1 1 Slide Chapter 6 Continuous Probability Distributions n Uniform Probability Distribution n Normal Probability Distribution n Exponential Probability.

1 1 Slide Slide

Chapter 6Chapter 6 Continuous Probability Distributions Continuous Probability Distributions

Uniform Probability DistributionUniform Probability Distribution Normal Probability DistributionNormal Probability Distribution Exponential Probability DistributionExponential Probability Distribution

f (x)f (x)

x x

UniformUniform

xx

f f ((xx)) NormalNormal

xx

f (x)f (x) ExponentialExponential

2 2 Slide Slide

6.1 Continuous Probability Distributions6.1 Continuous 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.

3 3 Slide Slide

Continuous Probability DistributionsContinuous Probability Distributions

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

f (x)f (x)

x x

UniformUniform

xx11 xx11 xx22 xx22

xx

f f ((xx)) NormalNormal

xx11 xx11 xx22 xx22

xx11 xx11 xx22 xx22

ExponentialExponential

xx

f (x)f (x)

xx11 xx11 xx22 xx22

4 4 Slide Slide

6.2 Uniform Probability Distribution6.2 Uniform Probability Distribution

where: where: aa = smallest value the variable can assume = smallest value the variable can assume bb = largest value the variable can assume = largest value the variable can assume

f f ((xx) = 1/() = 1/(bb – – aa) for) for aa << xx << bb = 0 elsewhere= 0 elsewhere f f ((xx) = 1/() = 1/(bb – – aa) for) for aa << xx << bb = 0 elsewhere= 0 elsewhere

A random variable is A random variable is uniformly distributeduniformly distributed whenever the probability is proportional to the whenever the probability is proportional to the intervalinterval’’s length. s length.

The The uniform probability density functionuniform probability density function is: is:

5 5 Slide Slide

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

E(E(xx) = () = (aa + + bb)/2)/2E(E(xx) = () = (aa + + bb)/2)/2

Uniform Probability DistributionUniform Probability Distribution

Expected Value of Expected Value of xx

Variance of Variance of xx

6 6 Slide Slide


Example: Slater's BuffetExample: Slater's BuffetSlater customers are chargedSlater customers are chargedfor the amount of salad they take. for the amount of salad they take. Sampling suggests that theSampling suggests that theamount of salad taken is amount of salad taken is uniformly distributeduniformly distributedbetween 5 ounces and 15 ounces.between 5 ounces and 15 ounces.

7 7 Slide Slide

Uniform Probability Density FunctionUniform Probability Density Function

ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 15 15 = 0 elsewhere= 0 elsewhere

ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 15 15 = 0 elsewhere= 0 elsewhere

where:where: xx = salad plate filling weight = salad plate filling weight


8 8 Slide Slide

Expected Value of Expected Value of xx

Variance of Variance of xx

E(E(xx) = () = (aa + + bb)/2)/2 = (5 + 15)/2= (5 + 15)/2 = 10= 10

E(E(xx) = () = (aa + + bb)/2)/2 = (5 + 15)/2= (5 + 15)/2 = 10= 10

Var(Var(xx) = () = (bb - - aa))22/12/12 = (15 – 5)= (15 – 5)22/12/12 = 8.33= 8.33

Var(Var(xx) = () = (bb - - aa))22/12/12 = (15 – 5)= (15 – 5)22/12/12 = 8.33= 8.33


9 9 Slide Slide

Uniform Probability DistributionUniform Probability Distributionfor Salad Plate Filling Weightfor Salad Plate Filling Weight

f(x)f(x)

x x55 1010 1515

1/101/10

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


10 10 Slide Slide

f(x)f(x)

x x55 1010 1515

1/101/10

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

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

What is the probability that a customerWhat is the probability that a customer will take between 12 and 15 ounces of salad?will take between 12 and 15 ounces of salad?

1212


11 11 Slide Slide

6.3 Normal Probability Distribution6.3 Normal Probability Distribution

The The normal probability distributionnormal probability distribution is the most is the most important distribution for describing a important distribution for describing a continuous random variable.continuous random variable.

It is widely used in statistical inference.It is widely used in statistical inference.

12 12 Slide Slide

Normal Probability DistributionNormal Probability Distribution

It has been used in a wide variety of applications:It has been used in a wide variety of applications:

TestTest scoresscores

AmountsAmountsof rainfallof rainfall

13 13 Slide Slide


Normal Probability Density FunctionNormal Probability Density Function

2 2( ) / 21( )

2xf x e

2 2( ) / 21( )

2xf x e

= mean= mean = standard deviation= standard deviation = 3.14159= 3.14159ee = 2.71828 = 2.71828

where:where:

14 14 Slide Slide

The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero. The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero.


CharacteristicsCharacteristics

xx

15 15 Slide Slide

The entire family of normal probabilityThe entire family of normal probability distributions is defined by itsdistributions is defined by its meanmean μμ and itsand its standard deviationstandard deviationσσ..

The entire family of normal probabilityThe entire family of normal probability distributions is defined by itsdistributions is defined by its meanmean μμ and itsand its standard deviationstandard deviationσσ..



Standard DeviationStandard Deviation

MeanMean xx

16 16 Slide Slide

The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the medianmedian and and modemode.. The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the medianmedian and and modemode..



xx

17 17 Slide Slide



-10-10 00 2020

The mean can be any numerical value: negative,The mean can be any numerical value: negative, zero, or positive.zero, or positive. The mean can be any numerical value: negative,The mean can be any numerical value: negative, zero, or positive.zero, or positive.

xx

18 18 Slide Slide



= 15= 15

= 25= 25

The standard deviation determines the width of theThe standard deviation determines the width of thecurve: larger values result in wider, flatter curves.curve: larger values result in wider, flatter curves.The standard deviation determines the width of theThe standard deviation determines the width of thecurve: larger values result in wider, flatter curves.curve: larger values result in wider, flatter curves.

xx

19 19 Slide Slide

Probabilities for the normal random variable areProbabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area. The total area under the curve is 1 (.5 to the left of the mean andunder the curve is 1 (.5 to the left of the mean and .5 to the right)..5 to the right).

Probabilities for the normal random variable areProbabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area. The total area under the curve is 1 (.5 to the left of the mean andunder the curve is 1 (.5 to the left of the mean and .5 to the right)..5 to the right).



.5.5 .5.5

xx

20 20 Slide Slide



of values of a normal random variableof values of a normal random variable are within of its mean.are within of its mean. of values of a normal random variableof values of a normal random variable are within of its mean.are within of its mean.68.26%68.26%68.26%68.26%

+/- 1 standard deviation+/- 1 standard deviation+/- 1 standard deviation+/- 1 standard deviation


+/- 2 standard deviations+/- 2 standard deviations+/- 2 standard deviations+/- 2 standard deviations


+/- 3 standard deviations+/- 3 standard deviations+/- 3 standard deviations+/- 3 standard deviations

21 21 Slide Slide



xx – – 33 – – 11

– – 22 + 1+ 1

+ 2+ 2 + 3+ 3

68.26%68.26%95.44%95.44%99.72%99.72%

22 22 Slide Slide

Standard Normal Probability DistributionStandard Normal Probability Distribution

A random variable having a normal distributionA random variable having a normal distribution with a mean of 0 and a standard deviation of 1 iswith a mean of 0 and a standard deviation of 1 is said to have a said to have a standard normal probabilitystandard normal probability distributiondistribution..

A random variable having a normal distributionA random variable having a normal distribution with a mean of 0 and a standard deviation of 1 iswith a mean of 0 and a standard deviation of 1 is said to have a said to have a standard normal probabilitystandard normal probability distributiondistribution..

23 23 Slide Slide

00zz

The letter The letter z z is used to designate the standardis used to designate the standard normal random variable.normal random variable. The letter The letter z z is used to designate the standardis used to designate the standard normal random variable.normal random variable.


24 24 Slide Slide

Converting to the Standard Normal DistributionConverting to the Standard Normal Distribution


zx

zx

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

25 25 Slide Slide


Standard Normal Density FunctionStandard Normal Density Function

2 / 21( )

2zf x e

2 / 21( )

2zf x e

z z = ( = (xx – – )/)/ = 3.14159= 3.14159ee = 2.71828 = 2.71828

where:where:

26 26 Slide Slide


Example: Pep ZoneExample: Pep ZonePep Zone sells auto parts and supplies includingPep Zone sells auto parts and supplies includinga popular multi-grade motor oil. When thea popular multi-grade motor oil. When thestock of this oil drops to 20 gallons, astock of this oil drops to 20 gallons, areplenishment order is placed.replenishment order is placed. Pep

Zone5w-20Motor Oil

27 27 Slide Slide

The store manager is concerned that sales are beingThe store manager is concerned that sales are beinglost due to stockouts while waiting for an order.lost due to stockouts while waiting for an order.It has been determined that demand duringIt has been determined that demand duringreplenishment lead-time is normallyreplenishment lead-time is normallydistributed with a mean of 15 gallons anddistributed with a mean of 15 gallons anda standard deviation of 6 gallons. a standard deviation of 6 gallons.

The manager would like to know theThe manager would like to know theprobability of a stockout, probability of a stockout, PP((xx > 20). > 20).


PepZone5w-20Motor Oil

Example: Pep ZoneExample: Pep Zone

28 28 Slide Slide

zz = ( = (xx - - )/)/ = (20 - 15)/6= (20 - 15)/6 = .83= .83

zz = ( = (xx - - )/)/ = (20 - 15)/6= (20 - 15)/6 = .83= .83

Solving for the Stockout ProbabilitySolving for the Stockout Probability

Step 1: Convert Step 1: Convert xx to the standard normal distribution. to the standard normal distribution.Step 1: Convert Step 1: Convert xx to the standard normal distribution. to the standard normal distribution.

PepZone5w-20

Motor Oil

Step 2: Find the area under the standard normalStep 2: Find the area under the standard normal curve to the left of z = .83.curve to the left of z = .83.Step 2: Find the area under the standard normalStep 2: Find the area under the standard normal curve to the left of z = .83.curve to the left of z = .83.


29 29 Slide Slide

Cumulative Probability Table for Cumulative Probability Table for the Standard Normal Distributionthe Standard Normal Distribution

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

. . . . . . . . . . .

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

. . . . . . . . . . .

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

. . . . . . . . . . .

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

. . . . . . . . . . .

PepZone5w-20

Motor Oil

PP((zz << .83) .83)


30 30 Slide Slide

PP((z z > .83) = 1 – > .83) = 1 – PP((zz << .83) .83) = 1- .7967= 1- .7967

= .2033= .2033

PP((z z > .83) = 1 – > .83) = 1 – PP((zz << .83) .83) = 1- .7967= 1- .7967

= .2033= .2033


Step 3: Compute the area under the standard normalStep 3: Compute the area under the standard normal curve to the right of curve to the right of zz = .83. = .83.Step 3: Compute the area under the standard normalStep 3: Compute the area under the standard normal curve to the right of curve to the right of zz = .83. = .83.

PepZone5w-20

Motor Oil

ProbabilityProbability of a stockoutof a stockout PP((xx > 20) > 20)


31 31 Slide Slide


00 .83.83

Area = .7967Area = .7967Area = 1 - .7967Area = 1 - .7967

= .2033= .2033

zz

PepZone5w-20

Motor Oil


32 32 Slide Slide

Standard Normal Probability DistributionStandard Normal Probability DistributionIf the manager of Pep Zone wants the If the manager of Pep Zone wants the

probability of a stockout to be no more than .05, probability of a stockout to be no more than .05, what should the reorder point be?what should the reorder point be?

PepZone5w-20

Motor Oil


33 33 Slide Slide

Solving for the Reorder PointSolving for the Reorder PointPep

Zone5w-20

Motor Oil

00

Area = .9500Area = .9500

Area = .0500Area = .0500

zzzz.05.05


34 34 Slide Slide

Solving for the Reorder PointSolving for the Reorder Point

PepZone5w-20

Motor Oil

Step 1: Find the Step 1: Find the zz-value that cuts off an area of .05-value that cuts off an area of .05 in the right tail of the standard normalin the right tail of the standard normal distribution.distribution.

Step 1: Find the Step 1: Find the zz-value that cuts off an area of .05-value that cuts off an area of .05 in the right tail of the standard normalin the right tail of the standard normal distribution.distribution.

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

. . . . . . . . . . .

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

. . . . . . . . . . .

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

. . . . . . . . . . .

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

. . . . . . . . . . .We look up the complement of We look up the complement of

the tail area (1 - .05 = .95)the tail area (1 - .05 = .95)


35 35 Slide Slide

Solving for the Reorder PointSolving for the Reorder Point

PepZone5w-20

Motor Oil

Step 2: Convert Step 2: Convert zz.05.05 to the corresponding value of to the corresponding value of xx..Step 2: Convert Step 2: Convert zz.05.05 to the corresponding value of to the corresponding value of xx..

xx = = μμ + + zz.05.05

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

= 24.87 or 25= 24.87 or 25

xx = = μμ + + zz.05.05

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

= 24.87 or 25= 24.87 or 25

A reorder point of 25 gallons will place the probabilityA reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than) .05.of a stockout during leadtime at (slightly less than) .05.


36 36 Slide Slide

Solving for the Reorder PointSolving for the Reorder PointPep

Zone5w-20

Motor Oil

By raising the reorder point from 20 gallons to By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout25 gallons on hand, the probability of a stockoutdecreases from about .20 to .05.decreases from about .20 to .05. This is a significant decrease in the chance that PepThis is a significant decrease in the chance that PepZone will be out of stock and unable to meet aZone will be out of stock and unable to meet acustomer’s desire to make a purchase.customer’s desire to make a purchase.


37 37 Slide Slide

Normal Approximation Normal Approximation of Binomial Probabilitiesof Binomial Probabilities

When the number of trials, When the number of trials, nn, becomes large,, becomes large,evaluating the binomial probability function byevaluating the binomial probability function byhand or with a calculator is difficulthand or with a calculator is difficult

The normal probability distribution provides anThe normal probability distribution provides aneasy-to-use approximation of binomial probabilitieseasy-to-use approximation of binomial probabilitieswhere where nn > 20, > 20, np np >> 5, and 5, and nn(1 - (1 - pp) ) >> 5. 5.

38 38 Slide Slide

Normal Approximation Normal Approximation of Binomial Probabilitiesof Binomial Probabilities

Set Set μμ = = npnp

(1 )np p (1 )np p

Add and subtract 0.5 (a Add and subtract 0.5 (a continuity correction factorcontinuity correction factor)) because a continuous distribution is being used tobecause a continuous distribution is being used to approximate a discrete distribution. For example, approximate a discrete distribution. For example, PP((xx = 10) is approximated by = 10) is approximated by PP(9.5 (9.5 << xx << 10.5). 10.5).

39 39 Slide Slide

End of Chapter 6End of Chapter 6

1 1 Slide Chapter 6 Continuous Probability Distributions n Uniform Probability Distribution n Normal Probability Distribution n Exponential Probability.

Documents