1 1 Slide Chapter 3-PART I Chapter 3-PART I John Kooti QUANTITATIVE METHODS FOR BUSINESS DISCREETE PROBABILITY DISTRIBUTION QUANTITATIVE METHODS FOR BUSINESS

1 1 Slide Slide

Chapter 3-PART IChapter 3-PART I Chapter 3-PART IChapter 3-PART I

John KootiJohn Kooti

QUANTITATIVE QUANTITATIVE METHODS FORMETHODS FOR

BUSINESSBUSINESSDISCREETE PROBABILITY DISCREETE PROBABILITY

DISTRIBUTIONDISTRIBUTION

QUANTITATIVE QUANTITATIVE METHODS FORMETHODS FOR

BUSINESSBUSINESSDISCREETE PROBABILITY DISCREETE PROBABILITY

DISTRIBUTIONDISTRIBUTION

2 2 Slide Slide

Chapter 3, Part IChapter 3, Part I Discrete Probability Distributions Discrete Probability Distributions

Random VariablesRandom Variables Discrete Random VariablesDiscrete Random Variables Expected Value and VarianceExpected Value and Variance Binomial Probability DistributionBinomial Probability Distribution Poisson Probability DistributionPoisson Probability Distribution

3 3 Slide Slide

Random VariablesRandom Variables

A A random variablerandom variable is a numerical description is a numerical description of the outcome of an experiment.of the outcome of an experiment.

A random variable can be classified as being A random variable can be classified as being either discrete or continuous depending on the either discrete or continuous depending on the numerical values it assumes.numerical values it assumes.

A A discrete random variablediscrete random variable may assume either may assume either a finite number of values or an infinite a finite number of values or an infinite sequence of values.sequence of values.

A A continuous random variablecontinuous random variable may assume may assume any numerical value in an interval or collection any numerical value in an interval or collection of intervals.of intervals.

4 4 Slide Slide

Example: JSL AppliancesExample: JSL Appliances

Discrete random variable with a finite number of Discrete random variable with a finite number of values:values:

Let Let xx = number of TV sets sold at the store in one day = number of TV sets sold at the store in one day

where where xx can take on 5 values (0, 1, 2, 3, 4) can take on 5 values (0, 1, 2, 3, 4)

Discrete random variable with an infinite sequence of Discrete random variable with an infinite sequence of values:values:

Let Let xx = number of customers arriving in one day = number of customers arriving in one day

where where xx can take on the values 0, 1, 2, . . . can take on the values 0, 1, 2, . . .

We can count the customers arriving, but there is no We can count the customers arriving, but there is no finite upper limit on the number that might arrive.finite upper limit on the number that might arrive.

5 5 Slide Slide

Discrete Probability DistributionsDiscrete Probability Distributions

The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed over the describes how probabilities are distributed over the values of the random variable.values of the random variable.

The probability distribution is defined by a The probability distribution is defined by a probabilityprobability functionfunction, denoted by , denoted by ff((xx), which provides the ), which provides the probability for each value of the random variable.probability for each value of the random variable.

The required conditions for a discrete probability The required conditions for a discrete probability function are:function are:

ff((xx) ) >> 0 0

ff((xx) = 1) = 1 We can describe a discrete probability distribution We can describe a discrete probability distribution

with a table, graph, or equation.with a table, graph, or equation.

6 6 Slide Slide


Using past data on TV sales (below left), a tabular Using past data on TV sales (below left), a tabular representation of the probability distribution for TV representation of the probability distribution for TV sales (below right) was developed.sales (below right) was developed.

NumberNumber

Units SoldUnits Sold of Daysof Days xx ff((xx))

00 8080 0 0 .40 .40

11 5050 1 1 .25 .25

22 4040 2 2 .20 .20

33 1010 3 3 .05 .05

44 2020 4 4 .10 .10

200200 1.00 1.00

7 7 Slide Slide


A graphical representation of the probability A graphical representation of the probability distribution for TV sales in one daydistribution for TV sales in one day

.10.10

.20.20

.30.30

.40.40

.50.50

0 1 2 3 40 1 2 3 4Values of Random Variable x (TV sales)Values of Random Variable x (TV sales)

Pro

bab

ilit

yP

rob

ab

ilit

y

8 8 Slide Slide

Expected Value and VarianceExpected Value and Variance

The The expected valueexpected value, or mean, of a random , or mean, of a random variable is a measure of its central location.variable is a measure of its central location.• Expected value of a discrete random Expected value of a discrete random

variable:variable:

EE((xx) = ) = = = xfxf((xx)) The The variancevariance summarizes the variability in the summarizes the variability in the

values of a random variable.values of a random variable.• Variance of a discrete random variable:Variance of a discrete random variable:

Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx)) The The standard deviationstandard deviation, , , is defined as the , is defined as the

positive square root of the variance.positive square root of the variance.

9 9 Slide Slide


Expected Value of a Discrete Random VariableExpected Value of a Discrete Random Variable

xx ff((xx)) xfxf((xx))

00 .40.40 .00 .00

11 .25.25 .25 .25

22 .20.20 .40 .40

33 .05.05 .15 .15

44 .10.10 .40.40

1.20 = 1.20 = EE((xx))

The expected number of TV sets sold in a day The expected number of TV sets sold in a day is 1.2is 1.2

10 10 Slide Slide

Variance and Standard DeviationVariance and Standard Deviation

of a Discrete Random Variableof a Discrete Random Variable

xx x - x - ( (x - x - ))22 ff((xx)) ((xx - - ))22ff((xx)) _____ _________ ___________ _______ ____________________ _________ ___________ _______ _______________

00 -1.2-1.2 1.44 1.44 .40.40 .576 .576

11 -0.2-0.2 0.04 0.04 .25.25 .010 .010

22 0.8 0.8 0.64 0.64 .20.20 .128 .128

33 1.8 1.8 3.24 3.24 .05.05 .162 .162

44 2.8 2.8 7.84 7.84 .10.10 .784 .784

1.660 = 1.660 =

The variance of daily sales is 1.66 TV sets The variance of daily sales is 1.66 TV sets squaredsquared..

The standard deviation of sales is 1.29 TV sets.The standard deviation of sales is 1.29 TV sets.


11 11 Slide Slide


Formula Spreadsheet for Computing Expected Formula Spreadsheet for Computing Expected Value and VarianceValue and Variance

A B C D

1 x f(x) xf(x) (x- ) 2f(x)2 0 0.40 =A2*B2 =(A2-C$7) 2̂*B23 1 0.25 =A3*B3 =(A3-C$7) 2̂*B34 2 0.20 =A4*B4 =(A4-C$7) 2̂*B45 3 0.05 =A5*B5 =(A5-C$7) 2̂*B56 4 0.10 =A6*B6 =(A6-C$7) 2̂*B67 =SUM(C2:C6) =SUM(D2:D6)8 Expected Value Variance

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Binomial Probability DistributionBinomial Probability Distribution

Properties of a Binomial ExperimentProperties of a Binomial Experiment• The experiment consists of a sequence of The experiment consists of a sequence of nn

identical trials.identical trials.• Two outcomes, Two outcomes, successsuccess and and failurefailure, are , are

possible on each trial. possible on each trial. • The probability of a success, denoted by The probability of a success, denoted by pp, ,

does not change from trial to trial.does not change from trial to trial.• The trials are independent.The trials are independent.

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Example: Evans ElectronicsExample: Evans Electronics


Evans is concerned about a low retention Evans is concerned about a low retention rate for employees. On the basis of past rate for employees. On the basis of past experience, management has seen a turnover of experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, any hourly employees chosen at random, management estimates a probability of 0.1 that management estimates a probability of 0.1 that the person will not be with the company next year.the person will not be with the company next year.

Choosing 3 hourly employees a random, Choosing 3 hourly employees a random, what is the probability that 1 of them will leave what is the probability that 1 of them will leave the company this year?the company this year?

LetLet: : pp = .10, = .10, nn = 3, = 3, xx = 1 = 1

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Binomial Probability FunctionBinomial Probability Function

wherewhere

ff((xx) = the probability of ) = the probability of xx successes in successes in nn trialstrials

nn = the number of trials = the number of trials

pp = the probability of success on any one = the probability of success on any one trialtrial

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

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Using the Binomial Probability FunctionUsing the Binomial Probability Function

= (3)(0.1)(0.81)= (3)(0.1)(0.81)

= .243 = .243

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

16 16 Slide Slide


Using the Tables of Binomial Probabilities Using the Tables of Binomial Probabilities

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

17 17 Slide Slide

Using a Tree DiagramUsing a Tree Diagram

FirstWorker

FirstWorker

Second

Worker

Second

Worker

ThirdWorke

r

ThirdWorke

r

Leaves (.1)Leaves (.1)

Stays (.9)Stays (.9)

Valueof x

Valueof x

33

22

00

22

22

11



S (.9)S (.9)



S (.9)S (.9)

S (.9)S (.9)

S (.9)S (.9)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1)

Probab.Probab..0010.0010

.0090.0090

.0090.0090

.7290.7290

.0090.0090

11

11

.0810.0810

.0810.0810

.0810.0810


18 18 Slide Slide


Using an Excel SpreadsheetUsing an Excel Spreadsheet• Step 1: Select a cell in the worksheet where youStep 1: Select a cell in the worksheet where you

want the binomial probabilities to want the binomial probabilities to appear.appear.

• Step 2: Select the Step 2: Select the InsertInsert pull-down menu. pull-down menu.• Step 3: Choose the Step 3: Choose the FunctionFunction option. option.• Step 4: When the Paste Function dialog box appears:Step 4: When the Paste Function dialog box appears:

Choose Statistical from the Function Category box.

Choose BINOMDIST from the Function Name box. Select OK.

continued

19 19 Slide Slide


Using an Excel Spreadsheet (continued)Using an Excel Spreadsheet (continued)• Step 5: When the BINOMDIST dialog box appears:Step 5: When the BINOMDIST dialog box appears:

Enter 1 in the Number_s box (value of x). Enter 3 in the Trials box (value of n). Enter .1 in the Probability_s box (value of p). Enter false in the Cumulative box. [Note: At this point the desired binomial probability of .243 is automatically computed and appears in the right center of the dialog box.] Select OK (and .243 will appear in the worksheet cell requested in Step 1).

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The Binomial Probability DistributionThe Binomial Probability Distribution

Expected ValueExpected Value

EE((xx) = ) = = = npnp VarianceVariance

Var(Var(xx) = ) = 22 = = npnp(1 - (1 - pp)) Standard DeviationStandard Deviation


EE((xx) = ) = = 3(.1) = .3 employees out of 3 = 3(.1) = .3 employees out of 3

Var(Var(xx) = ) = 22 = 3(.1)(.9) = .27 = 3(.1)(.9) = .27

SD( ) ( )x np p 1SD( ) ( )x np p 1

employees 52.)9)(.1(.3)(SD x employees 52.)9)(.1(.3)(SD x

21 21 Slide Slide

MINITAB APPLICATIONMINITAB APPLICATION

Minitab ProgramMinitab Program

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Poisson Probability DistributionPoisson Probability Distribution

Properties of a Poisson ExperimentProperties of a Poisson Experiment• The probability of an occurrence is the same The probability of an occurrence is the same

for any two intervals of equal length.for any two intervals of equal length.• The occurrence or nonoccurrence in any The occurrence or nonoccurrence in any

interval is independent of the occurrence or interval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

23 23 Slide Slide


Poisson Probability FunctionPoisson Probability Function

wherewhere

f(x) f(x) = probability of x occurrences in an = probability of x occurrences in an intervalinterval

= mean number of occurrences in an = mean number of occurrences in an intervalinterval

ee = 2.71828 = 2.71828

f xe

x

x( )

!

f x

e

x

x( )

!

24 24 Slide Slide

Example: Mercy HospitalExample: Mercy Hospital

Patients arrive at the emergency room of Patients arrive at the emergency room of MercyMercy

Hospital at the average rate of 6 per hour on Hospital at the average rate of 6 per hour on weekendweekend

evenings. What is the probability of 4 arrivals in evenings. What is the probability of 4 arrivals in 3030

minutes on a weekend evening?minutes on a weekend evening?

Using the Poisson Probability FunctionUsing the Poisson Probability Function

= 6/hour = 3/half-hour, = 6/hour = 3/half-hour, xx = 4 = 4f ( )

( . )!

.43 2 71828

41680

4 3

f ( )( . )

!.4

3 2 718284

16804 3

25 25 Slide Slide

Example: Mercy HospitalExample: Mercy Hospital

Using the Tables of Poisson ProbabilitiesUsing the Tables of Poisson Probabilities

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .00819 .0003 .0004 .0005 .0007 .0009 .0011 .0014 .0018 .0022 .002710 .0001 .0001 .0001 .0002 .0002 .0003 .0004 .0005 .0006 .000811 .0000 .0000 .0000 .0000 .0000 .0001 .0001 .0001 .0002 .000212 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0001

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .00819 .0003 .0004 .0005 .0007 .0009 .0011 .0014 .0018 .0022 .002710 .0001 .0001 .0001 .0002 .0002 .0003 .0004 .0005 .0006 .000811 .0000 .0000 .0000 .0000 .0000 .0001 .0001 .0001 .0002 .000212 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0001

26 26 Slide Slide


The Poisson probability distribution can be The Poisson probability distribution can be used as used as

an approximation of the binomial probabilityan approximation of the binomial probability

distribution when distribution when pp, the probability of success, is , the probability of success, is smallsmall

and and nn, the number of trials, is large., the number of trials, is large.• Approximation is good when Approximation is good when pp << .05 and .05 and nn

>> 20 20• Set Set = = npnp and use the Poisson tables. and use the Poisson tables.

27 27 Slide Slide

MINIAB APPLICATIONMINIAB APPLICATION


28 28 Slide Slide

The End of Discrete Probability DistributionThe End of Discrete Probability Distribution

29 29 Slide Slide

Chapter 3, Part IIChapter 3, Part II Continuous Random Variables Continuous Random Variables

Continuous Random VariablesContinuous Random Variables Normal Probability DistributionNormal Probability Distribution Exponential Probability DistributionExponential Probability Distribution

30 30 Slide Slide

Continuous Probability DistributionsContinuous Probability Distributions

A A continuous random variablecontinuous random variable can assume any can assume any value in an interval on the real line or in a value in an interval on the real line or in a collection of intervals.collection of intervals.

It is not possible to talk about the probability of It is not possible to talk about the probability of the random variable assuming a particular value.the random variable assuming a particular value.

Instead, we talk about the probability of the Instead, we talk about the probability of the random variable assuming a value within a given random variable assuming a value within a given interval.interval.

The probability of the random variable assuming a The probability of the random variable assuming a value within some given interval from value within some given interval from xx11 to to xx22 is is defined to be the defined to be the area under the grapharea under the graph of the of the probability density functionprobability density function betweenbetween x x11 andand x x22..

31 31 Slide Slide

Uniform Probability DistributionUniform Probability Distribution

A random variable is A random variable is uniformly distributeduniformly distributed wheneverwheneverthe probability is proportional to the length of thethe probability is proportional to the length of theinterval. interval. Uniform Probability Density FunctionUniform Probability Density Function

ff((xx) = 1/() = 1/(bb - - aa) for ) for aa << xx << bb = 0 elsewhere= 0 elsewhere

Expected Value of Expected Value of xxE(E(xx) = () = (aa + + bb)/2)/2

Variance of Variance of xx Var(Var(xx) = () = (bb - - aa))22/12/12

where: where: aa = smallest value the variable can = smallest value the variable can assumeassume

bb = largest value the variable can = largest value the variable can assumeassume

32 32 Slide Slide

Example: Slater's BuffetExample: Slater's Buffet

Slater customers are charged for the amount of Slater customers are charged for the amount of saladsalad

they take. Sampling suggests that the amount of they take. Sampling suggests that the amount of saladsalad

taken is uniformly distributed between 5 ounces taken is uniformly distributed between 5 ounces and 15and 15

ounces.ounces. Probability Density Function Probability Density Function

ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 15 15

= 0 elsewhere= 0 elsewhere

wherewhere

xx = salad plate filling weight = salad plate filling weight

33 33 Slide Slide

Example: Slater's BuffetExample: Slater's Buffet

What is the probability that a customer will What is the probability that a customer will taketake

between 12 and 15 ounces of salad?between 12 and 15 ounces of salad?f(x)f(x)

x x55 1010 15151212

1/101/10

Salad Weight (oz.)Salad Weight (oz.)

P(12 < x < 15) = 1/10(3) = .3P(12 < x < 15) = 1/10(3) = .3

34 34 Slide Slide

MINITAB APPLICATIONMINITAB APPLICATION


35 35 Slide Slide

The Normal Probability DistributionThe Normal Probability Distribution

Graph of the Normal Probability Density Graph of the Normal Probability Density FunctionFunction

xx

ff((xx))

36 36 Slide Slide

Normal Probability DistributionNormal Probability Distribution

The Normal CurveThe Normal Curve• The shape of the normal curve is often The shape of the normal curve is often

illustrated as a illustrated as a bell-shaped curvebell-shaped curve. . • The highest point on the normal curve is at The highest point on the normal curve is at

the the meanmean, which is also the , which is also the medianmedian and and modemode of the distribution. of the distribution.

• The normal curve is The normal curve is symmetricsymmetric..• The The standard deviationstandard deviation determines the determines the

width of the curve.width of the curve.• The total The total area under the curvearea under the curve is 1. is 1.• Probabilities for the normal random variable Probabilities for the normal random variable

are given by areas under the curve.are given by areas under the curve.

37 37 Slide Slide

Normal Probability DistributionNormal Probability Distribution

Normal Probability Density FunctionNormal Probability Density Function

wherewhere

= mean= mean

= standard deviation= standard deviation

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

f x e x( ) ( ) / 1

2

2 22

f x e x( ) ( ) / 1

2

2 22

38 38 Slide Slide

Standard Normal Probability DistributionStandard Normal Probability Distribution

A random variable that has a normal A random variable that has a normal distribution with a mean of zero and a distribution with a mean of zero and a standard deviation of one is said to have a standard deviation of one is said to have a standard normal probability distributionstandard normal probability distribution..

The letter The letter z z is commonly used to designate is commonly used to designate this normal random variable.this normal random variable.

Converting to the Standard Normal Converting to the Standard Normal Distribution Distribution

We can think of We can think of zz as a measure of the number as a measure of the number of standard deviations of standard deviations xx is from is from ..

zx

zx

39 39 Slide Slide

Example: Pep ZoneExample: Pep Zone

Pep Zone sells auto parts and supplies including aPep Zone sells auto parts and supplies including a

popular multi-grade motor oil. When the stock of thispopular multi-grade motor oil. When the stock of this

oil drops to 20 gallons, a replenishment order is placed.oil drops to 20 gallons, a replenishment order is placed.

The store manager is concerned that sales are being The store manager is concerned that sales are being

lost due to stockouts while waiting for an order. It haslost due to stockouts while waiting for an order. It has

been determined that leadtime demand is normallybeen determined that leadtime demand is normally

distributed with a mean of 15 gallons and a standarddistributed with a mean of 15 gallons and a standard

deviation of 6 gallons. deviation of 6 gallons.

The manager would like to know the probability of aThe manager would like to know the probability of a

stockout, P(stockout, P(xx > 20). > 20).

40 40 Slide Slide

Standard NormalStandard Normal

DistributionDistribution

zz = ( = (xx - - )/)/

= (20 - 15)/6= (20 - 15)/6

= .83= .83

The Standard Normal table shows an area The Standard Normal table shows an area of .2967 for the region between the of .2967 for the region between the zz = 0 line = 0 line and the and the z z = .83 line above. The shaded tail = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a area is .5 - .2967 = .2033. The probability of a stockout is .2033.stockout is .2033.

00 .83.83

Area = .2967Area = .2967

Area = .5Area = .5

Area = .2033Area = .2033

zz


41 41 Slide Slide

Using the Standard Normal Probability TableUsing the Standard Normal Probability Tablez .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549

.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852

.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133

.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549

.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852

.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133

.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389


42 42 Slide Slide


Using an Excel SpreadsheetUsing an Excel Spreadsheet• Step 1: Select a cell in the worksheet where Step 1: Select a cell in the worksheet where

you you want the normal probability to want the normal probability to appear.appear.

• Step 2: Select the Step 2: Select the InsertInsert pull-down menu. pull-down menu.• Step 3: Choose the Step 3: Choose the FunctionFunction option. option.• Step 4: When the Step 4: When the Paste FunctionPaste Function dialog box dialog box

appears: appears: Choose Choose StatisticalStatistical from the from the Function CategoryFunction Category box. box. Choose Choose NORMDISTNORMDIST from the from the Function NameFunction Name box. box. Select Select OKOK..

continue

43 43 Slide Slide


Using an Excel Spreadsheet (continued)Using an Excel Spreadsheet (continued)• Step 5: When the Step 5: When the NORMDISTNORMDIST dialog box dialog box

appears:appears:Enter 20 in the Enter 20 in the xx box. box.

Enter 15 in the Enter 15 in the meanmean box. box.Enter 6 in the Enter 6 in the standard deviationstandard deviation box. box.Enter true in the Enter true in the cumulativecumulative box. box.Select Select OKOK..

At this point, .7967 will appear in the cell selected At this point, .7967 will appear in the cell selected in Step 1, indicating that the probability of demand in Step 1, indicating that the probability of demand during lead time being less than or equal to 20 during lead time being less than or equal to 20 gallons is .7967. The probability that demand will gallons is .7967. The probability that demand will exceed 20 gallons is 1 - .7967 = .2033.exceed 20 gallons is 1 - .7967 = .2033.

44 44 Slide Slide

If the manager of Pep Zone wants the probability of aIf the manager of Pep Zone wants the probability of astockout to be no more than .05, what should thestockout to be no more than .05, what should thereorder point be?reorder point be?

Let Let zz.05.05 represent the represent the zz value cutting the tail area value cutting the tail area of .05. of .05.

Area = .05Area = .05

Area = .5 Area = .5 Area = .45 Area = .45

00 zz.05.05


45 45 Slide Slide

Using the Standard Normal Probability TableUsing the Standard Normal Probability TableWe now look-up the .4500 area in the We now look-up the .4500 area in the Standard Normal Probability table to find the Standard Normal Probability table to find the corresponding corresponding zz.05.05 value. value.

zz.05.05 = 1.645 is a reasonable = 1.645 is a reasonable estimate. estimate.

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 .

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 .


46 46 Slide Slide

The corresponding value of The corresponding value of xx is given by is given by

xx = = + + zz.05.05

= 15 + 1.645(6)= 15 + 1.645(6)

= 24.87= 24.87

A reorder point of 24.87 gallons will place A reorder point of 24.87 gallons will place thethe

probability of a stockout during leadtime at .05. probability of a stockout during leadtime at .05.

Perhaps Pep Zone should set the reorder Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability point at 25 gallons to keep the probability under .05.under .05.


47 47 Slide Slide

Exponential Probability DistributionExponential Probability Distribution

Exponential Probability Density FunctionExponential Probability Density Function

for for xx >> 0, 0, > 0 > 0

wherewhere = mean = mean

ee = 2.71828 = 2.71828 Cumulative Exponential Distribution FunctionCumulative Exponential Distribution Function

wherewhere xx00 = some specific value of = some specific value of xx

f x e x( ) / 1

f x e x( ) / 1

P x x e x( ) / 0 1 o P x x e x( ) / 0 1 o

48 48 Slide Slide

The time between arrivals of cars at Al’s The time between arrivals of cars at Al’s Carwash Carwash

follows an exponential probability distribution follows an exponential probability distribution with awith a

mean time between arrivals of 3 minutes. Al mean time between arrivals of 3 minutes. Al would likewould like

to know the probability that the time between to know the probability that the time between twotwo

successive arrivals will be 2 minutes or less.successive arrivals will be 2 minutes or less.

P(P(xx << 2) = 1 - 2.71828 2) = 1 - 2.71828-2/3-2/3 = 1 - .5134 = 1 - .5134 = .4866= .4866

Example: Al’s CarwashExample: Al’s Carwash

49 49 Slide Slide

Example: Al’s CarwashExample: Al’s Carwash

Graph of the Probability Density FunctionGraph of the Probability Density Function

xx

f(x)f(x)

.1.1

.3.3

.4.4

.2.2

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10

P(x < 2) = area = .4866P(x < 2) = area = .4866

50 50 Slide Slide

Relationship Between theRelationship Between thePoisson and Exponential DistributionsPoisson and Exponential Distributions

The continuous exponential probability The continuous exponential probability distribution is related to the discrete Poisson distribution is related to the discrete Poisson distribution.distribution.

The The Poisson distributionPoisson distribution provides an appropriate provides an appropriate description of the description of the number of occurrences per number of occurrences per intervalinterval..

The The exponential distributionexponential distribution provides a provides a description of the description of the length of the interval between length of the interval between occurrencesoccurrences..

51 51 Slide Slide

The End of Chapter 3The End of Chapter 3

Documents

1 1 Slide Chapter 3-PART I Chapter 3-PART I John Kooti QUANTITATIVE METHODS FOR BUSINESS DISCREETE PROBABILITY DISTRIBUTION QUANTITATIVE METHODS FOR BUSINESS