Transcript P(A and B)

Chapter 2
Probability
Concepts and
Applications
Prepared by Lee Revere and John Large
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-1
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Learning Objectives
Students will be able to:
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.
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Chapter Outline
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Introduction
Fundamental Concepts
Mutually Exclusive and
Collectively Exhaustive
Events
Statistically Independent
Events
Statistically Dependent Events
Revising Probabilities with
Bayes’ Theorem
Further Probability Revisions
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Chapter Outline
continued
2.8
2.9
2.10
2.11
2.12
Random Variables
Probability Distributions
The Binomial Distribution
The Normal Distribution
The Exponential
Distribution
2.13 The Poisson Distribution
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Introduction
 Life is uncertain!
 We must deal with risk!
 A probability is a numerical
statement about the likelihood
that an event will occur.
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Basic Statements about
Probability
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.
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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 frequencies of demand are
represented in the following table:
Qty Demanded
0
1
2
3
4
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No. of Days
40
80
50
20
10
Total 200
2-7
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Diversey Paint Example
(continued)
Probabilities of Demand
Probability
Quantity Freq.
(Relative Freq)
Demand (days)
(40/200) = 0.20
0
40
(80/200) = 0.40
1
80
(50/200) = 0.25
2
50
(20/200) = 0.10
3
20
(10/200) = 0.05
4
10
Total Prob =1.00
Total days = 200
Note: 0
 P(event)  1
and
 P(event) = 1
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Types of Probability
Objective probability is based on
logical observations:
P ( event ) =
Number of times event occurs
Total number of outcomes or occurrences
Determined by:
 Relative frequency – Obtained using
historical data (Diversey Paint)
 Classical method – Known
probability for each outcome (tossing
a coin)
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Types of Probability
Subjective probability is based on
personal experiences.
Determined by:
 Judgment of experts
 Opinion polls
 Delphi method
 Others
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Mutually Exclusive Events
 Events are said to be mutually
exclusive if only one of the
events can occur on any one trial.
Example: a fair coin toss results in
either a heads or a tails.
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Collectively Exhaustive
Events
 Events are said to be collectively
exhaustive if the list of outcomes
includes every possible outcome.
 Heads and tails as possible outcomes
of coin flip.
Example: a collectively exhaustive list of
possible outcomes for a fair coin toss
includes heads and tails.
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Die Roll Example
Outcome Probability
of Roll
1
1/6
2
1/6
This is a collectively
exhaustive list of
3
1/6
potential outcomes
4
1/6
for a single die roll.
5
1/6
6
1/6
Total = 1
The outcome is a mutually exclusive event
because only one event can occur (a 1, 2, 3, 4,
5, or 6) on any single roll.
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Twin Birth Example
A woman is pregnant with nonidentical twins. Following is a list of
collectively exhaustive, mutually
exclusive possible outcomes:
Outcome
of Birth
Probability
Boy/Boy
Boy/Girl
Girl/Girl
Girl/Boy
¼
¼
¼
¼
What is the probability that both
babies will be girls? / boys?
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In-Class Practice
Assuming a traditional 52-card deck, can
you identify if these outcomes are mutually
exclusive and/or collectively exhaustive ??
 Draw a spade and a club
 Draw a face card and a number
card
 Draw an ace and a 3
 Draw a club and a nonclub
 Draw a 5 and a diamond
 Draw a red card and a diamond
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Law of Addition:
Mutually Exclusive
P (event A or event B) =
P (event A) + P (event B)
or:
P (A or B) = P (A) + P (B)
Example:
P (spade or club) = P (spade) + P (club)
= 13/52 + 13/52
= 26/52 = 1/2 = 50%
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Law of Addition:
not Mutually Exclusive
P(event A or event B) =
P(event A) + P(event B) P(event A and event B both
occurring)
or
P(A or B) = P(A)+P(B) - P(A and B)
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Venn Diagram
P(A)
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P(B)
2-18
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Venn Diagram
P(A or B)
-
+
P(A)
P(B)
=
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P(A and B)
P(A or B)
2-19
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In-Class Example:
Specialized University
Specialized University offers four different
graduate degrees: business, education,
accounting, and science. Enrollment figures
show 25% of their graduate students are in
each specialty. Although 50% of the
students are female, only 15% are female
business majors. If a student is randomly
selected from the University’s registration
database:
 What is the probability the student is a
business or education major?
 What is the probability the student is a
female or a business major?
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Specialized University
Solution
The probability that the student is a business
or education major is mutually exclusive
event. Thus:
P(Bus or Edu) = P(Bus) + P(Edu)
= .25 + .25
= .50 or 50%
The probability that the student is a female
or a business major is not mutually exclusive
because the student could be a female
business major. Thus:
P(Fem or Bus) = P(Fem) + P(Bus)
– P(Fem and Bus)
= .50 + .25 - .15
= .60 or 60%
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Statistical Dependence
 Events are either
 statistically independent (the
occurrence of one event has no
effect on the probability of
occurrence of the other), or
statistically dependent (the
occurrence of one event gives
information about the occurrence
of the other).
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Which Are Independent?
(a) Your education
(b) Your income level
(a) Draw a jack of hearts from a
full 52-card deck
(b) Draw a jack of clubs from a
full 52-card deck
(a) Chicago Cubs win the
National League pennant
(b) Chicago Cubs win the World
Series
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Probabilities:
Independent Events
 Marginal probability: the probability of
an event occurring: P(A)
 Joint probability: the probability of
multiple, independent events, occurring
at the same time:
P(AB) = P(A)*P(B)
 Conditional probability (for
independent events):
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)
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Venn Diagram: P(A|B)
P(B)
P(B|A)
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P(A|B)
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Independent Events
Example
1. P(black ball drawn
on first draw)
• P(B) = 0.30
A bucket
contains 3 black
(marginal
balls and 7
probability)
green balls. We
2. P(two green balls
draw a ball
drawn)
from the
• P(GG) =
bucket, replace
P(G)*P(G) =
it, and draw a
second ball.
0.70*0.70 = 0.49
(joint probability
for two
independent
events)
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Independent Events Example
continued
1. P(black ball drawn on second
draw, first draw was green)
P(B|G) = P(B) = 0.30
(conditional probability)
2. P(green ball drawn on second
draw, first draw was green)
P(G|G) = 0.70
(conditional probability)
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Probabilities: Dependent
Events
 Marginal probability: probability of an
event occurring: P(A)
 Conditional probability (for dependent
events):
The probability of event B given that
event A has occurred:
P(B|A) = P(AB)/P(A)
The probability of event A given that
event B has occurred:
P(A|B) = P(AB)/P(B)
 Joint probability: The probability of
multiple events occurring at the same
time: P(AB) = P(B|A)*P(A)
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Venn Diagram: P(B|A)
P(A)
P(B)
/
P(B)
P(AB)
P(A)
P(B|A) = P(AB)/P(A)
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Venn Diagram: P(A|B)
P(A)
P(B)
/
P(A)
P(AB)
P(B)
P(A|B) = P(AB)/P(B)
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Dependent Events
Example
Then:
Assume that we
 P(WL) = 4/10 = 0.40
have an urn
 P(WN) = 2/10 = 0.20
containing 10 balls
of the following
 P(W) = 6/10 = 0.60
descriptions:
 P(YL) = 3/10 = 0.3
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)
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 P(YN) = 1/10 = 0.1
 P(Y) = 4/10 = 0.4
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Dependent Events
Example (continued)
Then:
 P(Y) = .4
- marginal probability
 P(L|Y) = P(YL)/P(Y)
= 0.3/0.4 = 0.75
- conditional probability
 P(W|L) = P(WL)/P(L)
= 0.4/0.7 = 0.57
- conditional probability
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Dependent Events: Joint
Probability Example
Your stockbroker informs you that if
the stock market reaches the 10,500
point level by January, there is a 70%
probability that Tubeless Electronics
will go up in value. Your own
feeling is that there is only a 40%
chance of the market reaching 10,500
by January.
What is the probability that both the
stock market will reach 10,500
points, and the price of Tubeless will
go up in value?
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Dependent Events: Joint
Probabilities Solution
Let M represent
the event of the
Then:
stock market
P(MT) =P(T|M)P(M)
reaching the
= (0.70)(0.40)
10,500 point
= 0.28
level, and T
represent the
event that
Tubeless goes
up.
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Revising Probabilities:
Bayes’ Theorem
Bayes’ theorem can be used to
calculate revised or posterior
probabilities.
Prior
Probabilities
Bayes’
Process
Posterior
Probabilities
New
Information
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General Form of
Bayes’ Theorem
P( AB)
P( A | B) =
P( B)
or
P( B | A) P( A)
P( A | B) =
P( B | A) P( A)  P( B | A ) P( A )
where A = complement of the event A
For example, if the event A is " fair" die,
then the event A is " unfair" die.
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Posterior Probabilities
Example
A cup contains two dice identical in
appearance. One, however, 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 have no idea which die
is which, but we select one by chance,
and toss it. The result is a 3.
What is the probability that the die
rolled was fair?
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Posterior Probabilities
Example (continued)
 We know that:
P(fair) = 0.50
P(3|fair) = 0.166
P(loaded) = 0.50
P(3|loaded) = 0.60
- marginal probability
 Then:
P(3 and fair) = P(3|fair)P(fair)
= (0.166)(0.50)
= 0.083
P(3 and loaded) = P(3|loaded)P(loaded)
= (0.60)(0.50)
= 0.300
- joint probability
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Posterior Probabilities
Example continued
A 3 can occur in combination with the
state “fair die” or in combination with
the state ”loaded die.” The sum of their
probabilities gives the marginal
probability of a 3 on a toss:
P(3) = 0.083 + 0.0300 = 0.383
- marginal probability
Then, the probability that the die rolled
was the fair one is given by:
P(Fair | 3) =
P(Fair and 3) 0.083
=
= 0.22
P(3)
0.383
- conditional probability
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Further Probability
Revisions
To obtain further information as to
whether the die just rolled is fair or
loaded, let’s roll it again….
Again we get a 3.
Given that we have now rolled two 3s,
what is the probability that the die
rolled is fair?
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Further Probability
Revisions continued
 We know from before that:
P(fair) = 0.50, P(loaded) = 0.50
 Then:
P(3,3|fair) = P(3|fair)*P(3|fair)
= (0.166)(0.166) = 0.027
P(3,3|loaded) = P(3|loaded)*P(3|loaded)
= (0.60)(0.60) = 0.36
 So:
P(3,3 and fair) = P(3,3|fair)*P(fair)
= (0.027)(0.05) = 0.013
P(3,3 and loaded) = P(3,3|loaded)P(loaded)
= (0.36)(0.5) = 0.18
Thus, the probability of getting two 3s is a marginal
probability obtained from the sum of the probability
of two joint probabilities:
P(3,3) = 0.013 + 0.18 = 0.193
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Further Probability
Revisions continued
 Using the probabilities from the
previous slide:
P(3,3 and Fair)
P(Fair | 3,3) =
P(3,3)
0.013
=
= 0.067
0.193
P(3,3 and Loaded)
P(Loaded | 3,3) =
P(3,3)
0.18
=
= 0.933
0.193
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Further Probability
Revisions continued
To give the final comparison:
P(fair|3) = 0.22
P(loaded|3) = 0.78
P(fair|3,3) = 0.067
P(loaded|3,3) = 0.933
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Random Variables
 Discrete random variable - can
assume only a finite or limited
set of values - i.e., the number
of automobiles sold in a year.
 Continuous random variable can assume any one of an
infinite set of values - i.e.,
temperature, product lifetime.
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Random Variables
(Numeric)
Stock 50
Xmas trees
Inspect 600
items
Send out
5,000 sales
letters
Build an
apartment
building
Test the
lifetime of a
light bulb
(minutes)
Outcome
Random Variable
Number of
trees sold
Number
acceptable
Number of
people
responding
% completed
after 4
months
Time bulb
lasts - up to
80,000
minutes
X = number of
trees sold
Y = number
acceptable
Z = number of
people responding
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R = % completed
after 4 months
S = time bulb
burns
2-45
Range of
Random
Variable
0,1,2,, 50
0,1,2,…,
600
0,1,2,…,
5,000
0R100
0S80,000
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Discrete
Experiment
Random
Variables (Non-numeric)
Experiment
Outcome
Random
Variable
Range of
Random
Variable
Students
Strongly agree (SA)
X = 5 if SA
1,2,3,4,5
respond to a Agree (A)
4 if A
questionnaire Neutral (N)
3 if N
Disagree (D)
2 if D
Strongly Disagree (SD)
1 if SD
One machine is Defective
Y = 0 if defective
0,1
inspected
Not defective
1 if not defective
Consumers
Good
Z = 3 if good
1,2,3
respond to how Average
2 if average
they like a
Poor
1 if poor
product
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Probability Distributions
Probability distribution – the set of all
possible values of a random variable
and their associated probabilities.
In a discrete probability distribution a
probability between 0 and 1 is assigned to
each discrete variable.The sum of the
probabilities sum to 1.
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Discrete Probability
Distribution Example
Dr. Shannon asked students to respond to the
statement, “The textbook was well written and
helped me acquire the necessary information.”
Shown below is the discrete probability
distribution of the respondents.
Outcome
SA
A
N
D
SD
X # Responding
5
10
4
20
3
30
2
30
1
10
P(X)
0.10
0.20
0.30
0.30
0.10
Sum of P(X) = 1.0
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Discrete Probability
Distribution Graph
A graphical display of a probability distribution
yields information about its shape, the central
tendency (expected values) and the spread of the
data (variance). Below is a graphical depiction
of Dr. Shannon’s student responses.
0.3
0.25
0.2
0.15
0.1
0.05
0
SA
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A
N
2-49
DA
SD
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Expected Value of a
Discrete Prob. Distribution
 The expected value of a discrete
probability distribution is:
n
E( X ) = 
X
i P( X i )
=1
i
For Dr. Shannon’s class:
E( X ) =
5
 X P( X
i
i)
i =1
= X 1P( X 1 )  X 2 P( X 2 )  X 3 P( X 3 )
 X 4 P( X 4 )  X 5 P( X 5 )
= (5)(0.1)  ( 4)(0.2)  (3)(0.3)
 ( 2)(0.3)  (1)(0.1)
= 2.9
Thus, the mean response to Dr. Shannon’s
question is between disagree (2) and neutral (3),
with the average being closer to neutral.
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Variance of a Discrete
Probability Distribution
 The variance of a discrete probability
distribution is:
n
s = [X i - E(X )] P(X i )
2
2
i= 1
For Dr. Shannon’s class:
s 2 = (5 - 2.9)2 (0.1)  (4 - 2.9)2 (0.2)
 (3 - 2.9) (0.3)  (2 - 2.9) 2 (0.3)
2
 (1 - 2.9) 2 (0.1)
= 0.44 - 0.242  0.003  0.243  0.361
= 1.29
Thus, the standard deviation for Dr.
Shannon’s question is 1.29 = 1.136
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-51
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Variance of a Discrete
Probability Distribution
 The standard deviation is a measure of
the dispersion or spread of the data that is
related to the variance. The formula for
the standard deviation of all probability
functions is:
s= s
2
For Dr. Shannon’s class:
s = 1.29,
2
so,
s = 1.358
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-52
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Binomial Distribution
 The binomial distribution is a
probability distribution with:
trials that follow a Bernoulli
process and have two possible
outcomes.
probabilities that stay the same
from one trial to the next.
trials that are statistically
independent.
a positive integer number of trials.
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-53
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Binomial Formulas
 The binomial formula can be
used to determine the
probability of r successes in n
trials.
n!
r n-r
=
pq
r!(n - r)!
Where,
n = number of trials
r = number of successes
p = probability of success
q = probability of failure (1-p)
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-54
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Binomial Formulas
(continued)
 For a binomial distribution, the
expected value, or mean, is:
m = np
 The variance is:
s 2 = np (1 - p)
 The standard deviation is:
s= s2
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-55
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Soda Selection:
Binomial Example
 Suppose 50% of your friends prefer
diet soda to regular soda.
 You decide to practice your new
binomial skills while studying with
five friends.
 You bring both diet and regular soda
to your next study session and offer
one to each of your friends.
 What is the probability that only one of
your friends will select a diet soda?
 What is the probability that three of your
friends will select the diet soda?
 What is the expected value, variance, and
standard deviation of your experiment?
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-56
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Soda Selection Solution
What is the probability that only one of your
friends will select a diet soda?
= .1563
What is the probability that three of your friends
will select the diet soda?
= .3125
These questions can be answered using the
binomial formula, where n = 5, r = 1 then 3, p = .5
and q = .5.
Below is a graphical depiction of the answers.
0.40
P(r=3)
P(r)
0.30
P(r=1)
0.20
0.10
0.00
0
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
1
2
3
(r) Number of Successes
2-57
4
5
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Soda Selection Solution
– Binomial Table
Now let’s answer these same
questions using the binomial table:
p= 50% success
n
r
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
1
0
0.9500
0.9000
0.8500
0.8000
0.7500
0.7000
0.6500
0.6000
0.5500
0.5000
1
0.0500
0.1000
0.1500
0.2000
0.2500
0.3000
0.3500
0.4000
0.4500
0.5000
0
0.9025
0.8100
0.7225
0.6400
0.5625
0.4900
0.4225
0.3600
0.3025
0.2500
1
0.0950
0.1800
0.2550
0.3200
0.3750
0.4200
0.4550
0.4800
0.4950
0.5000
2
0.0025
0.0100
0.0225
0.0400
0.0625
0.0900
0.1225
0.1600
0.2025
0.2500
0
0.8574
0.7290
0.6141
0.5120
0.4219
0.3430
0.2746
0.2160
0.1664
0.1250
1
0.1354
0.2430
0.3251
0.3840
0.4219
0.4410
0.4436
0.4320
0.4084
0.3750
2
0.0071
0.0270
0.0574
0.0960
0.1406
0.1890
0.2389
0.2880
0.3341
0.3750
3
0.0001
0.0010
0.0034
0.0080
0.0156
0.0270
0.0429
0.0640
0.0911
0.1250
0
0.8145
0.6561
0.5220
0.4096
0.3164
0.2401
0.1785
0.1296
0.0915
0.0625
1
0.1715
0.2916
0.3685
0.4096
0.4219
0.4116
0.3845
0.3456
0.2995
0.2500
2
0.0135
n= 5 friends
0.0486
0.0975
0.1536
0.2109
0.2646
0.3105
0.3456
0.3675
0.3750
3
0.0005
0.0036
0.0115
0.0256
0.0469
0.0756
0.1115
0.1536
0.2005
0.2500
4
0.0000
0.0001
0.0005
0.0016
0.0039
0.0081
0.0150
0.0256
0.0410
0.0625
0
0.7738
0.5905
0.4437
0.3277
0.2373
0.1681
0.1160
0.0778
0.0503
0.0313
1
0.2036
0.3281
0.3915
0.4096
0.3955
0.3602
0.3124
0.2592
0.2059
0.1563
2
0.0214
0.0729
0.1382
0.2048
0.2637
0.3087
0.3364
0.3456
0.3369
0.3125
3
0.0011
0.0081
0.0244
0.0512
0.0879
0.1323
0.1811
0.2304
0.2757
0.3125
4
0.0000
0.0005
0.0022
0.0064
0.0146
0.0284
0.0488
0.0768
0.1128
0.1563
5
0.0000
0.0000
0.0001
0.0003
0.0010
0.0024
0.0053
0.0102
0.0185
0.0313
0
0.7351
0.5314
0.3771
0.2621
0.1780
0.1176
0.0754
0.0467
0.0277
0.0156
1
0.2321
0.3543
0.3993
0.3932
0.3560
0.3025
0.2437
0.1866
0.1359
0.0938
2
0.0305
0.0984
0.1762
0.2458
0.2966
0.3241
0.3280
0.3110
0.2780
0.2344
3
0.0021
0.0146
0.0415
0.0819
0.1318
0.1852
0.2355
0.2765
0.3032
0.3125
0.0154
2-580.0330
© 20060.0951
by Prentice
Hall, Inc.,
0.0595
0.1382
0.1861
0.2344
0.0015
0.0044
0.0102
0.0205
0.0369
0.0609
0.0938
0.0001
0.0002
0.0007
0.0018
0.0041
0.0083
0.0156
2
3
4
5
6
To accompany
Quantitative
Analysis
4
0.0001
0.0012
0.0055
for Management, 9e
5
0.0000
0.0001
0.0004
by Render/Stair/Hanna
6
0.0000
0.0000
0.0000
Upper Saddle River, NJ 07458
Soda Selection Solution
continued
What is the expected value, variance, and
standard deviation of your experiment?
expected value (u) = np
= 5(.5) = 2.5
variance = np(1-p)
= 5(.5)(.5) = 1.25
standard deviation = variance
= 1.25 = 1.118
So, on average half of your five friends will select
diet soda – this is intuitive because 50% of your
friends prefer diet soda.
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-59
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Normal Distribution
 The normal distribution is the
most popular and useful
continuous probability
distribution.
Specified completely by the mean
and standard deviation
Symmetrical, with the midpoint
representing the mean
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.
To accompany Quantitative Analysis
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by Render/Stair/Hanna
2-60
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Normal Distribution
continued
Normal Distribution
Probability density function - f(X)
5
5.05
5.1
f (X ) =
5.15
5.2
5.3
5.35
-1 / 2 ( X - m ) 2
1
s 2
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
5.25
2-61
e
s2
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
5.4
Normal Distribution for
Different Values of m
Different values of the mean shift the curve, but
do not affect the shape of the distribution.
m=40
m=50
m=60
40
50
60
0
30
To accompany Quantitative Analysis
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by Render/Stair/Hanna
2-62
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
70
Normal Distribution for
Different Values of s
Different values of the standard deviation will
flatten the curve, but do not affect the mean.
m=1
s=0.1
s=0.3
0
0.5
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
s=0.2
1
2-63
1.5
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
2
Three Common Areas
under the Curve
Three
commonly used
areas under the
normal curve
are +/- 1, 2 and
3 standard
deviations.
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-64
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Relationship Between
Z and X
m=100
s=15
x-m
Z=
s
55
70
85
100
115
130
145
-3
-2
-1
0
1
2
3
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-65
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Example
Haynes Construction Company builds primarily
triplex and quadraplex apartment buildings for
investors, and it is believed that the total
construction time follows a normal distribution.
The mean time to construct a triplex is 100
days, and the standard deviation is 20 days.
Failure to complete the construction in 125
days results in penalty fees. However, early
completion of 75 days or less results in a
bonus.
What is the probability that Haynes will pay a
penalty fee?
What is the probability that Haynes will receive
a bonus?
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-66
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Solution
What is the probability Haynes will pay a
penalty fee?
Z=
x-m
= (125 – 100) / 20
s
= 1.25
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-67
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Standard Normal
Table
There is an 89.44% chance the construction will
be complete in 125 days or less, thus, there is a
1-.89435, or a 10.57% chance the construction
will take longer causing Haynes to pay a penalty!
Z
0
0.01
0.02
0.03
0.04
0.05
0.06
0.0
0.50000
0.50399
0.50798
0.51197
0.51595
0.51994
0.52392
0.1
0.53983
0.54380
0.54776
0.55172
0.55567
0.55962
0.56356
0.2
0.57926
0.58317
0.58706
0.59095
0.59483
0.59871
0.60257
0.3
0.61791
0.62172
0.62552
0.62930
0.63307
0.63683
0.64058
0.4
0.65542
0.65910
0.66276
0.66640
0.67003
0.67364
0.67724
0.5
0.69146
0.69497
0.69847
0.70194
0.70540
0.70884
0.71226
0.6
0.72575
0.72907
0.73237
0.73565
0.73891
0.74215
0.74537
0.7
0.75804
0.76115
0.76424
0.76730
0.77035
0.77337
0.77637
0.8
0.78814
0.79103
0.79389
0.79673
0.79955
0.80234
0.80511
0.9
0.81594
0.81859
0.82121
0.82381
0.82639
0.82894
0.83147
1.0
0.84134
0.84375
0.84614
0.84849
0.85083
0.85314
0.85543
1.1
0.86433
0.86650
0.86864
0.87076
0.87286
0.87493
0.87698
1.2
0.88493
0.88686
0.88877
0.89065
0.89251
0.89435
0.89617
1.3
0.90320
0.90490
0.90658
0.90824
0.90988
0.91149
0.91308
1.4
0.91924
0.92073
0.92220
0.92364
0.92507
0.92647
0.92785
To accompany Quantitative Analysis
0.93319
0.93448
for1.5
Management,
9e
by Render/Stair/Hanna
0.935742-68
0.93699
© 2006 by Prentice Hall, Inc.,
0.93822
0.93943
Upper Saddle
River, NJ 0.94062
07458
Haynes Construction
Company Solution
What is the probability Haynes will pay a
penalty fee?
Z=
x-m
s
= (75 – 100) / 20
= -1.25
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-69
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Standard Normal
Table
The standard normal table does not have negative
value, so we must look up the positive value and
subtract from one ~ this works because of the
symmetrical property of the normal. Thus, there is a
1-.89435, or a 10.57% chance the construction will
take less than 75 days and Haynes will get a bonus!
Z
0
0.01
0.02
0.03
0.04
0.05
0.06
0.0
0.50000
0.50399
0.50798
0.51197
0.51595
0.51994
0.52392
0.1
0.53983
0.54380
0.54776
0.55172
0.55567
0.55962
0.56356
0.2
0.57926
0.58317
0.58706
0.59095
0.59483
0.59871
0.60257
0.3
0.61791
0.62172
0.62552
0.62930
0.63307
0.63683
0.64058
0.4
0.65542
0.65910
0.66276
0.66640
0.67003
0.67364
0.67724
0.5
0.69146
0.69497
0.69847
0.70194
0.70540
0.70884
0.71226
0.6
0.72575
0.72907
0.73237
0.73565
0.73891
0.74215
0.74537
0.7
0.75804
0.76115
0.76424
0.76730
0.77035
0.77337
0.77637
0.8
0.78814
0.79103
0.79389
0.79673
0.79955
0.80234
0.80511
0.9
0.81594
0.81859
0.82121
0.82381
0.82639
0.82894
0.83147
1.0
0.84134
0.84375
0.84614
0.84849
0.85083
0.85314
0.85543
1.1
0.86433
0.86650
0.86864
0.87076
0.87286
0.87493
0.87698
1.2
0.88493
0.88686
0.88877
0.89065
0.89251
0.89435
0.89617
1.3
0.90320
0.90490
0.90658
0.90824
0.90988
0.91149
0.91308
1.4
0.91924
0.92073
0.92220
0.92364
0.92507
0.92647
0.92785
To accompany Quantitative Analysis
0.93319
0.93448
for1.5
Management,
9e
by Render/Stair/Hanna
0.935742-70
0.93699
© 2006 by Prentice Hall, Inc.,
0.93822
0.93943
Upper Saddle
River, NJ 0.94062
07458
Haynes Construction
Company Example
Other questions, such as the probability that the
construction will be completed between 110 and
125 days can also be answered.
Z=
x-m
s
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
= (125 – 100) / 20 = 1.25
and
= (110 – 100) / 20 = .5
2-71
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Example
The area under the curve must be found for
both z values and subtracted from one another.
Z (.5) = .69146
Z (1.25) = .89435
P(.5 < Z < 1.25) = .89435 - .69146
= .20289 or 20.29%
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
2-72
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Negative
Exponential Distribution
 The negative exponential distribution is
a continuous distribution that is often
used to describe the time required to
service a customer.
6
f ( X ) = me
5
m=5
4
- mx
Expected value = 1/m
Variance = 1/m2
3
2
1
0
0
0.2
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
0.4
0.6
2-73
0.8
1
1.2
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Poisson Distribution
 The Poisson distribution is a
discrete distribution that is often
used to describe arrival rates.
=2
0.30
P( X ) =
0.25
0.20
e
x -
X!
Expected value = 
Variance = 
0.15
0.10
0.05
0.00
1
2
To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna
3
4
2-74
5
6
7
8
9
© 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458