PPT Render Ch 2

7/28/2019 PPT Render Ch 2

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2008 Prentice-Hall, Inc.

Chapter 2

To accompanyQuant i tat ive Analysis for Management, Tenth Edit io n,

by Render, Stair, and HannaPower Point slides created by Jeff Heyl

Probab i li ty Concep ts and

Appl icat ions

2009 Prentice-Hall, Inc.


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2008 Prentice-Hall, Inc. 2 2

Learn ing Object ives

1. Understand the basic foundations ofprobability analysis

2. Describe statistically dependent andindependent events

3. Use Bayes theorem to establish posteriorprobabilities

4. Describe and provide examples of both

discrete and continuous random variables5. Explain the difference between discrete and

continuous probability distributions

6. Calculate expected values and variances

and use the normal table

After completing this chapter, students will be able to:


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Chapter Ou tl ine

2.1 Introduction

2.2 Fundamental Concepts

2.3 Mutually Exclusive and Collectively

Exhaustive Events2.4 Statistically Independent Events

2.5 Statistically Dependent Events

2.6 Revising Probabilities with BayesTheorem

2.7 Further Probability Revisions


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Chapter Ou tl ine

2.8 Random Variables

2.9 Probability Distributions

2.10 The Binomial Distribution

2.11 The Normal Distribution2.12 The FDistribution

2.13 The Exponential Distribution

2.14 The Poisson Distribution


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In t roduct ion

Life is uncertain, we are not surewhat the future will bring

Risk and probability is a part ofour daily lives

Probabi l i tyis a numerical

statement about the likelihoodthat an event will occur


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Fundamental Concepts

1. The probability, P, of any event orstate of nature occurring is greaterthan or equal to 0 and less than or

equal to 1. That is:

0 P(event) 1

2. The sum of the simple probabilitiesfor all possible outcomes of anactivity must equal 1


7/91 2008 Prentice-Hall, Inc. 2 7

Chapters in This BookThat Use Probabi l i ty

CHAPTER TITLE

3 Decision Analysis

4 Regression Models

5 Forecasting

6 Inventory Control Models

13 Project Management

14 Waiting Lines and Queuing Theory Models

15 Simulation Modeling

16 Markov Analysis17 Statistical Quality Control

Module 3 Decision Theory and the Normal Distribution

Module 4 Game Theory

Table 2.1



Diversey Paint Examp le

Demand for white latex paint at Diversey Paintand Supply has always been either 0, 1, 2, 3, or 4gallons per day

Over the past 200 days, the owner has observed

the following frequencies of demandQUANTITY

DEMANDEDNUMBER 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)



Diversey Paint Examp le

Demand for white latex paint at Diversey Paintand Supply has always been either 0, 1, 2, 3, or 4gallons per day

Over the past 200 days, the owner has observed

the following frequencies of demandQUANTITY

DEMANDEDNUMBER 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)

Notice the individual probabilitiesare all between 0 and 1

0 P(event) 1

And the total of all eventprobabilities equals 1

P(event) = 1.00



Determining ob ject ive probabi l i ty

Relative frequency Typically based on historical data

Types o f Probab i l ity

P(event) =Number of occurrences of the event

Total number of trials or outcomes

Classical or logical method

Logically determine probabilities withouttrials

P(head) =1

2

Number of ways of getting a head

Number of possible outcomes (head or tail)



Types o f Probabi l i ty

Subject ive probabi l i tyis based onthe experience and judgment of theperson making the estimate

Opinion polls

Judgment of experts

Delphi method

Other methods



Mutual ly Exc lus ive Even ts

Events are said to be mutual lyexclus iveif only one of the events canoccur on any one trial

Tossing a coin will resultin eithera head or a tail

Rolling a die will result inonly one of six possibleoutcomes



Col lect ively Exhaus t ive Events

Events are said to be collectivelyexhaustive if the list of outcomesincludes every possible outcome

Both heads andtails as possibleoutcomes ofcoin flips

All six possibleoutcomesof the rollof a die

OUTCOMEOF ROLL

PROBABILITY

1 1/6

2 1/6

31

/64 1/6

5 1/6

6 1/6

Total 1



Draw ing a Card

Draw one card from a deck of 52 playing cards

P(drawing a 7) = 4/52 =1/13

P(drawing a heart) = 13/52 = 1/4

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

These two events are not collectivelyexhaustive since there are other cards in thedeck besides 7s and hearts


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Tab le o f Differences

DRAWSMUTUALLYEXCLUSIVE

COLLECTIVELYEXHAUSTIVE

1. Draws a spade and a club Yes No

2. Draw a face card and anumber card Yes Yes

3. Draw an ace and a 3 Yes No

4. Draw a club and a nonclub Yes Yes

5. Draw a 5 and a diamond No No

6. Draw a red card and adiamond

No No


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Add ing Mutually Exc lusive Events

We often want to know whether one or asecond event will occur

When two events are mutually

exclusive, the law of addition is

P(event A or event B) = P(event A) + P(event B)

P(spade or club) = P(spade) + P(club)

= 13/52 +13/52

= 26/52 =1/2 = 0.50 = 50%


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Add ing Not Mutually Exclus ive Events

P(event A or event B) = P(event A) + P(event B)

P(event Aand event Bbothoccurring)

P(A orB) = P(A) + P(B)P(A and B)

P(five or diamond) =P(five) + P(diamond)P(five and diamond)

= 4/52 +13/52

1/52= 16/52 =

4/13

The equation must be modified to accountfor double counting

The probability is reduced bysubtracting the chance of both eventsoccurring together


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Venn Diagrams

P(A) P(B)

Events that are mutuallyexclusive

P(A orB) = P(A) + P(B)

Figure 2.1

Events that are notmutually exclusive

P(A orB) = P(A) + P(B)P(A and B)

Figure 2.2

P(A) P(B)

P(A and B)


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Stat ist ical ly Independen t Even ts

Events may be either independent ordependent For independent events, the occurrence

of one event has no effect on theprobability of occurrence of the secondevent


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Which Sets o f Events A re Independent?

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

Dependent events

Dependent

events

Independent

events

Ind ependent events


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Three Types o f Probab i l i t ies

Marginal(ors imple) probability is just theprobability of an event occurring

P(A)

Jo in tprobability is the probability of two or more

events occurring and is the product of theirmarginal probabilities for independent events

P(AB) = P(A) x P(B)

Condi t ionalprobability is the probability of event

Bgiven that event A has occurredP(B| A) = P (B)

Or the probability of event A given that event Bhas occurred

P(A | B) = P(A)


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Jo int Probabi l i ty Example

The probability of tossing a 6 on the firstroll of the die and a 2 on the second roll

P(6 on first and 2 on second)

= P(tossing a 6) x P(tossing a 2)

= 1/6 x1/6 =

1/36 = 0.028


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Independen t Even ts

1. A black ball drawn on first drawP(B) = 0.30 (a marginal prob abi l ity)

2. Two green balls drawn

P(GG) = P(G) x P(G) = 0.7 x 0.7 = 0.49

(a joint p robabi l i ty for two ind ependent events)

A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and

draw a second ball


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Independen t Even ts

3. A black ball drawn on second draw if the firstdraw is green

P(B|G) = P(B) = 0.30(a cond it ional probabi l i ty but equal to the marginalbecause the two d raws are independent events)

4. A green ball is drawn on the second if the firstdraw was green

P(G|G) = P(G) = 0.70(a cond it ional probabi l i ty as in event 3)

A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and

draw a second ball


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Statist ical ly Dependen t Events

The marginalprobability of an event occurring iscomputed the same

P(A)

The formula for thejo in tprobability of two events is

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

P(A | B) =P(AB)

P(B)

Calculating condi t ionalprobabilities is slightly morecomplicated. The probability of event A given thatevent Bhas occurred is


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When Events Are Dependent

Assume that we have an urn containing 10 balls ofthe following descriptions

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)

P(WL

) = 4/10

= 0.4P

(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


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4 ballsWhite (W)

andLettered (L )

2 ballsWhite (W)

andNumbered (N)

3 ballsYellow (Y)

andLettered (L )

1 ball Yellow (Y)and Numbered (

N)

Probability (WL ) =4

10

Probability (YN) =1

10

Probability (YL ) =310

Probability (WN) =210

Urn contains10 balls

Figure 2.3


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

P(L| Y) = = = 0.75P(YL )

P(Y)

0.3

0.4

Verify P(YL ) using the joint probability formula

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


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Jo int Probabi l i t iesfor Dependent Events

P(MT) = P(T| M) x P(M) = (0.70)(0.40) = 0.28

If the stock market reaches 12,500 point by January,there is a 70% probability that Tubeless Electronicswill go up

There is a 40% chance the stock market will

reach 12,500 Let Mrepresent the event of the stock market

reaching 12,500 and let Tbe the event thatTubeless goes up in value


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PosteriorProbabilities

BayesProcess

Revis ing Probabi l i t ies w ith

Bayes Theorem

Bayes theorem is used to incorporate additionalinformation and help create pos ter ior prob abi l i t ies

PriorProbabilities

NewInformation

Figure 2.4


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Pos terio r Probab i l i t ies

A cup contains two dice identical in appearance butone 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 loadeddie is 0.60

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

What is the probability thatthe die rolled was fair?

What is the probability that

the loaded die was rolled?


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We know the probability of the die being fair orloaded is

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

And that

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

We compute the probabilities ofP(3 and fair)and P(3 and loaded)

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


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We know the probability of the die being fair orloaded is

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

And that

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

We compute the probabilities ofP(3 and fair)and P(3 and loaded)

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 sum of these probabilitiesgives us the unconditionalprobability of tossing a 3

P(3) = 0.083 + 0.300 = 0.383


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P(loaded | 3) = = = 0.78P(loaded and 3)

P(3)

0.300

0.383

The probability that the die was loaded is

P(fair | 3) = = = 0.22P(fair and 3)

P(3)

0.083

0.383

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

These are the revisedorposter iorprobabi l i t iesfor thenext roll of the die

We use these to revise ourpr ior probabi l i tyestimates


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Bayes Calcu lat ionsGiven event Bhas occurred

STATE OFNATURE

P(B| STATEOF NATURE)

PRIORPROBABILITY

JOINTPROBABILITY

POSTERIORPROBABILITY

A P(B| A) x P(A) = P(Band A) P(Band A)/P(B) = P(A|B)

A P(B| A) x P(A) = P(Band A) P(Band A)/P(B) = P(A|B)

P(B)

Table 2.2Given a 3 was rolled

STATE OF

NATURE

P(B| STATE

OF NATURE)

PRIOR

PROBABILITY

JOINT

PROBABILITY

POSTERIOR

PROBABILITYFair 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

Table 2.3


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General Form of Bayes Theorem

)()|()()|(

)()|(

)|( APABPAPABP

APABP

BAP

We can compute revised probabilities moredirectly by using

where

the complement of the event ;for example, if is the event fair die,

then is loaded die

AA

AA


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General Form of Bayes Theorem

This is basically what we did in the previous example

If we replace with fair dieReplace with loaded die

Replace with 3 rolledWe get

AAB

)|( rolled3diefairP

)()|()()|()()|(

loadedloaded3fairfair3fairfair3

PPPPPP

2203830

0830

5006005001660

5001660.

.

.

).)(.().)(.(

).)(.(


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FurtherProbab i l ity Revis ions

We can obtain additional information by performingthe experiment a second time

If you can afford it, perform experimentsseveral times

We roll the die again and again get a 3

500loadedand500fair .)(.)( PP

3606060loaded33

027016601660fair33

.).)(.()|,(

.).)(.()|,(

P

P


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500loadedand500fair .)(.)( PP

3606060loaded33

027016601660fair33

.).)(.()|,(

.).)(.()|,(

P

P

)()|,()( fairfair33fairand3,3 PPP 0130500270 .).)(.(

)()|,()( loadedloaded33loadedand3,3 PPP 18050360 .).)(.(


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50.0)loaded(and50.0)fair( PP

36.0)6.0)(6.0()loaded|3,3(

027.0)166.0)(166.0()fair|3,3(

P

P

)()|,()( fairfair33fairand3,3 PPP 0130500270 .).)(.(

)()|,()( loadedloaded33loadedand3,3 PPP 18050360 .).)(.(

06701930

0130

33

fairand3,333fair .

.

.

),(

)(),|(

P

PP

93301930

180

33

loadedand3,333loaded .

.

.

),(

)(),|(

P

PP


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After the first roll of the die

probability the die is fair = 0.22

probability the die is loaded = 0.78

After the second roll of the die

probability the die is fair = 0.067probability the die is loaded = 0.933


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Random Variab les

Discrete random variablescan assume onlya finite or limited set of values

Cont inuous random var iablescan assumeany one of an infinite set of values

A random variableassigns a real numberto every possible outcome or event in anexperiment

X= number of refrigerators sold during the day


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Random Variab les Numbers

EXPERIMENT OUTCOMERANDOM

VARIABLES

RANGE OFRANDOM

VARIABLES

Stock 50Christmas trees

Number of Christmastrees sold

X 0, 1, 2,, 50

Inspect 600

items

Number of acceptable

items

Y 0, 1, 2,, 600

Send out 5,000sales letters

Number of peopleresponding to theletters

Z 0, 1, 2,, 5,000

Build an

apartmentbuilding

Percent of building

completed after 4months

R 0 R 100

Test the lifetimeof a lightbulb(minutes)

Length of time thebulb lasts up to 80,000minutes

S 0 S 80,000

Table 2.4


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Random Variab les Not Numbers

EXPERIMENT OUTCOMERANDOM

VARIABLES

RANGE OFRANDOM

VARIABLES

Studentsrespond to aquestionnaire

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 machineis inspected

Defective Y=Not defective

0 if defective

1 if not defective

0, 1

Consumersrespond tohow they likea product

GoodAveragePoor

3 if good.Z= 2 if average

1 if poor..

1, 2, 3

Table 2.5


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Probabi l i ty Distr ibu t ion o f a

Discrete Random Variable

Dr. Shannon asked studentsto respond to the statement,

The textbook was wellwritten and helped meacquire the necessaryinformation.

Selecting the right probability distributionis important

Fordisc rete random variablesa

probability is assigned to each event

5. Strongly agree4. Agree

3. Neutral2. Disagree1. Strongly disagree


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Discrete Random Variable

OUTCOMERANDOMVARIABLE (X)

NUMBERRESPONDING

PROBABILITYP(X)

Strongly agree 5 10 0.1 = 10/100

Agree 4 20 0.2 = 20/100

Neutral 3 30 0.3 = 30/100Disagree 2 30 0.3 = 30/100

Strongly disagree 1 10 0.1 = 10/100

Total 100 1.0 = 100/100

Distribution follows all three rules1. 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


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Probabi l i ty Distr ibut ion fo r

Dr. Shannons Class

P(X)

0.4

0.3

0.2

0.1

0| | | | | |

1 2 3 4 5

XFigure 2.5


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Probabi l i ty Distr ibut ion fo r

Dr. Shannons Class

P(X)

0.4

0.3

0.2

0.1

0| | | | | |

1 2 3 4 5

XFigure 2.5

Central tendency of thedistribution is the meanorexpected value

Amount of variability orspread is the variance


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Expected Value of a Disc rete

Probabi l i ty Distr ibu t ion

n

iii XPXXE

1 )(...)( 2211 nn XPXXPXXPX

The expected value is a measure of the centraltendencyof the distribution and is a weightedaverage of the values of the random variable

where

iX

)(iXP

n

i 1

)(XE

= random variables possible values

= probability of each of the random variablespossible values

= summation sign indicating we are adding all npossible values

= expected value or mean of the random sample


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Variance of a

Discrete Probabi l i ty Distr ibu t ion

For a discrete probability distribution thevariance can be computed by

)()]([ n

iii XPXEX1

22

Variance

where

iX

)(XE

)(i

XP

= random variables possible values

= expected value of the random variable

= difference between each value of the randomvariable and the expected mean

= probability of each possible value of therandom sample

)]([ XEXi


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Variance of a


For Dr. Shannons class

)()]([variance5

1

2

i

iiXPXEX

).().().().(variance 2092410925 22).().().().( 3092230923 22

).().( 10921 2

291

36102430003024204410

.

.....


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Variance of a


A related measure of dispersion is thestandard deviation

2

Variance where

= square root

= standard deviation


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Variance of a


A related measure of dispersion is thestandard deviation

2

Variance where

= square root

= standard deviation

For the textbook question

Variance 141291 ..


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Cont inuous Random Var iable

Since random variables can take on an infinitenumber of values, the fundamental rules forcontinuous random variables must be modified

The sum of the probability values must still

equal 1 But the probability of each value of the

random variable must equal 0 or the sumwould be infinitely large

The probability distribution is defined by a

continuous mathematical function called theprobability density function or just the probabilityfunction

Represented by f(X)


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Cont inuous Random Var iable

Probability

| | | | | | |

5.06 5.10 5.14 5.18 5.22 5.26 5.30

Weight (grams)

Figure 2.6


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The Binom ial Distr ibut ion

Many business experiments can becharacterized by the Bernoulli process

The Bernoulli process is described by the

binomial probability distribution1. Each trial has only two possible outcomes

2. The probability stays the same from one trialto the next

3. The trials are statistically independent4. The number of trials is a positive integer


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The binomial distribution is used to find theprobability of a specific number of successesout ofntrials

We need to known= number of trials

p= the probability of success on anysingle trial

We letr= number of successes

q= 1p= the probability of a failure


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The binomial formula is

rnrqp

rnr

nnr

)!(!!

trialsinsuccessesofyProbabilit

The symbol ! means factorial, and

n! = n(n 1)(n2)(1)

For example

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

1! = 1 and 0! = 1


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NUMBER OFHEADS (r) Probability = (0.5)

r(0.5)5r5!

r!(5r)!

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


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Solv ing Problems w i th the

Binom ial 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 5050

454

5trials5insuccesses4 ..)!(!

!)(P

156250500625011234

12345.).)(.(

)!)()()((

))()()((

Or about 16%

S


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Solv ing Problems w i th the

Binom ial Formula

Probability

P(r)

| | | | | | |1 2 3 4 5 6

Values ofr(number of successes)

0.4

0.3

0.2

0.1

0

Figure 2.7

S l i P bl i h


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Solv ing Problems w ith

B inom ial TablesMSA Electronics is experimenting with themanufacture of a new transistor

Every hour a sample of 5 transistors is taken

The probability of one transistor being

defective is 0.15What 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,but using the table is easier

S l i P bl i th


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B inom ial TablesP

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 tableforn= 5, p= 0.15, and r= 3, 4, and 5 andadd them together

S l i P bl i th


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Table 2.8 (partial)

We find the three probabilities in the tableforn= 5, p= 0.15, and r= 3, 4, and 5 andadd them together


B inom ial TablesP

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

)()()()( 543defectsmoreor3 PPPP 02670000100022002440 ....

S l i P bl i th


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B inom ial TablesIt is easy to find the expected value (or mean) andvariance of a binomial distribution

Expected value (mean) = np

Variance = np(1p)

For the MSA example

6375085015051Variance7501505valueExpected

.).)(.()(.).( pnp np


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The Normal Distr ibu t ion

The no rmal dis t ribut ionis the most popularand useful continuous probabilitydistribution

The formula for the probability density

function is rather complex

2

2

2

2

1

)(

)(

x

eXf

The normal distribution is specifiedcompletely when we know the mean, ,and the standard deviation,


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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 Xaxis are measured in the

number of standard deviations away fromthe mean

As the standard deviation becomes larger,the curve flattens

As the standard deviation becomessmaller, the curve becomes steeper


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| | |

40 = 50 60

| | |

= 40 50 60

Smaller, same

| | |

40 50 = 60

Larger, same

Figure 2.8


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Figure 2.9

Same , smaller

Same

, larger


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Figure 2.10

68%16% 16%

1 +1a b

95.4%2.3% 2.3%

2 +2a b

99.7%0.15% 0.15%

3 +3a b


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If IQs in the United States were normally distributedwith = 100 and = 15, then

1. 68% of the population would have IQs

between 85 and 115 points (1)2. 95.4% of the people have IQs between 70

and 130 (2)3. 99.7% of the population have IQs in the

range from 55 to 145 points (

3)4. Only 16% of the people have IQs greaterthan 115 points (from the first graph, thearea to the right of +1)


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Using the Standard Normal Tab le

Step 1Convert the normal distribution into a standardnormal dis t r ibut ion

A standard normal distribution has a meanof 0 and a standard deviation of 1

The new standard random variable is Z

XZ

whereX= value of the random variable we want to measure

= mean of the distribution

= standard deviation of the distributionZ= number of standard deviations from Xto the mean,


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For example, = 100, = 15, and we want to findthe probability that Xis less than 130

15

100130 X

Z

devstd215

30

| | | | | | |

55 70 85 100 115 130 145

| | | | | | |

3 2 1 0 1 2 3

X= IQ

XZ

= 100 = 15P(X< 130)

Figure 2.11


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Step 2Look up the probability from a table of normalcurve areas

Use Appendix A or Table 2.9 (portion below)

The column on the left has Zvalues The row at the top has second decimal

places for the Zvalues

AREA UNDER THE NORMAL CURVE

Z 0.00 0.01 0.02 0.031.8 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 2.9

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

= 97.7%


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Haynes Cons truct ion Company

Haynes builds apartment buildings

Total construction time follows a normaldistribution

For triplexes, = 100 days and = 20 days Contract calls for completion in 125 days Late completion will incur a severe penalty

fee

What is the probability of completing in 125

days?


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From Appendix A, forZ= 1.25 the area is 0.89435

There is about an 89% probability Hayneswill not violate the contract

20

100125 X

Z

25120

25.

= 100 days X= 125 days

= 20 daysFigure 2.12


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For triplexes, = 100 days and = 20 days Completion in 75 days or less will earn a

bonus of $5,000

What is the probability of getting thebonus?


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But Appendix A has only positive Zvalues, theprobability we are looking for is in the negative tail

20

10075 X

Z

25120

25.

Figure 2.12

= 100 daysX= 75 days

P(X < 75 days)

Area ofInterest


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Because the curve is symmetrical, we can lookat the probability in the positive tail for thesame distance away from the mean

20

10075 X

Z

25120

25.


P(X > 125 days)Area of

Interest


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We know the probabilitycompleting in125 days is 0.89435

So the probabilitycompleting in more

than 125 days is1 0.89435 = 0.10565


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= 100 daysX= 75 days

The probability of completing in less than75 days is 0.10565 or about 11%

Going back tothe left tail of thedistribution

The probabilitycompleting in morethan 125 days is1 0.89435 = 0.10565


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For triplexes, = 100 days and = 20 days What is the probability of completing

between 110 and 125 days?

We know the probability of completing in 125

days, P(X< 125) = 0.89435 We have to complete the probability of

completing in 110 days and find the areabetween those two events

C C


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From Appendix A, forZ= 0.5 the area is 0.69146

P(110 < X< 125) = 0.89435 0.69146 = 0.20289or about 20%

20

100110 X

Z

5020

10.

Figure 2.14

= 100days

125days

= 20 days

110days


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TheFDistr ibut ion

A continuous probability distribution

The Fstatistic is the ratio of two sample variances

Fdistributions 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 numeratordf2 = degrees of freedom for the denominator

Th F Di ib i


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TheFDistr ibut ion

df1 = 5

df2 = 6= 0.05Consider the example:

From Appendix D, we get

F, df1, df2 = F0.05, 5, 6 = 4.39

This means

P(F> 4.39) = 0.05

There is only a 5% probability that Fwill exceed 4.39

Th F Di t ib t i


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F

TheFDistr ibut ion

Figure 2.15

Th F Di t ib t i


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TheFDistr ibut ion

Figure 2.16

F= 4.390.05

Fvalue for 0.05 probabilitywith 5 and 6 degrees offreedom

Th E ti l Di t ib t i


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The Exponen t ial Distr ibu t ion

The exponent ial dist r ibu t ion(also calledthe negat ive exponent ial d ist r ibut ion) is acontinuous distribution often used inqueuing models to describe the time

required to service a customerx

eXf )(

where

X= random variable (service times)= average number of units the service facility can

handle in a specific period of time

e= 2.718 (the base of natural logarithms)

Th E ti l Di t ib t i


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The Exponen t ial Distr ibu t ion

timeserviceAverage1valueExpected 2

1Variance

f(X)

XFigure 2.17

Th P i Di t ib t i


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The Poisson Distr ibut ion

The Poissondis t r ibut ionis a discretedistribution that is often used in queuingmodels to describe arrival rates over time

!)(

XeXP

x

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= specific value (0, 1, 2, 3, and so on) of the randomvariable

Th P i Di t ib t i


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The Poisson Distr ibut ion

The mean and variance of the distribution areboth

Expected value = Variance =

0.3

0.2

0.1

0| | | | | | | | | |

P(X)

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