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Quantitative Analysis for Management

Thirteenth Edition

Chapter 2

Probability Concepts and Applications

Learning Outcomes

Estimate an unknown population proportion with the standard normal distribution.

THE NORMAL DISTRIBUTION (1 OF 4)

One of the most popular and useful continuous probability distributions

The probability density function

Completely specified by the mean, μ, and the standard deviation, σ

The Normal Distribution (2 of 4)

FIGURE 2.7 Normal Distribution with Different Values for μ

The Normal Distribution (3 of 4)

FIGURE 2.8 Normal Distribution with Different Values for σ

The Normal Distribution (4 of 4)

Symmetrical with the midpoint representing the mean

Shifting the mean does not change the shape

Values on the X axis measured in the number of standard deviations away from the mean

As standard deviation becomes larger, curve flattens

As standard deviation becomes smaller, curve becomes steeper

Using the Standard Normal Table (1 of 4)

Step 1

Convert the normal distribution into a standard normal distribution

Mean of 0 and a standard deviation of 1

The new standard random variable is Z

where

X = value of the random variable we want to measure

μ = mean of the distribution

σ = standard deviation of the distribution

Z = number of standard deviations from X to the mean, μ

Using the Standard Normal Table (2 of 4)

For μ = 100, σ = 15, find the probability that X is less than 130

FIGURE 2.9 Normal Distribution Showing the Relationship Between Z Values and X Values

Using the Standard Normal Table (3 of 4)

Step 2

Look up the probability from a table of normal curve areas

Use Appendix A or Table 2.10

Column on the left is Z value

Row at the top has second decimal places for Z values

Using the Standard Normal Table (4 of 4)

TABLE 2.10 (partial) Standardized Normal Distribution Function

AREA UNDER THE NORMAL CURVE

Z	0.00	0.01	0.02	0.03
1.8	0.96407	0.96485	0.96562	0.96638
1.9	0.97128	0.97193	0.97257	0.97320
2.0	0.97725	0.97784	0.97831	0.97882
2.1	0.98214	0.98257	0.98300	0.98341
2.2	0.98610	0.98645	0.98679	0.98713

For Z = 2.00

P(X < 130) = P(Z < 2.00) = 0.97725

P(X > 130) = 1 − P(X ≤ 130) = 1 − P(Z ≤ 2)

= 1 − 0.97725 = 0.02275

Haynes Construction Company (1 of 7)

Builds three- and four-unit apartment buildings

Total construction time follows a

normal distribution

For triplexes, μ = 100 days

and σ = 20 days

Contract calls for

completion in 125 days

Late completion will

incur a severe

penalty fee

Probability of completing in 125 days?

FIGURE 2.10 Normal Distribution for Haynes Construction

Haynes Construction Company (2 of 7)

Compute Z

From Appendix A, for Z = 1.25

area = 0.89435

FIGURE 2.10 Normal Distribution for Haynes Construction

Haynes Construction Company (3 of 7)

Compute Z

From Appendix A, for Z = 1.25

area = 0.89435

FIGURE 2.10 Normal Distribution for Haynes Construction

The probability is about 0.89

that Haynes will not violate the contract

Haynes Construction Company (4 of 7)

If finished in 75 days or less, bonus = $5,000

Probability of bonus?

Because the distribution is symmetrical, equivalent to Z = 1.25

so area = 0.89435

FIGURE 2.11 Probability That Haynes Will Receive the Bonus by Finishing in 75 Days or Less

Haynes Construction Company (5 of 7)

If finished in 75 days or less, bonus = $5,000

Probability of bonus?

Because the distribution is symmetrical, equivalent to Z = 1.25

so area = 0.89435

FIGURE 2.11 Probability That Haynes Will Receive the Bonus by Finishing in 75 Days or Less

P(X > 125) = 1.0 − P(X ≤ 125)

= 1.0 − 0.89435 = 0.10565

The probability of completing the contract in 75 days or less is about 11%

Haynes Construction Company (6 of 7)

Probability of completing between 110 and 125 days?

P(110 < X < 125) = P(X ≤ 125) − P(X < 110)

P(X ≤ 125) = 0.89435

For Z = 0.5 area = 0.69146

FIGURE 2.12 Probability That Haynes Will Complete in 110 to 125 Days

Haynes Construction Company (7 of 7)

Probability of completing between 110 and 125 days?

P(110 < X < 125) = P(X ≤ 125) − P(X < 110)

P(X ≤ 125) = 0.89435

For Z = 0.5 area = 0.69146

FIGURE 2.12 Probability That Haynes Will Complete in 110 to 125 Days

P(110 ≤ X < 125) = 0.89435 − 0.69146

= 0.20289

The probability of completing between 110 and 125 days is about 20%

Standard Normal Distribution

TABLE 2.10 (partial) Standardized Normal Distribution Function

AREA UNDER THE NORMAL CURVE

Z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.5	.69146	.69497	.69847	.70194	.70540	.70884	.71226	.71566	.71904	.72240
0.6	.72575	.72907	.73237	.73536	.73891	.74215	.74537	.74857	.75175	.75490
0.7	.75804	.76115	.76424	.76730	.77035	.77337	.77637	.77935	.78230	.78524
0.8	.78814	.79103	.79389	.79673	.79955	.80234	.80511	.80785	.81057	.81327
0.9	.81594	.81859	.82121	.82381	.82639	.82894	.83147	.83398	.83646	.83891
1.0	.84134	.84375	.84614	.84849	.85083	.85314	.85543	.85769	.85993	.86214
1.1	.86433	.86650	.86864	.87076	.87286	.87493	.87698	.87900	.88100	.88298
1.2	.88493	.88686	.88877	.89065	.89251	.89435	.89617	.89796	.89973	.90147
1.3	.90320	.90490	.90658	.90824	.90988	.91149	.91309	.91466	.91621	.91774
1.4	.91924	.92073	.92220	.92364	.92507	.92647	.92785	.92922	.93056	.93189
1.5	.93319	.93448	.93574	.93699	.93822	.93943	.94062	.94179	.94295	.94408

Using Excel

PROGRAM 2.3A Excel 2016 Output for the Normal Distribution Example

PROGRAM 2.3B Function in an Excel 2016 Spreadsheet for the Normal Distribution Example

The Empirical Rule (1 of 2)

For a normally distributed random variable with mean μ and standard deviation σ

Approximately 68% of values will be within ±1σ of the mean

Approximately 95% of values will be within ±2σ of the mean

Almost all (99.7%) of values will be within ±3σ of the mean

The Empirical Rule (2 of 2)

FIGURE 2.13 Approximate Probabilities from the Empirical Rule

THE F DISTRIBUTION (1 OF 4)

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

df1 = degrees of freedom for the numerator

df2 = degrees of freedom for the denominator

The probabilities of large values of F are very small

The F Distribution (2 of 4)

FIGURE 2.14 The F Distribution

The F Distribution (3 of 4)

Consider the example

df1 = 5

df2 = 6

 = 0.05

From Appendix D, we get

F, df1, df2 = F0.05, 5, 6 = 4.39

This means

P(F > 4.39) = 0.05

The probability is only 0.05 F will exceed 4.39

The F Distribution (4 of 4)

FIGURE 2.15 F Value for 0.05 Probability with 5 and 6 Degrees of Freedom

Using Excel

PROGRAM 2.4A Excel 2016 Output for the F Distribution

PROGRAM 2.4B Functions in an Excel 2016 Spreadsheet for the F Distribution

The Exponential Distribution (1 of 2)

Also called the negative exponential distribution

A continuous distribution often used in queuing models

Probability function given by

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)

The Exponential Distribution (2 of 2)

FIGURE 2.16 Exponential Distribution

Arnold’s Muffler Shop (1 of 3)

Installs new mufflers on automobiles and small trucks

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?

X = Exponentially distributed service time

μ = average number of units the served per time period = 3 per hour

t = ½ hour = 0.5 hour

P(X ≤ 0.5) = 1 − e−3(0.5) = 1 − e −1.5 = 1 = 0.2231 = 0.7769

Arnold’s Muffler Shop (2 of 3)

FIGURE 2.17 Probability That the Mechanic Will Install a Muffler in 0.5 Hour

Arnold’s Muffler Shop (3 of 3)

Similarly

And

P(X > 0.5) = 1 − P(X ≤ 0.5) = 1 − 0.7769 = 0.2231

Using Excel

PROGRAM 2.5A Excel 2016 Output for the Exponential Distribution

PROGRAM 2.5B Function in an Excel 2016 Spreadsheet for the Exponential Distribution

The Poisson Distribution (1 of 3)

A discrete probability distribution

Often used in queuing models to describe arrival rates over time

Probability function given by

where

P(X) = probability of exactly X arrivals or occurrences

 = average number of arrivals per unit of time (the mean arrival rate)

e = 2.718, the base of natural logarithms

X = number of occurrences (0, 1, 2, 3, …)

The Poisson Distribution (2 of 3)

From Appendix C for λ = 2

The Poisson Distribution (3 of 3)

FIGURE 2.18 Sample Poisson Distributions with λ = 2 and λ = 4

Using Excel

PROGRAM 2.6A Excel 2016 Output for the Poisson Distribution

PROGRAM 2.6B Functions in an Excel 2016 Spreadsheet for the Poisson Distribution

IN-CLASS EXERCISE

Summary

Estimated an unknown population proportion with the standard normal distribution.

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125 – 100

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125 – 100

==1.25

75 – 100

–25

–1.25

110 – 100

0.5

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fXe

Expected value = = Average service time

Variance =

(X)1

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