments.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 5
An Experiment and Its Sample Space
An experiment is any process that generates welldefined outcomes.
The sample space for an experiment is the set of
all experimental outcomes.
An experimental outcome is also called a sample
point.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 6
An Experiment and Its Sample Space
Experiment
Experiment Outcomes
Toss a coin
Inspection a part
Conduct a sales call
Roll a die
Play a football game
Head, tail
Defective, non-defective
Purchase, no purchase
1, 2, 3, 4, 5, 6
Win, lose, tie
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 7
An Experiment and Its Sample Space
 Example: Bradley Investments
Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows.
Investment Gain or Loss
in 3 Months (in $000)
Markley Oil Collins Mining
8
10
-2
5
0
-20
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 8
A Counting Rule for
Multiple-Step Experiments
 If an experiment consists of a sequence of k steps
in which there are n1 possible results for the first step,
n2 possible results for the second step, and so on,
then the total number of experimental outcomes is
given by (n1)(n2) . . . (nk).
 A helpful graphical representation of a multiple-step
experiment is a tree diagram.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 9
A Counting Rule for
Multiple-Step Experiments
 Example: Bradley Investments
Bradley Investments can be viewed as a two-step
experiment. It involves two stocks, each with a set of
experimental outcomes.
Markley Oil:
Collins Mining:
Total Number of
Experimental Outcomes:
n1 = 4
n2 = 2
n1n2 = (4)(2) = 8
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 10
Tree Diagram
 Example: Bradley Investments
Markley Oil
(Stage 1)
Collins Mining
(Stage 2)
Gain 8
Gain 10
Gain 8
Gain 5
Lose 2
Lose 2
Gain 8
Even
Lose 20
Gain 8
Lose 2
Lose 2
Experimental
Outcomes
(10, 8)
Gain $18,000
(10, -2) Gain
$8,000
(5, 8)
Gain $13,000
(5, -2)
Gain
$3,000
(0, 8)
Gain
$8,000
(0, -2)
Lose
$2,000
(-20, 8) Lose $12,000
(-20, -2) Lose $22,000
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 11
Counting Rule for Combinations
 Number of Combinations of N Objects
Taken n at a Time
A second useful counting rule enables us to count
the number of experimental outcomes when n objects
are to be selected from a set of N objects.
CnN
where:
N!
 N
  
 n  n !(N - n )!
N! = N(N - 1)(N - 2) . . . (2)(1)
n! = n(n - 1)(n - 2) . . . (2)(1)
0! = 1
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Slide 12
Counting Rule for Permutations
 Number of Permutations of N Objects
Taken n at a Time
A third useful counting rule enables us to count
the number of experimental outcomes when n
objects are to be selected from a set of N objects,
where the order of selection is important.
PnN
where:
N!
 N
 n !  
 n  (N - n )!
N! = N(N - 1)(N - 2) . . . (2)(1)
n! = n(n - 1)(n - 2) . . . (2)(1)
0! = 1
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Slide 13
Assigning Probabilities
 Basic Requirements for Assigning Probabilities
1. The probability assigned to each experimental
outcome must be between 0 and 1, inclusively.
0 < P(Ei) < 1 for all i
where:
Ei is the ith experimental outcome
and P(Ei) is its probability
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Slide 14
Assigning Probabilities
 Basic Requirements for Assigning Probabilities
2. The sum of the probabilities for all experimental
outcomes must equal 1.
P(E1) + P(E2) + . . . + P(En) = 1
where:
n is the number of experimental outcomes
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 15
Assigning Probabilities
Classical Method
Assigning probabilities based on the assumption
of equally likely outcomes
Relative Frequency Method
Assigning probabilities based on experimentation
or historical data
Subjective Method
Assigning probabilities based on judgment
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 16
Classical Method
 Example: Rolling a Die
If an experiment has n possible outcomes, the
classical method would assign a probability of 1/n
to each outcome.
Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a
1/6 chance of occurring
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Slide 17
Relative Frequency Method
 Example: Lucas Tool Rental
Lucas Tool Rental would like to assign probabilities
to the number of car polishers it rents each day.
Office records show the following frequencies of daily
rentals for the last 40 days.
Number of
Polishers Rented
0
1
2
3
4
Number
of Days
4
6
18
10
2
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Slide 18
Relative Frequency Method
 Example: Lucas Tool Rental
Each probability assignment is given by dividing
the frequency (number of days) by the total frequency
(total number of days).
Number of
Polishers Rented
0
1
2
3
4
Number
of Days
4
6
18
10
2
40
Probability
.10
.15
.45
4/40
.25
.05
1.00
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 19
Subjective Method
 When economic conditions and a company’s
circumstances change rapidly it might be
inappropriate to assign probabilities based solely on
historical data.
 We can use any data available as well as our
experience and intuition, but ultimately a probability
value should express our degree of belief that the
experimental outcome will occur.
 The best probability estimates often are obtained by
combining the estimates from the classical or relative
frequency approach with the subjective estimate.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 20
Subjective Method
 Example: Bradley Investments
An analyst made the following probability estimates.
Exper. Outcome Net Gain or Loss Probability
(10, 8)
.20
$18,000 Gain
(10, -2)
.08
$8,000 Gain
(5, 8)
.16
$13,000 Gain
(5, -2)
.26
$3,000 Gain
(0, 8)
.10
$8,000 Gain
(0, -2)
.12
$2,000 Loss
(-20, 8)
.02
$12,000 Loss
(-20, -2)
.06
$22,000 Loss
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Slide 21
Events and Their Probabilities
An event is a collection of sample points.
The probability of any event is equal to the sum of
the probabilities of the sample points in the event.
If we can identify all the sample points of an
experiment and assign a probability to each, we
can compute the probability of an event.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 22
Events and Their Probabilities
 Example: Bradley Investments
Event M = Markley Oil Profitable
M = {(10, 8), (10, -2), (5, 8), (5, -2)}
P(M) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2)
= .20 + .08 + .16 + .26
= .70
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 23
Events and Their Probabilities
 Example: Bradley Investments
Event C = Collins Mining Profitable
C = {(10, 8), (5, 8), (0, 8), (-20, 8)}
P(C) = P(10, 8) + P(5, 8) + P(0, 8) + P(-20, 8)
= .20 + .16 + .10 + .02
= .48
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 24
Some Basic Relationships of Probability
There are some basic probability relationships that
can be used to compute the probability of an event
without knowledge of all the sample point probabilities.
Complement of an Event
Union of Two Events
Intersection of Two Events
Mutually Exclusive Events
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Slide 25
Complement of an Event
The complement of event A is defined to be the event
consisting of all sample points that are not in A.
The complement of A is denoted by Ac.
Event A
Ac
Sample
Space S
Venn
Diagram
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 26
Union of Two Events
The union of events A and B is the event containing
all sample points that are in A or B or both.
The union of events A and B is denoted by A B
Event A
Event B
Sample
Space S
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 27
Union of Two Events
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
M C = Markley Oil Profitable
or Collins Mining Profitable (or both)
M C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)}
P(M C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2)
+ P(0, 8) + P(-20, 8)
= .20 + .08 + .16 + .26 + .10 + .02
= .82
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 28
Intersection of Two Events
The intersection of events A and B is the set of all
sample points that are in both A and B.
The intersection of events A and B is denoted by A 
Event A
Event B
Sample
Space S
Intersection of A and B
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 29
Intersection of Two Events
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
M C = Markley Oil Profitable
and Collins Mining Profitable
M C = {(10, 8), (5, 8)}
P(M C) = P(10, 8) + P(5, 8)
= .20 + .16
= .36
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 30
Addition Law
The addition law provides a way to compute the
probability of event A, or B, or both A and B occurring.
The law is written as:
P(A B) = P(A) + P(B) - P(A  B
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 31
Addition Law
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
M C = Markley Oil Profitable
or Collins Mining Profitable
We know: P(M) = .70, P(C) = .48, P(M C) = .36
Thus: P(M  C) = P(M) + P(C) - P(M  C)
= .70 + .48 - .36
= .82
(This result is the same as that obtained earlier
using the definition of the probability of an event.)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 32
Mutually Exclusive Events
Two events are said to be mutually exclusive if the
events have no sample points in common.
Two events are mutually exclusive if, when one event
occurs, the other cannot occur.
Event A
Event B
Sample
Space S
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 33
Mutually Exclusive Events
If events A and B are mutually exclusive, P(A  B = 0.
The addition law for mutually exclusive events is:
P(A B) = P(A) + P(B)
There is no need to
include “- P(A  B”
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 34
Conditional Probability
The probability of an event given that another event
has occurred is called a conditional probability.
The conditional probability of A given B is denoted
by P(A|B).
A conditional probability is computed as follows :
P( A  B)
P( A|B) 
P( B)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 35
Conditional Probability
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
P(C | M ) = Collins Mining Profitable
given Markley Oil Profitable
We know: P(M C) = .36, P(M) = .70
P(C  M ) .36
Thus: P(C | M ) 

 .5143
P( M )
.70
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 36
Multiplication Law
The multiplication law provides a way to compute the
probability of the intersection of two events.
The law is written as:
P(A B) = P(B)P(A|B)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 37
Multiplication Law
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
M C = Markley Oil Profitable
and Collins Mining Profitable
We know: P(M) = .70, P(C|M) = .5143
Thus: P(M  C) = P(M)P(M|C)
= (.70)(.5143)
= .36
(This result is the same as that obtained earlier
using the definition of the probability of an event.)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 38
Joint Probability Table
Collins Mining
Profitable (C) Not Profitable (Cc)
Markley Oil
Total
Profitable (M)
.36
.34
.70
Not Profitable (Mc)
.12
.18
.30
Total
.48
.52
1.00
Joint Probabilities
(appear in the body
of the table)
Marginal Probabilities
(appear in the margins
of the table)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 39
Independent Events
If the probability of event A is not changed by the
existence of event B, we would say that events A
and B are independent.
Two events A and B are independent if:
P(A|B) = P(A)
or
P(B|A) = P(B)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 40
Multiplication Law
for Independent Events
The multiplication law also can be used as a test to see
if two events are independent.
The law is written as:
P(A B) = P(A)P(B)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 41
Multiplication Law
for Independent Events
 Example: Bradley Investments
Event M = Markley Oil Profitable
Event C = Collins Mining Profitable
Are events M and C independent?
Does P(M  C) = P(M)P(C) ?
We know: P(M  C) = .36, P(M) = .70, P(C) = .48
But: P(M)P(C) = (.70)(.48) = .34, not .36
Hence: M and C are not independent.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 42
Mutual Exclusiveness and Independence
Do not confuse the notion of mutually exclusive
events with that of independent events.
Two events with nonzero probabilities cannot be
both mutually exclusive and independent.
If one mutually exclusive event is known to occur,
the other cannot occur.; thus, the probability of the
other event occurring is reduced to zero (and they
are therefore dependent).
Two events that are not mutually exclusive, might
or might not be independent.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 43
Bayes’ Theorem
 Often we begin probability analysis with initial or
prior probabilities.
 Then, from a sample, special report, or a product
test we obtain some additional information.
 Given this information, we calculate revised or
posterior probabilities.
 Bayes’ theorem provides the means for revising the
prior probabilities.
Prior
Probabilities
New
Information
Application
of Bayes’
Theorem
Posterior
Probabilities
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 44
Bayes’ Theorem
 Example: L. S. Clothiers
A proposed shopping center will provide strong
competition for downtown businesses like L. S.
Clothiers. If the shopping center is built, the owner
of L. S. Clothiers feels it would be best to relocate to
the shopping center.
The shopping center cannot be built unless a
zoning change is approved by the town council.
The planning board must first make a
recommendation, for or against the zoning change,
to the council.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 45
Prior Probabilities
 Example: L. S. Clothiers
Let:
A1 = town council approves the zoning change
A2 = town council disapproves the change
Using subjective judgment:
P(A1) = .7, P(A2) = .3
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 46
New Information
 Example: L. S. Clothiers
The planning board has recommended against
the zoning change. Let B denote the event of a
negative recommendation by the planning board.
Given that B has occurred, should L. S. Clothiers
revise the probabilities that the town council will
approve or disapprove the zoning change?
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 47
Conditional Probabilities
 Example: L. S. Clothiers
Past history with the planning board and the town
council indicates the following:
Hence:
P(B|A1) = .2
P(B|A2) = .9
P(BC|A1) = .8
P(BC|A2) = .1
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 48
Tree Diagram
 Example: L. S. Clothiers
Town Council Planning Board
P(B|A1) = .2
P(A1) = .7
c
P(B |A1) = .8
P(B|A2) = .9
P(A2) = .3
c
P(B |A2) = .1
Experimental
Outcomes
P(A1  B) = .14
P(A1  Bc) = .56
P(A2  B) = .27
P(A2  Bc) = .03
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 49
Bayes’ Theorem
 To find the posterior probability that event Ai will
occur given that event B has occurred, we apply
Bayes’ theorem.
P( Ai )P( B| Ai )
P( Ai |B) 
P( A1 )P( B| A1 )  P( A2 )P( B| A2 )  ...  P( An )P( B| An )
 Bayes’ theorem is applicable when the events for
which we want to compute posterior probabilities
are mutually exclusive and their union is the entire
sample space.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 50
Posterior Probabilities
 Example: L. S. Clothiers
Given the planning board’s recommendation not
to approve the zoning change, we revise the prior
probabilities as follows:
P( A1 |B) 

=
P( A1 )P( B| A1 )
P( A1 )P( B| A1 )  P( A2 )P( B| A2 )
(. 7 )(. 2 )
(. 7 )(. 2 )  (. 3)(. 9)
.34
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Slide 51
Posterior Probabilities
 Example: L. S. Clothiers
The planning board’s recommendation is good
news for L. S. Clothiers. The posterior probability of
the town council approving the zoning change is .34
compared to a prior probability of .70.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 52
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 1
Prepare the following three columns:
Column 1 - The mutually exclusive events for
which posterior probabilities are desired.
Column 2 - The prior probabilities for the events.
Column 3 - The conditional probabilities of the
new information given each event.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 53
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 1
(1)
(2)
(3)
(4)
(5)
Conditional
Prior
Events Probabilities Probabilities
Ai
P(Ai)
P(B|Ai)
A1
.7
.2
A2
.3
.9
1.0
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Slide 54
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 2
Prepare the fourth column:
Column 4
Compute the joint probabilities for each event and
the new information B by using the multiplication
law.
Multiply the prior probabilities in column 2 by
the corresponding conditional probabilities in
column 3. That is, P(Ai IB) = P(Ai) P(B|Ai).
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Slide 55
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 2
(1)
(2)
(3)
(4)
(5)
Conditional
Prior
Joint
Events Probabilities Probabilities Probabilities
Ai
P(Ai)
P(B|Ai)
P(Ai I B)
A1
.7
.2
.14
A2
.3
.9
.27
1.0
.7 x .2
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 56
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 2 (continued)
We see that there is a .14 probability of the town
council approving the zoning change and a
negative recommendation by the planning board.
There is a .27 probability of the town council
disapproving the zoning change and a negative
recommendation by the planning board.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 57
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 3
Sum the joint probabilities in Column 4. The
sum is the probability of the new information,
P(B). The sum .14 + .27 shows an overall
probability of .41 of a negative recommendation
by the planning board.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 58
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 3
(1)
(2)
(3)
(4)
(5)
Conditional
Prior
Joint
Events Probabilities Probabilities Probabilities
Ai
P(Ai)
P(B|Ai)
P(Ai I B)
A1
.7
.2
.14
A2
.3
.9
.27
1.0
P(B) = .41
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 59
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 4
Prepare the fifth column:
Column 5
Compute the posterior probabilities using the
basic relationship of conditional probability.
P( Ai  B)
P( Ai | B) 
P( B)
The joint probabilities P(Ai I B) are in column 4
and the probability P(B) is the sum of column 4.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 60
Bayes’ Theorem: Tabular Approach
 Example: L. S. Clothiers
• Step 4
(1)
(2)
(3)
(4)
(5)
Prior
Joint
Posterior
Conditional
Events Probabilities Probabilities Probabilities Probabilities
P(B|Ai)
P(Ai)
P(Ai I B)
P(Ai |B)
Ai
A1
.7
.2
.14
.3415
A2
.3
.9
.27
.6585
P(B) = .41
1.0000
1.0
.14/.41
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 61

Example: Incidence of a rare disease
Only 1 in 1000 adults is afflicted with a rare
disease for which a diagnostic test has been
developed. The test is such that when an
individual actually has a disease, a positive
result will occur 99% of the time, whereas an
individual without the disease will show a
positive test result only 2% of the time. If a
randomly selected individual is tested and the
result is positive, what is the probability that the
individual has the disease?
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or duplicated, or posted to a publicly accessible website, in whole or in part.
62 Slide 62
End of Chapter 4
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 63
Chapter 5
Discrete Probability Distributions






Random Variables
Discrete Probability Distributions
Expected Value and Variance
Binomial Probability Distribution
Poisson Probability Distribution .40
Hypergeometric Probability
.30
Distribution
.20
.10
0
1
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or duplicated, or posted to a publicly accessible website, in whole or in part.
2
3
4
Slide 64
Random Variables
A random variable is a numerical description of the
outcome of an experiment.
A discrete random variable may assume either a
finite number of values or an infinite sequence of
values.
A continuous random variable may assume any
numerical value in an interval or collection of
intervals.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 65
Discrete Random Variable
with a Finite Number of Values

Example: JSL Appliances
Let x = number of TVs sold at the store in one day,
where x can take on 5 values (0, 1, 2, 3, 4)
We can count the TVs sold, and there is a finite
upper limit on the number that might be sold (which
is the number of TVs in stock).
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 66
Discrete Random Variable
with an Infinite Sequence of Values

Example: JSL Appliances
Let x = number of customers arriving in one day,
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is
no finite upper limit on the number that might arrive.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 67
Random Variables
Question
Family
size
Type
Random Variable x
x = Number of dependents
reported on tax return
Discrete
Distance from x = Distance in miles from
home to store
home to the store site
Continuous
Own dog
or cat
Discrete
x = 1 if own no pet;
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 68
Discrete Probability Distributions
The probability distribution for a random variable
describes how probabilities are distributed over
the values of the random variable.
We can describe a discrete probability distribution
with a table, graph, or formula.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 69
Discrete Probability Distributions
The probability distribution is defined by a
probability function, denoted by f(x), which provides
the probability for each value of the random variable.
The required conditions for a discrete probability
function are:
f(x) > 0
f(x) = 1
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Slide 70
Discrete Probability Distributions

Example: JSL Appliances
• Using past data on TV sales, …
• a tabular representation of the probability
distribution for TV sales was developed.
Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
x
0
1
2
3
4
f(x)
.40
.25
.20
.05
.10
1.00
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or duplicated, or posted to a publicly accessible website, in whole or in part.
80/200
Slide 71
Discrete Probability Distributions

Example: JSL Appliances
Graphical
representation
of probability
distribution
Probability
.50
.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 72
Discrete Uniform Probability Distribution
The discrete uniform probability distribution is the
simplest example of a discrete probability
distribution given by a formula.
The discrete uniform probability function is
f(x) = 1/n
the values of the
random variable
are equally likely
where:
n = the number of values the random
variable may assume
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 73
Expected Value
The expected value, or mean, of a random variable
is a measure of its central location.
E(x) =  = xf(x)
The expected value is a weighted average of the
values the random variable may assume. The
weights are the probabilities.
The expected value does not have to be a value the
random variable can assume.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 74
Variance and Standard Deviation
The variance summarizes the variability in the
values of a random variable.
Var(x) =  2 = (x - )2f(x)
The variance is a weighted average of the squared
deviations of a random variable from its mean. The
weights are the probabilities.
The standard deviation, , is defined as the positive
square root of the variance.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 75
Expected Value

Example: JSL Appliances
x
0
1
2
3
4
f(x)
xf(x)
.40
.00
.25
.25
.20
.40
.05
.15
.10
.40
E(x) = 1.20
expected number of
TVs sold in a day
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 76
Variance

Example: JSL Appliances
x
x-
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
(x - )2
f(x)
(x - )2f(x)
1.44
0.04
0.64
3.24
7.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
TVs
squared
Variance of daily sales = 2 = 1.660
Standard deviation of daily sales = 1.2884 TVs
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 77
Binomial Probability Distribution

Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
2. Two outcomes, success and failure, are possible
on each trial.
3. The probability of a success, denoted by p, does
not change from trial to trial.
stationarity
assumption
4. The trials are independent.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 78
Binomial Probability Distribution
Our interest is in the number of successes
occurring in the n trials.
We let x denote the number of successes
occurring in the n trials.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 79
Binomial Probability Distribution

Binomial Probability Function
n!
f (x) 
p x (1 - p )( n - x )
x !(n - x )!
where:
x = the number of successes
p = the probability of a success on one trial
n = the number of trials
f(x) = the probability of x successes in n trials
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 80
Binomial Probability Distribution

Binomial Probability Function
n!
f (x) 
p x (1 - p )( n - x )
x !(n - x )!
Number of experimental
outcomes providing exactly
x successes in n trials
Probability of a particular
sequence of trial outcomes
with x successes in n trials
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 81
Binomial Probability Distribution

Example: Evans Electronics
Evans Electronics is concerned about a low
retention rate for its employees. In recent years,
management has seen a turnover of 10% of the
hourly employees annually.
Thus, for any hourly employee chosen at random,
management estimates a probability of 0.1 that the
person will not be with the company next year.
Choosing 3 hourly employees at random, what is
the probability that 1 of them will leave the company
this year?
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 82
Binomial Probability Distribution

Example: Evans Electronics
The probability of the first employee leaving and the
second and third employees staying, denoted (S, F, F),
is given by
p(1 – p)(1 – p)
With a .10 probability of an employee leaving on any
one trial, the probability of an employee leaving on
the first trial and not on the second and third trials is
given by
(.10)(.90)(.90) = (.10)(.90)2 = .081
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 83
Binomial Probability Distribution

Example: Evans Electronics
Two other experimental outcomes also result in one
success and two failures. The probabilities for all
three experimental outcomes involving one success
follow.
Experimental
Outcome
Probability of
Experimental Outcome
(S, F, F)
(F, S, F)
(F, F, S)
p(1 – p)(1 – p) = (.1)(.9)(.9) = .081
(1 – p)p(1 – p) = (.9)(.1)(.9) = .081
(1 – p)(1 – p)p = (.9)(.9)(.1) = .081
Total = .243
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 84
Binomial Probability Distribution

Example: Evans Electronics
Let: p = .10, n = 3, x = 1
Using the
probability
function
n!
f ( x) 
p x (1 - p ) (n - x )
x !( n - x )!
3!
f (1) 
(0.1)1 (0.9)2  3(.1)(.81)  .243
1!(3 - 1)!
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 85
Binomial Probability Distribution

Example: Evans Electronics
1st Worker
2nd Worker
Leaves (.1)
Leaves
(.1)
Using a tree diagram
3rd Worker
L (.1)
x
3
Prob.
.0010
S (.9)
2
.0090
L (.1)
2
.0090
S (.9)
1
.0810
L (.1)
2
.0090
S (.9)
1
.0810
1
.0810
0
.7290
Stays (.9)
Leaves (.1)
Stays
(.9)
L (.1)
Stays (.9)
S (.9)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 86
Binomial Probabilities
and Cumulative Probabilities
Statisticians have developed tables that give
probabilities and cumulative probabilities for a
binomial random variable.
These tables can be found in some statistics
textbooks.
With modern calculators and the capability of
statistical software packages, such tables are
almost unnecessary.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 87
Binomial Probability Distribution

Expected Value
E(x) =  = np

Variance
Var(x) =  2 = np(1 - p)

Standard Deviation
  np(1 - p )
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 88
Binomial Probability Distribution

Example: Evans Electronics
• Expected Value
E(x) = np = 3(.1) = .3 employees out of 3
•
Variance
Var(x) = np(1 – p) = 3(.1)(.9) = .27
• Standard Deviation
  3(.1)(.9)  .52 employees
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 89
Chapter 6
Continuous Probability Distributions

Uniform Probability Distribution
Normal Probability Distribution
Normal Approximation of Binomial Probabilities

Exponential Probability Distribution


f (x)
f (x) Exponential
Uniform
f (x)
Normal
x
x
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 90
Continuous Probability Distributions

A continuous random variable can assume any value
in an interval on the real line or in a collection of
intervals.

It is not possible to talk about the probability of the
random variable assuming a particular value.

Instead, we talk about the probability of the random
variable assuming a value within a given interval.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 91
Continuous Probability Distributions

f (x)
The probability of the random variable assuming a
value within some given interval from x1 to x2 is
defined to be the area under the graph of the
probability density function between x1 and x2.
f (x) Exponential
Uniform
f (x)
x1 x2
Normal
x1 xx12 x2
x
x1 x2
x
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 92
Uniform Probability Distribution

A random variable is uniformly distributed
whenever the probability is proportional to the
interval’s length.

The uniform probability density function is:
f (x) = 1/(b – a) for a < x < b
=0
elsewhere
where: a = smallest value the variable can assume
b = largest value the variable can assume
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 93
Uniform Probability Distribution

Expected Value of x
E(x) = (a + b)/2

Variance of x
Var(x) = (b - a)2/12
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 94
Uniform Probability Distribution

Example: Slater's Buffet
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.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 95
Uniform Probability Distribution

Uniform Probability Density Function
f(x) = 1/10 for 5 < x < 15
=0
elsewhere
where:
x = salad plate filling weight
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 96
Uniform Probability Distribution

Expected Value of x
E(x) = (a + b)/2
= (5 + 15)/2
= 10

Variance of x
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 97
Uniform Probability Distribution

Uniform Probability Distribution
for Salad Plate Filling Weight
f(x)
1/10
0
5
10
Salad Weight (oz.)
x
15
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 98
Uniform Probability Distribution
What is the probability that a customer
will take between 12 and 15 ounces of salad?
f(x)
P(12 < x < 15) = 1/10(3) = .3
1/10
0
5
10 12
Salad Weight (oz.)
x
15
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 99
Area as a Measure of Probability

The area under the graph of f(x) and probability are
identical.

This is valid for all continuous random variables.

The probability that x takes on a value between some
lower value x1 and some higher value x2 can be found
by computing the area under the graph of f(x) over
the interval from x1 to x2.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 100
Normal Probability Distribution



The normal probability distribution is the most
important distribution for describing a continuous
random variable.
It is widely used in statistical inference.
It has been used in a wide variety of applications
including:
• Heights of people
• Rainfall amounts
• Test scores
• Scientific measurements

Abraham de Moivre, a French mathematician,
published The Doctrine of Chances in 1733.

He derived the normal distribution.
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Slide 101
Normal Probability Distribution

Normal Probability Density Function
1
- ( x -  )2 /2 2
f (x) 
e
 2
where:
 = mean
 = standard deviation
 = 3.14159
e = 2.71828
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Slide 102
Normal Probability Distribution

Characteristics
The distribution is symmetric; its skewness
measure is zero.
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 103
Normal Probability Distribution

Characteristics
The entire family of normal probability
distributions is defined by its mean  and its
standard deviation  .
Standard Deviation 
Mean 
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 104
Normal Probability Distribution

Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 105
Normal Probability Distribution

Characteristics
The mean can be any numerical value: negative,
zero, or positive.
x
-10
0
25
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 106
Normal Probability Distribution

Characteristics
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
 = 15
 = 25
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 107
Normal Probability Distribution

Characteristics
Probabilities for the normal random variable are
given by areas under the curve. The total area
under the curve is 1 (.5 to the left of the mean and
.5 to the right).
.5
.5
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 108
Normal Probability Distribution

Characteristics (basis for the empirical rule)
68.26% of values of a normal random variable
are within +/- 1 standard deviation of its mean.
95.44% of values of a normal random variable
are within +/- 2 standard deviations of its mean.
99.72% of values of a normal random variable
are within +/- 3 standard deviations of its mean.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 109
Normal Probability Distribution

Characteristics (basis for the empirical rule)
99.72%
95.44%
68.26%
 – 3
 – 1
 – 2

 + 3
 + 1
 + 2
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or duplicated, or posted to a publicly accessible website, in whole or in part.
x
Slide 110
Standard Normal Probability Distribution

Characteristics
A random variable having a normal distribution
with a mean of 0 and a standard deviation of 1 is
said to have a standard normal probability
distribution.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 111
Standard Normal Probability Distribution

Characteristics
The letter z is used to designate the standard
normal random variable.
1
z
0
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Slide 112
Standard Normal Probability Distribution

Converting to the Standard Normal Distribution
z
x-

We can think of z as a measure of the number of
standard deviations x is from .
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 113
Standard Normal Probability Distribution

Example: Pep Zone
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 stockouts while waiting for a
replenishment order.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 114
Standard Normal Probability Distribution

Example: Pep Zone
It has been determined that demand during
replenishment lead-time is normally distributed
with a mean of 15 gallons and a standard deviation
of 6 gallons.
The manager would like to know the probability
of a stockout during replenishment lead-time. In
other words, what is the probability that demand
during lead-time will exceed 20 gallons?
P(x > 20) = ?
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 115
Standard Normal Probability Distribution

Solving for the Stockout Probability
Step 1: Convert x to the standard normal distribution.
z = (x - )/
= (20 - 15)/6
= .83
Step 2: Find the area under the standard normal
curve to the left of z = .83.
see next slide
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 116
Standard Normal Probability Distribution

Cumulative Probability Table for
the Standard Normal Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7
.7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
.
.
.
.
.
.
.
.
.
.
.
P(z < .83)
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 117
Standard Normal Probability Distribution

Solving for the Stockout Probability
Step 3: Compute the area under the standard normal
curve to the right of z = .83.
P(z > .83) = 1 – P(z < .83)
= 1- .7967
= .2033
Probability
of a stockout
P(x > 20)
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 118
Standard Normal Probability Distribution

Solving for the Stockout Probability
Area = 1 - .7967
Area = .7967
= .2033
0
.83
z
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 119
Standard Normal Probability Distribution

Standard Normal Probability Distribution
If the manager of Pep Zone wants the probability
of a stockout during replenishment lead-time to be
no more than .05, what should the reorder point be?
--------------------------------------------------------------(Hint: Given a probability, we can use the standard
normal table in an inverse fashion to find the
corresponding z value.)
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 120
Standard Normal Probability Distribution

Solving for the Reorder Point
Area = .9500
Area = .0500
0
z.05
z
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 121
Standard Normal Probability Distribution

Solving for the Reorder Point
Step 1: Find the z-value that cuts off an area of .05
in the right tail of the standard normal
distribution.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
up.9706
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 We
.9693look
.9699
the.9756
complement
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750
.9761 .9767
.
.
.
.
.
.
.
.
of the
tail. area .
.
(1 - .05 = .95)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 122
Standard Normal Probability Distribution

Solving for the Reorder Point
Step 2: Convert z.05 to the corresponding value of x.
x =  + z.05
= 15 + 1.645(6)
= 24.87 or 25
A reorder point of 25 gallons will place the probability
of a stockout during leadtime at (slightly less than) .05.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 123
Normal Probability Distribution

Solving for the Reorder Point
Probability of no
stockout during
replenishment
lead-time = .95
Probability of a
stockout during
replenishment
lead-time = .05
15
24.87
x
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 124
Standard Normal Probability Distribution

Solving for the Reorder Point
By raising the reorder point from 20 gallons to
25 gallons on hand, the probability of a stockout
decreases from about .20 to .05.
This is a significant decrease in the chance that
Pep Zone will be out of stock and unable to meet a
customer’s desire to make a purchase.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 125
Normal Approximation of Binomial Probabilities
When the number of trials, n, becomes large,
evaluating the binomial probability function by hand
or with a calculator is difficult.
The normal probability distribution provides an
easy-to-use approximation of binomial probabilities
where np > 5 and n(1 - p) > 5.
In the definition of the normal curve, set
 = np and   np (1 - p )
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 126
Normal Approximation of Binomial Probabilities
Add and subtract a continuity correction factor
because a continuous distribution is being used to
approximate a discrete distribution.
For example, P(x = 12) for the discrete binomial
probability distribution is approximated by
P(11.5 < x < 12.5) for the continuous normal
distribution.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 127
Normal Approximation of Binomial Probabilities

Example
Suppose that a company has a history of making
errors in 10% of its invoices. A sample of 100
invoices has been taken, and we want to compute
the probability that 12 invoices contain errors.
In this case, we want to find the binomial
probability of 12 successes in 100 trials. So, we set:
 = np = 100(.1) = 10
  np (1 - p ) = [100(.1)(.9)] ½ = 3
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 128
Normal Approximation of Binomial Probabilities

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
=3
P(11.5 < x < 12.5)
(Probability
of 12 Errors)
 = 10
11.5
12.5
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 129
Normal Approximation of Binomial Probabilities

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
P(x < 12.5) = .7967
10 12.5
x
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 130
Normal Approximation of Binomial Probabilities

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
P(x < 11.5) = .6915
10
x
11.5
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 131
Normal Approximation of Binomial Probabilities

The Normal Approximation to the Probability
of 12 Successes in 100 Trials is .1052
P(x = 12)
= .7967 - .6915
= .1052
10
11.5
12.5
x
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 132