1 1 slide © 2016 cengage learning. all rights reserved. chapter 6 continuous probability...

1 1 Slide Slide© 2016 Cengage Learning. All Rights Reserved. © 2016 Cengage Learning. All Rights Reserved.

Chapter 6 Continuous Probability Distributions

Normal Probability DistributionNormal Probability Distribution

Uniform Probability DistributionUniform Probability Distribution


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 It is not possible to talk about the probability of the random variable assuming a particular of the random variable assuming a particular value.value. Instead, we talk about the probability of the Instead, we talk about the probability of the random variable assuming a value within a random variable assuming a value within a given interval.given interval.


Continuous Probability Distributions

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


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 interval’s length. interval’s length.

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


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


Expected Value of Expected Value of xx

Variance of Variance of xx

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


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

Uniform Probability Distribution Example

Slater buffet customers are charged for the Slater buffet customers are charged for the amount of salad they take. Sampling suggests amount of salad they take. Sampling suggests that the amount of salad taken is uniformly that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces.distributed between 5 ounces and 15 ounces.


Expected Value of Expected Value 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


Variance of Variance of xx



Uniform Probability Distribution for Salad Plate Filling Weight

f(x)f(x)

x x

1/101/10

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

55 1010 151500


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

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


1212


Area as a Measure of Probability

The area under the graph of The area under the graph of ff((xx) and ) and probability are identical.probability are identical.

This is valid for all continuous random This is valid for all continuous random variables.variables.

The probability that The probability that xx takes on a value takes on a value between some lower value between some lower value xx11 and some and some higher value higher value xx22 can be found by computing can be found by computing the area under the graph of the area under the graph of ff((xx) over the ) over the interval from interval from xx11 to to xx22. .


Normal Probability Distribution

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

It is widely used in statistical inference.It is widely used in statistical inference. It has been used in a wide variety of It has been used in a wide variety of

applicationsapplications

including:including:• Heights of Heights of peoplepeople• Rainfall Rainfall amountsamounts

• Test scoresTest scores• Scientific Scientific measurementsmeasurements Abraham de Moivre, a French mathematician, Abraham de Moivre, a French mathematician,

published published The Doctrine of ChancesThe Doctrine of Chances in 1733. in 1733. He derived the normal distribution.He derived the normal distribution.


Normal Probability Distribution Normal Probability Density Function

2 2( ) / 21( )

2xf x e

2 2( ) / 21( )

2xf x e

= mean= mean

= standard deviation= standard deviation

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

where:where:


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


The entire family of normal probability distributionsThe entire family of normal probability distributions is defined by itsis defined by its meanmean and its and its standard deviationstandard deviation . . The entire family of normal probability distributionsThe entire family of normal probability distributions is defined by itsis defined by its meanmean and its and its standard deviationstandard deviation . .


CharacteristicsCharacteristics

Standard Deviation Standard Deviation

Mean Mean xx


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..

Normal Probability Distribution CharacteristicsCharacteristics

xx



-10-10 00 2525

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



= 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


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



Characteristics (basis for the empirical rule)Characteristics (basis for the empirical rule)

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.95.44%95.44%95.44%95.44%

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



Characteristics (basis for the empirical rule)Characteristics (basis for the empirical rule)

xx – – 33 – – 11

– – 22 + 1+ 1

+ 2+ 2 + 3+ 3

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


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.

Standard Normal Probability Distribution

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


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


Standard Normal Probability Distribution Example: Pep Zone

The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? PP((xx > 20) = ? > 20) = ?

Pep Zone sells auto parts and supplies including a Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that is placed. The store manager is concerned that sales are being lost due to stockouts while waiting sales are being lost due to stockouts while waiting for a replenishment order.for a replenishment order.It has been determined that demand during It has been determined that demand during replenishment lead-time is normally distributed replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation with a mean of 15 gallons and a standard deviation of 6 gallons.of 6 gallons.


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

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

Solving 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.

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

see next slidesee next slide see next slidesee next slide



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

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



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

Solving for the Stockout ProbabilitySolving for the Stockout Probability

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.

ProbabilityProbability of a of a

stockoutstockoutPP((xx > > 20)20)



Solving for the Stockout ProbabilitySolving for the Stockout Probability

00 .83.83

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

= .2033= .2033

zz



Standard Normal Probability DistributionStandard Normal Probability Distribution


If the manager of Pep Zone wants the If the manager of Pep Zone wants the probabilityprobability

of a stockout during replenishment lead-time of a stockout during replenishment lead-time to beto be

no more than .05, what should the reorder no more than .05, what should the reorder point be?point be?

------------------------------------------------------------------------------------------------------------------------------

(Hint: Given a probability, we can use (Hint: Given a probability, we can use the standard normal table in an the standard normal table in an inverse fashion to find the inverse fashion to find the corresponding corresponding zz value.) value.)


Solving for the Reorder PointSolving for the Reorder Point

00

Area = .9500Area = .9500

Area = .0500Area = .0500

zzzz.05.05




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 = = + + zz.05.05

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

= 24.87 or 25= 24.87 or 25

xx = = + + 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.


zz = ( = (xx - - )/)/ zzx - x -

x x = = z z

zz = ( = (xx - - )/)/ zzx - x -

x x = = z z




1515xx

24.8724.87

Probability of Probability of aa

stockout stockout duringduring

replenishmenreplenishmentt

lead-time lead-time = .05= .05

Probability of Probability of aa


replenishmenreplenishmentt

lead-time lead-time = .05= .05

Probability of Probability of nono


replenishmentreplenishmentlead-time lead-time

= .95= .95

Probability of Probability of nono


replenishmentreplenishmentlead-time lead-time

= .95= .95



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 thatThis is a significant decrease in the chance thatPep Zone will be out of stock and unable to meet aPep Zone will be out of stock and unable to meet acustomer’s desire to make a purchase.customer’s desire to make a purchase.


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