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Probability Concepts and Applications

To accompany Quantitative Analysis for Management, Twelfth Edition,

by Render, Stair, Hanna and Hale

Power Point slides created by Jeff Heyl

After completing this chapter, students will be able to:

LEARNING OBJECTIVES

2 – 2

Understand the basic foundations of probability analysis.

Describe statistically dependent and independent events.

Use Bayes’ theorem to establish posterior probabilities.

Describe and provide examples of both discrete and continuous random variables.

Explain the difference between discrete and continuous probability distributions.

Calculate expected values and variances and use the normal table.

2 – 3

2.1 Introduction

2.2 Fundamental Concepts

2.3 Revising Probabilities with Bayes’ Theorem

2.4 Further Probability Revisions

2.5 Random Variables

2.6 Probability Distributions

2.7 The Binomial Distribution

2.8 The Normal Distribution

2.9 The F Distribution

2.10 The Exponential Distribution

2.11 The Poisson Distribution

CHAPTER OUTLINE

Introduction

Life is uncertain; we are not sure what the future will bring

Probability is a numerical statement about the likelihood that an event will occur

2 – 4

Chapters in This Book That Use Probability

2 – 5

CHAPTER	TITLE
3	Decision Analysis
4	Regression Models
5	Forecasting
6	Inventory Control Models
11	Project Management
12	Waiting Lines and Queuing Theory Models
13	Simulation Modeling
14	Markov Analysis
15	Statistical Quality Control
Module 3	Decision Theory and the Normal Distribution
Module 4	Game Theory

TABLE 2.1

Types of Probability

Objective Approach

Relative frequency approach

2 – 6

P (event) =

Number of occurrences of the event

Total number of trials or outcomes

P (head) =

Number of ways of getting a head

Number of possible outcomes (head or tail)

Classical or logical method

P (spade) =

Number of chances of drawing a spade

Number of possible outcomes

Diversey Paint Example

Historical demand for white latex paint at = 0, 1, 2, 3, or 4 gallons per day

Observed frequencies over the past 200 days

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QUANTITY DEMANDED (GALLONS)	NUMBER OF DAYS			PROBABILITY
0		40		0.20 (= 40/200)
1		80		0.40 (= 80/200)
2		50		0.25 (= 50/200)
3		20		0.10 (= 20/200)
4		10		0.05 (= 10/200)
	Total	200	Total	1.00 (= 200/200)

TABLE 2.2

Historical demand for white latex paint at = 0, 1, 2, 3, or 4 gallons per day

Observed frequencies over the past 200 days

Diversey Paint Example

2 – 8

QUANTITY DEMANDED (GALLONS)	NUMBER OF DAYS			PROBABILITY
0		40		0.20 (= 40/200)
1		80		0.40 (= 80/200)
2		50		0.25 (= 50/200)
3		20		0.10 (= 20/200)
4		10		0.05 (= 10/200)
	Total	200	Total	1.00 (= 200/200)

TABLE 2.2

Individual probabilities are all between 0 and 1

0 ≤ P (event) ≤ 1

Total of all event probabilities equals 1

∑ P (event) = 1.00

Types of Probability

Subjective Approach

Based on the experience and judgment of the person making the estimate

Opinion polls

Judgment of experts

Delphi method

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Mutually Exclusive and Collectively Exhaustive Events

Events are said to be mutually exclusive if only one of the events can occur on any one trial

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Tossing a coin will result in either a head or a tail

Rolling a die will result in only one of six possible outcomes

Mutually Exclusive and Collectively Exhaustive Events

Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome

2 – 11

Both heads and tails as possible outcomes of coin flips

All six possible outcomes of the roll of a die

OUTCOME OF ROLL	PROBABILITY
1		1/6
2		1/6
3		1/6
4		1/6
5		1/6
6		1/6
		Total 1

Venn Diagrams

2 – 12

Events that are mutually exclusive

FIGURE 2.1

FIGURE 2.2

Events that are not mutually exclusive

Drawing a Card

Draw one card from a deck of 52 playing cards

A = event that a 7 is drawn

B = event that a heart is drawn

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P (a 7 is drawn) = P(A)= 4/52 = 1/13

P (a heart is drawn) = P(B) = 13/52 = 1/4

These two events are not mutually exclusive since a 7 of hearts can be drawn

These two events are not collectively exhaustive since there are other cards in the deck besides 7s and hearts

Differences

2 – 14

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

Unions and Intersections of Events

Intersection – the set of all outcomes that are common to both events

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Intersection of event A and event B = A and B

= A ∩ B

= AB

Probability notation

P(Intersection of event A and event B) = P(A and B)

= P(A ∩ B)

= P(AB)

Sometimes called joint probability

Unions and Intersections of Events

Union – the set of all outcomes that are contained in either of two events

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Union of event A and event B = A or B

Probability notation

P(Union of event A and event B) = P(A or B)

= P(A ∪ B)

Unions and Intersections of Events

In the previous example

Intersection of event A and event B

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Union of event A and event B

(A and B)	= the 7 of hearts is drawn
P(A and B)	= P(7 of hearts is drawn) = 1/52

(A or B)	= either a 7 or a heart is drawn
P(A or B)	= P(any 7 or any heart is drawn) = 16/52

Probability Rules

General rule for union of two events, additive rule

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P(A or B) = P(A) + P(B) – P(A and B)

Union of two events, a 7 or a heart

P(A or B) = P(A) + P(B) – P(A and B)

= 4/52 + 13/52 – 1/52

= 16/52

Probability Rules

Conditional probability – probability that an event occurs given another event has already happened

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P(A | B) =

P(AB)

P(B)

P(AB) = P(A | B) P(B)

Probability of a 7 given a heart has been drawn

P(A | B) = = = 1/13

P(AB)

P(B)

1/52

13/52

Probability Rules

Independent one event has no effect on the other event

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

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Which sets are independent?

Dependent events

Independent events

Dependent events

Independent events

Independent one event has no effect on the other event

Probability Rules

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P(A | B) = P(A)

P(A and B) = P(A)P(B)

For a fair coin tossed twice

A = event that a head is the result of the first toss

B = event that a head is the result of the second toss

P(A) = 0.5 and P(B) = 0.5

P(AB) = P(A)P(B) = 0.5(0.5) = 0.25

Independent Events

A bucket contains 3 black balls and 7 green balls

Draw a ball from the bucket, replace it, and draw a second ball

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The probability of a black ball drawn on first draw is: P(B) = 0.30

The probability of two green balls drawn is: P(GG) = P(G) x P(G) = 0.7 x 0.7 = 0.49

Independent Events

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A bucket contains 3 black balls and 7 green balls

Draw a ball from the bucket, replace it, and draw a second ball

The probability of a black ball drawn on the second draw if the first draw is green is: P(B | G) = P(B) = 0.30

The probability of a green ball drawn on the second draw if the first draw is green is: P(G | G) = P(G) = 0.70

Dependent Events

An urn contains the following 10 balls:

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)

2 – 24

P(WL) = 4/10 = 0.4 P(YL) = 3/10 = 0.3

P(WN) = 2/10 = 0.2 P(YN) = 1/10 = 0.1

P(W) = 6/10 = 0.6 P(L) = 7/10 = 0.7

P(Y) = 4/10 = 0.4 P(N) = 3/10 = 0.3

Dependent Events

2 – 25

4 balls White (W) and Lettered (L)

2 balls White (W) and Numbered (N)

3 balls Yellow (Y) and Lettered (L)

1 ball Yellow (Y) and Numbered (N)

Probability (WL) =

Probability (YN) =

Probability (YL) =

Probability (WN) =

The urn contains 10 balls

Dependent Events

The conditional probability that the ball drawn is lettered, given that it is yellow

2 – 26

P(L | Y) = = = 0.75

P(YL)

P(Y)

0.3

0.4

P(YL) = P(L | Y) x P(Y) = (0.75)(0.4) = 0.3

We can verify P(YL) using the joint probability formula

Revising Probabilities with Bayes’ Theorem

Bayes’ theorem is used to incorporate additional information and help create posterior probabilities from original or prior probabilities

2 – 27

Posterior Probabilities

Bayes’ Process

Prior Probabilities

New Information

FIGURE 2.3

Revising Probabilities with Bayes’ Theorem

A cup contains two dice identical in appearance but one is fair (unbiased) and the other is loaded (biased)

The probability of rolling a 3 on the fair die is 1/6 or 0.166

The probability of tossing the same number on the loaded die is 0.60

We select one by chance, toss it, and get a 3

What is the probability that the die rolled was fair?

What is the probability that the loaded die was rolled?

2 – 28

Revising Probabilities with Bayes’ Theorem

The probability of the die being fair or loaded is

P(fair) = 0.50 P(loaded) = 0.50

And that

P(3 | fair) = 0.166 P(3 | loaded) = 0.60

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P(3 and fair) = P(3 | fair) x P(fair)

= (0.166)(0.50) = 0.083

P(3 and loaded) = P(3 | loaded) x P(loaded)

= (0.60)(0.50) = 0.300

The probabilities of P(3 and fair) and P(3 and loaded) are

The probability of the die being fair or loaded is:

P(fair) = 0.50 P(loaded) = 0.50

And that

P(3 | fair) = 0.166 P(3 | loaded) = 0.60

Revising Probabilities with Bayes’ Theorem

2 – 30

P(3 and fair) = P(3 | fair) x P(fair)

= (0.166)(0.50) = 0.083

P(3 and loaded) = P(3 | loaded) x P(loaded)

= (0.60)(0.50) = 0.300

The probabilities of P(3 and fair) and P(3 and loaded) are:

The sum of these probabilities gives us the unconditional probability of tossing a 3

P(3) = 0.083 + 0.300 = 0.383

Revising Probabilities with Bayes’ Theorem

If a 3 does occur, the probability that the die rolled was the fair one is

2 – 31

P(fair | 3) = = = 0.22

P(fair and 3)

P(3)

0.083

0.383

P(loaded | 3) = = = 0.78

P(loaded and 3)

P(3)

0.300

0.383

The probability that the die was loaded is

If a 3 does occur, the probability that the die rolled was the fair one is

Revising Probabilities with Bayes’ Theorem

2 – 32

P(fair | 3) = = = 0.22

P(fair and 3)

P(3)

0.083

0.383

P(loaded | 3) = = = 0.78

P(loaded and 3)

P(3)

0.300

0.383

The probability that the die was loaded is

These are the revised or posterior probabilities for the next roll of the die

We use these to revise our prior probability estimates

Revising Probabilities with Bayes’ Theorem

2 – 33

TABLE 2.3 – Event B has occurred

TABLE 2.4 – A 3 is rolled

STATE OF NATURE	P (B \| STATE OF NATURE)	PRIOR PROBABILITY	JOINT PROBABILITY	POSTERIOR PROBABILITY
A	P(B \| A)	x P(A)	= P(B and A)	P(B and A)/P(B) = P(A \| B)
A’	P(B \| A’)	x P(A’)	= P(B and A’)	P(B and A’)/P(B) = P(A’ \| B)
			P(B)

STATE OF NATURE	P (B \| STATE OF NATURE)	PRIOR PROBABILITY	JOINT PROBABILITY	POSTERIOR PROBABILITY
Fair die	0.166	x 0.5	= 0.083	0.083 / 0.383 = 0.22
Loaded die	0.600	x 0.5	= 0.300	0.300 / 0.383 = 0.78
			P(3) = 0.383

General Form of Bayes’ Theorem

We can compute revised probabilities more directly by using

2 – 34

where

A’ = the complement of the event A; for example, if A is the event “fair die”, then A’ is “loaded die”

General Form of Bayes’ Theorem

Conditional probability –

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P(A | B) =

P(AB)

P(B)

From the previous example

Replace A with “fair die”, A’ with “loaded die”, B with “3 rolled”

P(fair die | 3 rolled)

Further Probability Revisions

Additional information from a second experiment

If you can afford it, perform experiments several times

We roll the die again and again get a 3

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Further Probability Revisions

Additional information from a second experiment

If you can afford it, perform experiments several times

We roll the die again and again get a 3

2 – 37

Further Probability Revisions

Additional information from a second experiment

If you can afford it, perform experiments several times

We roll the die again and again get a 3

2 – 38

Further Probability Revisions

After the first roll of the die

probability the die is fair = 0.22

probability the die is loaded = 0.78

2 – 39

After the second roll of the die

probability the die is fair = 0.067

probability the die is loaded = 0.933

Random Variables

A random variable assigns a real number to every possible outcome or event in an experiment

X = number of refrigerators sold during the day

Discrete random variables can assume only a finite or limited set of values

Continuous random variables can assume any one of an infinite set of values

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Random Variables

2 – 41

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

TABLE 2.5 – Random Variable that are Numbers

EXPERIMENT	OUTCOME	RANGE OF RANDOM VARIABLES	RANDOM VARIABLES
Students respond to a questionnaire	Strongly agree (SA) Agree (A) Neutral (N) Disagree (D) Strongly disagree (SD)		1, 2, 3, 4, 5
One machine is inspected	Defective Not defective		0, 1
Consumers respond to how they like a product	Good Average Poor		1, 2, 3

Random Variables

2 – 42

TABLE 2.6 – Random Variables not Numbers

Probability Distributions

For discrete random variables, probability value assigned to each event

Algebra class of 100 students

Quiz with five problems with 1 point for each correct answer

Lowest score = 1, highest score = 5

Examples follows the three rules

Events are mutually exclusive and collectively exhaustive

Individual probability values between 0 and 1

Total probability sums to 1

2 – 43

Probability Distributions

2 – 44

RANDOM VARIABLE (X) – SCORE	NUMBER		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

TABLE 2.7 – Quiz Scores

Developed using relative frequency approach

Probability Distributions

2 – 45

FIGURE 2.4 – Class Distribution

P (X)

0.4 –

0.3 –

0.2 –

0.1 –

0 –

| | | | | |

1 2 3 4 5

Probability Distributions

2 – 46

FIGURE 2.4 – Class Distribution

P (X)

0.4 –

0.3 –

0.2 –

0.1 –

0 –

| | | | | |

1 2 3 4 5

Central tendency of the distribution is the mean or expected value

Amount of variability is the variance

Expected Value of a Discrete Probability Distribution

Expected value is a measure of the central tendency of the distribution

2 – 47

where

Xi = random variable’s possible values

P(Xi) = probability of each of the random variable’s possible values

= summation sign indicating we are adding all n possible values

E(X) = expected value or mean of the random variable

Expected Value of a Discrete Probability Distribution

For the quiz scores

2 – 48

Variance of a Discrete Probability Distribution

2 – 49

where

Xi = random variable’s possible values

E(Xi) = expected value of the random variable

[Xi – E(X)] = difference between each value of the random variable and the expected value

E(X) = probability of each possible value of the random variable

Variance of a Discrete Probability Distribution

For quiz scores

2 – 50

Variance = (5 – 2.9)2(0.1) + (4 – 2.9)2(0.2) + (3 – 2.9)2(0.3) + (2 – 2.9)2(0.3) + (1 – 2.9)2(0.1)

= (2.1)2(0.1) + (1.1)2(0.2) + (0.1)2(0.3) + (–0.9)2(0.3) + (–1.9)2(0.1)

= 0.441 + 0.242 + 0.003 + 0.243 + 0.361

= 1.29

Variance of a Discrete Probability Distribution

Standard deviation is the square root of the variance

2 – 51

where

Standard deviation is the square root of the variance

Variance of a Discrete Probability Distribution

2 – 52

where

For this example

Using Excel

2 – 53

PROGRAM 2.1A – Excel Output

Using Excel

2 – 54

PROGRAM 2.1B – Excel Formulas

Probability Distribution of a Continuous Random Variable

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 represented by f (X)

2 – 55

Probability Distribution of a Continuous Random Variable

2 – 56

Probability

| | | | | | |

5.06 5.10 5.14 5.18 5.22 5.26 5.30

Weight (grams)

FIGURE 2.5 – Sample Density Function

The Binomial Distribution

Many business experiments can be characterized by the Bernoulli process

The Bernoulli process is described by the binomial probability distribution

Each trial has only two possible outcomes

The probability of each outcome stays the same from one trial to the next

The trials are statistically independent

The number of trials is a positive integer

2 – 57

The Binomial Distribution

The binomial distribution is used to find the probability of a specific number of successes in n trials

2 – 58

We need to know

n = number of trials

p = the probability of success on any single trial

We let

r = number of successes

q = 1 – p = the probability of a failure

The Binomial Distribution

The binomial formula is

2 – 59

Probability of r success in n trials

The symbol ! means factorial, and n! = n(n – 1)(n – 2)…(1)

4! = (4)(3)(2)(1) = 24

Also, 1! = 1 and 0! = 0 by definition

Solving Problems with the Binomial Formula

Find the probability of getting 4 heads in 5 tosses of a coin

2 – 60

n = 5, r = 4, p = 0.5, and q = 1 – 0.5 = 0.5

P(4 success in 5 trials)

Solving Problems with the Binomial Formula

2 – 61

NUMBER OF

HEADS (r)

PROBABILITY = (0.5)r(0.5)5 – r

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

0!(5 – 0)!

1!(5 – 1)!

2!(5 – 2)!

3!(5 – 3)!

4!(5 – 4)!

5!(5 – 5)!

TABLE 2.8 – Binomial Distribution for n = 5, p = 0.50

Solving Problems with the Binomial Formula

2 – 62

Probability P (r)

| | | | | | |

1 2 3 4 5 6

Values of r (number of successes)

0.4 –

0.3 –

0.2 –

0.1 –

0 –

FIGURE 2.6 – Binomial Distribution for n = 5, p = 0.50

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?

2 – 63

n = 5, p = 0.15, and r = 3, 4, or 5

Solving Problems with Binomial Tables

2 – 64

		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 2.9 (partial) – Table for Binomial Distribution

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

2 – 65

		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 2.9 (partial) – Table for Binomial Distribution

We find the three probabilities in the table for n = 5, p = 0.15, and r = 3, 4, and 5 and add them together

P(3 or more defects) = P(3) + P(4) + P(5)

= 0.0244 + 0.0022 + 0.0001

= 0.0267

Solving Problems with Binomial Tables

Expected value is

2 – 66

Expected value (mean) = np

Variance = np(1 – p)

For the MSA example

Expected value = np 5(0.15) = 0.75

Variance = np(1 – p) = 5(0.15)(0.85) = 0.6375

Using Excel

2 – 67

PROGRAM 2.2A – Excel Output

Using Excel

2 – 68

PROGRAM 2.2B – Excel Functions

The Normal Distribution

One of the most popular and useful continuous probability distributions

The probability density function

2 – 69

Completely specified by the mean, m, and the standard deviation, s

The Normal Distribution

2 – 70

FIGURE 2.7 – Normal Distribution with Different ms

| | |

40 m = 50 60

| | |

m = 40 50 60

| | |

40 50 m = 60

Smaller m, same s

Larger m, same s

The Normal Distribution

2 – 71

FIGURE 2.8 – Normal

Distribution with Different s s

Same m, smaller s

Same m, larger s

The Normal Distribution

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

2 – 72

Using the Standard Normal Table

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

2 – 73

where

X = value of the random variable we want to measure

m = mean of the distribution

s = standard deviation of the distribution

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

Using the Standard Normal Table

For m = 100, s = 15, find the probability that X is less than 130

2 – 74

FIGURE 2.9 – Normal Distribution

| | | | | | |

55 70 85 100 115 130 145

| | | | | | |

–3 –2 –1 0 1 2 3

X = IQ

m = 100

 = 15

P(X < 130)

Using the Standard Normal Table

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

2 – 75

Using the Standard Normal Table

2 – 76

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

TABLE 2.10 – Standardized Normal Distribution (partial)

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

Builds three- and four-unit apartment buildings

Total construction time follows a normal distribution

For triplexes, m = 100 days and m = 20 days

Contract calls for completion in 125 days

Late completion will incur a severe penalty fee

Probability of completing in 125 days?

2 – 77

m = 100 days

X = 125 days

 = 20 days

FIGURE 2.10

Haynes Construction Company

Compute Z

2 – 78

m = 100 days

X = 125 days

 = 20 days

FIGURE 2.10

From Appendix A, for Z = 1.25 area = 0.89435

Z = 1.25

Compute Z

Haynes Construction Company

2 – 79

m = 100 days

X = 125 days

 = 20 days

FIGURE 2.10

From Appendix A, for Z = 1.25 area = 0.89435

The probability is about 0.89 that Haynes will not violate the contract

Z = 1.25

Haynes Construction Company

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

Probability of bonus?

2 – 80

FIGURE 2.11

Because the distribution is symmetrical, equivalent to Z = 1.25 so area = 0.89435

Z = –1.25

m = 100 days

X = 75 days

0.89435

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

Probability of bonus?

Haynes Construction Company

2 – 81

FIGURE 2.11

Z = –1.25

m = 100 days

X = 75 days

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%

Because the distribution is symmetrical, equivalent to Z = 1.25 so area = 0.89435

0.89435

Haynes Construction Company

Probability of completing between 110 and 125 days?

2 – 82

FIGURE 2.12

125 days

m =

110 days

100 days

s = 20 days

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

P(X ≤ 125) = 0.89435

For Z = 0.5 area = 0.69146

Probability of completing between 110 and 125 days?

Haynes Construction Company

2 – 83

FIGURE 2.12

125 days

m =

110 days

100 days

s = 20 days

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

P(X ≤ 125) = 0.89435

For Z = 0.5 area = 0.69146

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

2 – 84

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

TABLE 2.10 - partial

Using Excel

2 – 85

PROGRAM 2.3A – Excel Output

PROGRAM 2.3B – Excel Functions

The Empirical Rule

For a normally distributed random variable with mean m and standard deviation 

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

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

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

2 – 86

The Empirical Rule

2 – 87

FIGURE 2.13 – Approximate Probabilities

68%

16%

–1

+1

95%

2.5%

–2

+2

99.7%

0.15%

–3

+3

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

df1 = degrees of freedom for the numerator

df2 = degrees of freedom for the denominator

The probabilities of large values of F are very small

2 – 88

The F Distribution

2 – 89

F

FIGURE 2.14

The F Distribution

Consider the example

df1 = 5

df2 = 6

 = 0.05

2 – 90

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

F = 4.39

0.05

The F Distribution

2 – 91

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

Using Excel

2 – 92

PROGRAM 2.4A – Excel Output

PROGRAM 2.4B – Excel Functions

The Exponential Distribution

Also called the negative exponential distribution

A continuous distribution often used in queuing models

Probability function given by

2 – 93

where

X = random variable (service times)

m = 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 – 94

f(X)

Figure 2.16

Arnold’s Muffler Shop

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?

2 – 95

X = Exponentially distributed service time

m = 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 – 96

FIGURE 2.17 – Probability of Installation in 0.5 Hour

0.7769

P(service time ≤ 0.5) = 0.7769

| | | | | |

0 0.5 1 1.2 2 2.5

2.5 –

2 –

1.5 –

1 –

0.5 –

0 –

Arnold’s Muffler Shop

Similarly

2 – 97

And

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

Using Excel

2 – 98

PROGRAM 2.5A – Excel Output

PROGRAM 2.5B – Excel Functions

The Poisson Distribution

A discrete probability distribution

Often used in queuing models to describe arrival rates over time

Probability function given by

2 – 99

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

From Appendix C for l = 2

2 – 100

The Poisson Distribution

2 – 101

Figure 2.18 – Sample Poisson Distributions with l = 2 and l = 4

| | | | | | | | | |

0 1 2 3 4 5 6 7 8 9

0.30 –

0.25 –

0.20 –

0.15 –

0.10 –

0.05 –

0.00 –

l = 2 Distribution

Probability

| | | | | | | | | |

0 1 2 3 4 5 6 7 8 9

0.25 –

0.20 –

0.15 –

0.10 –

0.05 –

0.00 –

l = 4 Distribution

Probability

Using Excel

2 – 102

PROGRAM 2.6A – Excel Output

PROGRAM 2.6B – Excel Functions

2-103

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1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

1.7 Possible Problems in the Quantitative Analysis Approach

1.8 Implementation—Not Just the Final Step

1.1 Introduction 1.2 What Is Quantitative Analysis? 1.3 Business Analytics 1.4 The Quantitative Analysis Approach 1.5 How to Develop a Quantitative Analysis Model

CHAPTER OUTLINE

5. Use computers and spreadsheet models to perform quantitative analysis.

6. Discuss possible problems in using quantitative analysis.

7. Perform a break-even analysis.

1. Describe the quantitative analysis approach. 2. Understand the application of quantitative analysis

in a real situation. 3. Describe the three categories of business analytics. 4. Describe the use of modeling in quantitative

analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

1CHAPTER

LEARNING OBJECTIVES

M01_REND7331_12_SE_C01_pp2.indd 1 01/10/13 9:50 AM

1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

1.7 Possible Problems in the Quantitative Analysis Approach

1.8 Implementation—Not Just the Final Step

1.1 Introduction 1.2 What Is Quantitative Analysis? 1.3 Business Analytics 1.4 The Quantitative Analysis Approach 1.5 How to Develop a Quantitative Analysis Model

CHAPTER OUTLINE

5. Use computers and spreadsheet models to perform quantitative analysis.

6. Discuss possible problems in using quantitative analysis.

7. Perform a break-even analysis.

1. Describe the quantitative analysis approach. 2. Understand the application of quantitative analysis

in a real situation. 3. Describe the three categories of business analytics. 4. Describe the use of modeling in quantitative

analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

1CHAPTER

LEARNING OBJECTIVES

M01_REND7331_12_SE_C01_pp2.indd 1 01/10/13 9:50 AM

P(A | B)= P(B | A)P(A) P(B | A)P(A)+P(B | ʹ′A )P( ʹ′A )

P(A | B)=

P(B | A)P(A)

P(B | A)P(A)+P(B |

A)P(

= P(3 | fair)P(fair)

P(3 | fair)P(fair)+P(3 | loaded)P(loaded)

P(3 | fair)P(fair)

P(3 | fair)P(fair)+P(3 | loaded)P(loaded)

= (0.166)(0.50)

(0.166)(0.50)+(0.60)(0.50) = 0.083 0.383

=0.22

(0.166)(0.50)

(0.166)(0.50)+(0.60)(0.50)

0.083

0.383

=0.22

P(fair)= 0.50 and P(loaded)= 0.50

P(fair)=0.50 and P(loaded)=0.50

P(3,3 | fair)= (0.166)(0.166)=0.027 P(3,3 | loaded)= (0.6)(0.6)=0.36

P(3,3|fair)=(0.166)(0.166)=0.027

P(3,3|loaded)=(0.6)(0.6)=0.36

P(3,3 and fair)= P(3,3 | fair)×P(fair) = (0.027)(0.5)= 0.013

P(3,3 and fair)=P(3,3|fair)´P(fair)

=(0.027)(0.5)=0.013

P(3,3 and loaded)= P(3,3 | loaded)×P(loaded) = (0.36)(0.5)= 0.18

P(3,3 and loaded)=P(3,3|loaded)´P(loaded)

=(0.36)(0.5)=0.18

P(fair |3,3)= P(3,3 and fair) P(3,3)

= 0.013 0.193

= 0.067

P(fair|3,3)=

P(3,3 and fair)

P(3,3)

0.013

0.193

=0.067

P(loaded |3,3)= P(3,3 and loaded) P(3,3)

= 0.18 0.193

= 0.933

P(loaded|3,3)=

P(3,3 and loaded)

P(3,3)

0.18

0.193

=0.933

Z = 3 if good 2 if average 1 if poor

⎧

⎨ ⎪

⎩⎪

3 if good

2 if average

1 if poor

X =

5 if SA 4 if A 3 if N 2 if D 1 if SD

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

5 if SA

4 if A

3 if N

2 if D

1 if SD

Y = 0 if defective 1 if not defective{

0 if defective

1 if not defective

{

E X( ) = XiP Xi( ) i=1

∑

= X1P(X1)+ X2P(X2)+...+ XnP(Xn)

()

i=1

P(X

)+X

P(X

)+...+X

P(X

)

i=1

∑

i=1

E X( ) = XiP Xi( ) i=1

∑

= X1P(X1)+ X2P(X2)+ X3P(X3)+ X4P(X4)+ X5P(X5)

()

i=1

P(X

)+X

P(X

)+X

P(X

)+X

P(X

)+X

P(X

)

= (5)(0.1) + (4)(0.2) + (3)(0.3) + (2)(0.3) + (1)(0.1) = 2.9

=(5)(0.1) + (4)(0.2) + (3)(0.3) + (2)(0.3) + (1)(0.1)

=2.9

σ 2 = Variance= [Xi −E(X)] 2P(Xi

i=1

∑ )

=Variance=[X

-E(X)]

P(X

i=1

)

Variance= [Xi −E(X)] 2P(Xi

i=1

∑ )

Variance=[X

-E(X)]

P(X

i=1

)

σ = Variance = σ 2

s=Variance=s

= square root σ = standard deviation

=square root

s=standard deviation

σ = Variance

= 1.29 =1.14

s=Variance

=1.29=1.14

= n!

r!(n−r)! prqn−r

r!(n-r)!

n-r

= 5!

4!(5-4)! 0.54 0.55-4

= 5(4)(3)(2)(1) 4(3)(2)(1)1!

(0.0625)(0.5) = 0.15625

4!(5-4)!

0.5

5-4

5(4)(3)(2)(1)

4(3)(2)(1)1!

(0.0625)(0.5) = 0.15625

Using the cell references eliminates the need to retype the formula if you change a parameter such as p or r.

The function BINOM.DIST (r,n,p,TRUE) returns the cumulative probability.

f (X)= 1 σ 2π

e −(x−µ)2

2σ 2

f(X)=

s2p

-(x-m)

Z = X −µ σ

X-m

Z = X −µ σ

= 130-100

= 30 15

= 2 std dev

X-m

130-100

=2 std dev

Z = X −µ σ

X-m

Z = X −µ σ

= 125 – 100

= 25 20

=1.25

X-m

125 – 100

=1.25

Z = X −µ σ

= 75 – 100

= –25 20

= –1.25

X-m

75 – 100

–25

=–1.25

Z = X −µ σ

= 110 – 100

= 10 20

= 0.5

X-m

110 – 100

=0.5

f (X)= µe−µx

f(X)=me

-mx

Expected value = 1 µ

= Average service time

Variance = 1 µ2

Expected value =

= Average service time

Variance =

P(X ≤ t)=1−e−µt

P(X£t)=1-e

-mt

P X ≤ 1 3

⎛

⎝ ⎜

⎞

⎠ ⎟=1−e

−3 1 3 ⎛

⎝ ⎜ ⎞

⎠ ⎟ =1−e−1 =1−0.3679=0.6321

P X ≤ 2 3

⎛

⎝ ⎜

⎞

⎠ ⎟=1−e

−3 2 3 ⎛

⎝ ⎜ ⎞

⎠ ⎟ =1−e−2 =1−0.1353=0.8647

PX£

=1-e

-3

=1-e

-1

=1-0.3679=0.6321

PX£

=1-e

-3

=1-e

-2

=1-0.1353=0.8647

P(X)= λ xe−λ

P(X)=

-l

P(X)= e −λλx

P(0)= e −220

0! = (0.1353)1

1 =0.1353 ≈14%

P(1)= e −221

1! = e−22 1

= 0.1353(2)

1 =0.2706 ≈ 27%

P(2)= e −222

2! = e−24 2(1)

= 0.1353(4)

2 =0.2706 ≈ 27%

P(X)=

-l

P(0)=

-2

(0.1353)1

=0.1353»14%

P(1)=

-2

0.1353(2)

=0.2706»27%

P(2)=

-2

2(1)

0.1353(4)

=0.2706»27%