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Chapter 2
To accompanyQuantitative 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 2-٢
Learning Objectives
1. Understand the basic foundations of probability analysis.
2. Describe statistically dependent and independent events.
3. Describe and provide examples of both discrete and continuous random variables.
4. Explain the difference between discrete and continuous probability distributions.
5. Calculate expected values and variances and use the normal table.
After completing this chapter, students will be able to:After completing this chapter, students will be able to:
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Chapter Outline
2.1 Introduction2.2 Fundamental Concepts2.8 Random Variables2.9 Probability Distributions2.10 The Binomial Distribution2.11 The Normal Distribution2.12 The F Distribution2.13 The Exponential Distribution2.14 The Poisson Distribution
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤
Introduction
� Life is uncertain; we are not sure what the future will bring.
�� ProbabilityProbability is a numerical statement about the likelihood that an event will occur.
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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.
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Random Variables
Discrete random variablesDiscrete random variables can assume only a finite or limited set of values.
Continuous random variablesContinuous random variables can assume any one of an infinite set of values.
A random variablerandom variable assigns a real number to every possible outcome or event in an experiment.
X = number of refrigerators sold during the day
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Random Variables – Numbers
EXPERIMENT OUTCOMERANDOM
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
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Random Variables – Not Numbers
EXPERIMENT OUTCOME RANDOM VARIABLESRANGE OF RANDOM
VARIABLES
Students respond to a questionnaire
Strongly agree (SA)Agree (A)Neutral (N)Disagree (D)Strongly disagree (SD)
5 if SA4 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
GoodAveragePoor
3 if good….Z = 2 if average
1 if poor…..
1, 2, 3
Table 2.5
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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 variablesdiscrete random variables a probability is assigned to each event.
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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
AP6
Slide 10
AP6 11e Table 2.6 does not have the Outcome column because it's about quiz scoresAnnie Puciloski; 01/02/2011
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Expected Value of a Discrete Probability Distribution
(((( )))) (((( ))))∑∑∑∑====
====n
i
ii XPXXE1
(((( )))) )(...)( 2211 nn XPXXPXXPX ++++++++++++====
The expected value is a measure of the central central tendencytendency of the distribution and is a weighted average of the values of the random variable.
where
iX)( iXP
∑∑∑∑====
n
i 1
)(XE
= random variable’s possible values
= probability of each of the random variable’s possible values
= summation sign indicating we are adding all npossible values
= expected value or mean of the random sample AP7
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Expected Value of a Discrete Probability Distribution
(((( )))) (((( ))))∑∑∑∑====
====n
i
ii XPXXE1
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:
Slide 11
AP7 Text has random "variable" not "sample"; see p. 35Annie Puciloski; 01/02/2011
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Variance of a Discrete Probability Distribution
For a discrete probability distribution the variance can be computed by
)()]([∑∑∑∑====
−−−−========n
i
ii XPXEX1
22 Varianceσ
where
iX)(XE
)( iXP
= 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
)]([ XEX i −−−−
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Variance of a Discrete Probability Distribution
For Dr. Shannon’s class:
)()]([variance5
1
2
∑∑∑∑====
−−−−====i
ii XPXEX
++++−−−−++++−−−−==== ).().().().(variance 2092410925 22
++++−−−−++++−−−− ).().().().( 3092230923 22
).().( 10921 2−−−−
291
36102430003024204410
.
.....
====
++++++++++++++++====
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Variance of a Discrete Probability Distribution
A related measure of dispersion is the standard deviation.
2σVarianceσ ========
where
σ
= square root
= standard deviation
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Variance of a Discrete Probability Distribution
A related measure of dispersion is the standard deviation.
2σVarianceσ ========
where
σ
= square root
= standard deviation
For Dr. Shannon’s class:Varianceσ ====
141291 .. ========
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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).
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Probability Distribution of a Continuous Random Variable
Pro
bab
ilit
y
| | | | | | |
5.06 5.10 5.14 5.18 5.22 5.26 5.30
Weight (grams)
Figure 2.6
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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.
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The Binomial Distribution
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 letr = number of successes
q = 1 – p = the probability of a failure
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The Binomial Distribution
The binomial formula is:
rnrqprnr
nnr −−−−
−−−−====
)!(!
!trials in successes of yProbabilit
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
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The Binomial Distribution
NUMBER OFHEADS (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.
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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
Thus454
5.05.0)!45(!4
!5) trials5in successes 4(
−
−==P
156250500625011234
12345.).)(.(
)!)()()((
))()()((========
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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 5So
We could use the formula to solve this problem, We could use the formula to solve this problem, but using the table is easier.but using the table is easier.
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Solving Problems with Binomial Tables
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.
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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
Solving Problems with Binomial Tables
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
)()()()( 543defects more or 3 PPPP ++++++++====
02670000100022002440 .... ====++++++++====
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Solving Problems with Binomial Tables
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:
6375085015051Variance
7501505value Expected
.).)(.()(
.).(
========−−−−====
============
pnp
np
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The Normal Distribution
The normal distributionnormal 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
1σσσσ
µµµµ
ππππσσσσ
)(
)(
−−−−−−−−
====
x
eXf
� The normal distribution is specified completely when we know the mean, µ, and the standard deviation, σσσσ .
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The Normal Distribution
� 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.
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The Normal Distribution
| | |
40 µ = 50 60
| | |
µ = 40 50 60
Smaller µ, same σσσσ
| | |
40 50 µ = 60
Larger µ, same σσσσ
Figure 2.8
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µ
The Normal Distribution
Figure 2.9
Same µ, smaller σσσσ
Same µ, larger σσσσ
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The Empirical Rule
For a normally distributed random variable with mean µ and standard deviation σσσσ , then
1. About 68% of values will be within �1σσσσ of the mean.
2. About 95.4% of values will be within �2σσσσ of the mean.
3. About 99.7% of values will be within �3σσσσ of the mean.
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The Empirical Rule
Figure 2.14
68%16% 16%
–1σσσσ +1σσσσa µ b
95.4%2.3% 2.3%
–2σσσσ +2σσσσa µ b
99.7%0.15% 0.15%
–3σσσσ +3σσσσa µ b
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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
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Fα
The F Distribution
Figure 2.15
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The F Distribution
df1 = 5
df2 = 6
αααα = 0.05
Consider the example:
From Appendix D, we get
Fαααα, 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.
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The F Distribution
Figure 2.16
F = 4.39
0.05
F value for 0.05 probability with 5 and 6 degrees of freedom
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The Exponential Distribution
� The exponential distributionexponential distribution (also called the negative exponential distributionnegative 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:
xeXf
µµµµµµµµ −−−−====)(
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)
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The Exponential Distribution
time service Average1
value Expected ========µµµµ
2
1Variance
µµµµ====
f(X)
X
Figure 2.17
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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?
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Arnold’s Muffler Shop
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
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Arnold’s Muffler Shop
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
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Arnold’s Muffler Shop
Probability That the Mechanic Will Install a Muffler in 0.5 Hour
Figure 2.18
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The Poisson Distribution
� The PoissonPoisson distributiondistribution is a discretediscretedistribution that is often used in queuing models to describe arrival rates over time. Its probability function is given by:
!)(
X
eXP
x λλλλλλλλ −−−−
====
where
P(X) = probability of exactly X arrivals or occurrences
λλλλ = 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
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The Poisson Distribution
The mean and variance of the distribution are both λλλλ.
Expected value = λλλλ
Variance = λλλλ
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Poisson Distribution
We can use Appendix C to find Poisson probabilities.Suppose that λ = 2. Some probability calculations are:
2706.02
)1353.0(4
!2
2)2(
2706.01
)1353.0(2
!1
2)1(
1353.01
)1353.0(1
!0
2)0(
!)(
22
21
20
===
===
===
=
−
−
−
−
eP
eP
eP
X
eXP
x λλ
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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).
