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Chapter 6Chapter 6Continuous Probability DistributionsContinuous Probability Distributions
■■ Uniform Probability DistributionUniform Probability Distribution
■■ Normal Probability DistributionNormal Probability Distribution
■■ Exponential Probability DistributionExponential Probability Distribution
f (x)f (x)
xx
UniformUniform
xx
f f ((xx))NormalNormal
xx
f (x)f (x) ExponentialExponential
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Continuous Probability DistributionsContinuous Probability Distributions
■■ A A continuous random variablecontinuous random variable can assume any value can assume any value in an interval on the real line or in a collection of in an interval on the real line or in a collection of intervals.intervals.
■■ It is not possible to talk about the probability of the It is not possible to talk about the probability of the random variable assuming a particular value.random variable assuming a particular value.
■■ Instead, we talk about the probability of the random Instead, we talk about the probability of the random variable assuming a value within a given interval.variable assuming a value within a given interval.
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Continuous Probability DistributionsContinuous Probability Distributions
■■ 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 xx11 andand xx22..
f (x)f (x)
xx
UniformUniform
xx11xx11 xx22xx22
xx
f f ((xx))NormalNormal
xx11xx11 xx22xx22
xx11xx11 xx22xx22
ExponentialExponential
xx
f (x)f (x)
xx11xx11 xx22xx22
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Uniform Probability DistributionUniform 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
■■ A random variable is A random variable is uniformly distributeduniformly distributedwhenever 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:
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Var(Var(xx) = () = (bb -- aa))22/12/12
E(E(xx) = () = (aa + + bb)/2)/2
Uniform Probability DistributionUniform Probability Distribution
■■ Expected Value of Expected Value of xx
■■ Variance of Variance of xx
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Uniform Probability DistributionUniform Probability Distribution
■■ Example: Slater's BuffetExample: Slater's Buffet
Slater customers are chargedSlater customers are charged
for the amount of salad they take. for the amount of salad they take.
Sampling suggests that theSampling suggests that the
amount of salad taken is amount of salad taken is
uniformly distributeduniformly distributed
between 5 ounces and 15 ounces.between 5 ounces and 15 ounces.
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■■ Uniform Probability Density FunctionUniform Probability Density Function
ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 1515
= 0 elsewhere= 0 elsewhere
where:where:
xx = salad plate filling weight= salad plate filling weight
Uniform Probability DistributionUniform Probability Distribution
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■■ 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
Var(Var(xx) = () = (bb -- aa))22/12/12
= (15 = (15 –– 5)5)22/12/12
= 8.33= 8.33
Uniform Probability DistributionUniform Probability Distribution
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■■ Uniform Probability DistributionUniform Probability Distribution
for Salad Plate Filling Weightfor Salad Plate Filling Weight
f(x)f(x)
xx55 1010 1515
1/101/10
Salad Weight (oz.)Salad Weight (oz.)
Uniform Probability DistributionUniform Probability Distribution
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f(x)f(x)
xx55 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?
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Uniform Probability DistributionUniform Probability Distribution
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Normal Probability DistributionNormal Probability Distribution
■■ The The normal probability distributionnormal probability distribution is the most is the most important distribution for describing a continuous important distribution for describing a continuous random variable.random variable.
■■ It is widely used in statistical inference.It is widely used in statistical inference.
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HeightsHeightsof peopleof people
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:
ScientificScientificmeasurementsmeasurements
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AmountsAmounts
of rainfallof rainfall
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:
TestTestscoresscores
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Normal Probability DistributionNormal Probability Distribution
■■ Normal Probability Density FunctionNormal Probability Density Function
2 2( ) /21( )
2
xf x e µ σ
σ π− −=
µµ = mean= mean
σσ = standard deviation= standard deviation
ππ = 3.14159= 3.14159
ee = 2.71828= 2.71828
where:where:
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The distribution is The distribution is symmetricsymmetric; its skewness; its skewnessmeasure is zero.measure is zero.
Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
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The entire family of normal probabilityThe entire family of normal probability
distributions is defined by itsdistributions is defined by its meanmean µµ and itsand itsstandard deviationstandard deviation σσ ..
Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
Standard Deviation Standard Deviation σσ
Mean Mean µµxx
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The The highest pointhighest point on the normal curve is at theon the normal curve is at themeanmean, which is also the , which is also the medianmedian and and modemode..
Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
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Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
--1010 00 2020
The mean can be any numerical value: negative,The mean can be any numerical value: negative,zero, or positive.zero, or positive.
xx
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Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
σσσσσσσσ = 15= 15
σ σ σ σ σ σ σ σ = 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.
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Probabilities for the normal random variable areProbabilities for the normal random variable aregiven by given by areas under the curveareas under the curve. The total area. The total areaunder 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).
Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
.5.5 .5.5
xx
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Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
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%
+/+/-- 1 standard deviation1 standard deviation
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%
+/+/-- 2 standard deviations2 standard deviations
of values of a normal random variableof values of a normal random variable
are within of its mean.are within of its mean.
99.72%99.72%
+/+/-- 3 standard deviations3 standard deviations
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Normal Probability DistributionNormal Probability Distribution
■■ CharacteristicsCharacteristics
xxµµ –– 33σσ µµ –– 11σσ
µµ –– 22σσµµ + 1+ 1σσ
µµ + 2+ 2σσµµ + 3+ 3σσµµ
68.26%68.26%
95.44%95.44%
99.72%99.72%
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Standard Normal Probability DistributionStandard Normal Probability Distribution
A random variable having a normal distributionA random variable having a normal distributionwith a mean of 0 and a standard deviation of 1 iswith a mean of 0 and a standard deviation of 1 issaid to have a said to have a standard normal probabilitystandard normal probabilitydistributiondistribution..
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σ σ = 1= 1
00zz
The letter The letter z z is used to designate the standardis used to designate the standardnormal random variable.normal random variable.
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Converting to the Standard Normal DistributionConverting to the Standard Normal Distribution
Standard Normal Probability DistributionStandard Normal Probability Distribution
zx= − µ
σ
We can think of We can think of zz as a measure of the number ofas a measure of the number ofstandard deviations standard deviations xx is from is from µµ..
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Standard Normal Probability DistributionStandard Normal Probability Distribution
■■ Standard Normal Density FunctionStandard Normal Density Function
π−=
2 /21( )
2
zf x e
z z = (= (xx –– µµ)/)/σσππ = 3.14159= 3.14159
ee = 2.71828= 2.71828
where:where:
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Standard Normal Probability DistributionStandard Normal Probability Distribution
■■ Example: Pep ZoneExample: Pep Zone
Pep Zone sells auto parts and supplies includingPep Zone sells auto parts and supplies including
a popular multia popular multi--grade motor oil. When thegrade motor oil. When the
stock of this oil drops to 20 gallons, astock of this oil drops to 20 gallons, a
replenishment order is placed.replenishment order is placed.Pep
Zone
5w-20Motor Oil
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The store manager is concerned that sales are beingThe store manager is concerned that sales are being
lost 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 during
replenishment leadreplenishment lead--time is normallytime is normally
distributed with a mean of 15 gallons anddistributed with a mean of 15 gallons and
a standard deviation of 6 gallons. a standard deviation of 6 gallons.
The manager would like to know theThe manager would like to know the
probability of a stockout, probability of a stockout, PP((xx > 20). > 20).
Standard Normal Probability DistributionStandard Normal Probability Distribution
PepZone
5w-20Motor Oil
■■ Example: Pep ZoneExample: Pep Zone
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zz = (= (xx -- µµ)/)/σσ= (20 = (20 -- 15)/615)/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.
PepZone
5w-20Motor Oil
Step 2: Find the area under the standard normalStep 2: Find the area under the standard normalcurve to the left of curve to the left of zz = .83.= .83.
see next slidesee next slide
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ 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
. . . . . . . . . . .
PepZone
5w-20Motor Oil
PP((zz <<.83).83)
Standard Normal Probability DistributionStandard Normal Probability Distribution
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PP((z z > .83) = 1 > .83) = 1 –– PP((zz << .83) .83)
= 1= 1-- .7967.7967
= .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 normalcurve to the right of curve to the right of zz = .83.= .83.
PepZone
5w-20Motor Oil
ProbabilityProbabilityof a stockoutof a stockout PP((xx > 20)> 20)
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Solving for the Stockout ProbabilitySolving for the Stockout Probability
00 .83.83
Area = .7967Area = .7967Area = 1 Area = 1 -- .7967.7967
= .2033= .2033
zz
PepZone
5w-20Motor Oil
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Standard Normal Probability DistributionStandard Normal Probability Distribution
If the manager of Pep Zone wants the probability If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the of a stockout to be no more than .05, what should the reorder point be?reorder point be?
PepZone
5w-20Motor Oil
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Solving for the Reorder PointSolving for the Reorder Point
PepZone
5w-20Motor Oil
00
Area = .9500Area = .9500
Area = .0500Area = .0500
zzzz.05.05
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Solving for the Reorder PointSolving for the Reorder Point
PepZone
5w-20Motor Oil
Step 1: Find the Step 1: Find the zz--value that cuts off an area of .05value that cuts off an area of .05in the right tail of the standard normalin the right tail of the standard normaldistribution.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
. . . . . . . . . . .We look up the complement We look up the complement of the tail area (1 of the tail area (1 -- .05 = .95).05 = .95)
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Solving for the Reorder PointSolving for the Reorder Point
PepZone
5w-20Motor Oil
Step 2: Convert Step 2: Convert zz.05.05 to the corresponding value of to the corresponding value of xx..
xx = = µµ + + zz.05.05σσ = 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 probabilityof a stockout during leadtime at (slightly less than) .05.of a stockout during leadtime at (slightly less than) .05.
Standard Normal Probability DistributionStandard Normal Probability Distribution
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■■ Solving for the Reorder PointSolving for the Reorder Point
PepZone
5w-20Motor 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.
Standard Normal Probability DistributionStandard Normal Probability Distribution
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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 aneasyeasy--toto--use approximation of binomial probabilitiesuse approximation of binomial probabilitieswhere where nn > 20, > 20, np np >> 5, and 5, and nn(1 (1 -- pp) ) >> 5.5.
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Normal Approximation Normal Approximation of Binomial Probabilitiesof Binomial Probabilities
■■ Set Set µµ = = npnp
σ = −(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).
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Exponential Probability DistributionExponential Probability Distribution
■■ The exponential probability distribution is useful in The exponential probability distribution is useful in describing the time it takes to complete a task.describing the time it takes to complete a task.
■■ The exponential random variables can be used to The exponential random variables can be used to describe:describe:
Time betweenTime betweenvehicle arrivalsvehicle arrivalsat a toll boothat a toll booth
Time requiredTime requiredto completeto complete
a questionnairea questionnaire
Distance betweenDistance betweenmajor defectsmajor defectsin a highwayin a highway
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■■ Density FunctionDensity Function
Exponential Probability DistributionExponential Probability Distribution
where: where: µµ = mean= mean
ee = 2.71828= 2.71828
f x e x( ) /= −1µ
µfor for xx >> 0, 0, µµ > 0> 0
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■■ Cumulative ProbabilitiesCumulative Probabilities
Exponential Probability DistributionExponential Probability Distribution
P x x e x( ) /≤ = − −0 1 o µ
where:where:
xx00 = some specific value of = some specific value of xx
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Exponential Probability DistributionExponential Probability Distribution
■■ Example: Al’s FullExample: Al’s Full--Service PumpService Pump
The time between arrivals of carsThe time between arrivals of cars
at Al’s fullat Al’s full--service gas pump followsservice gas pump follows
an exponential probability distributionan exponential probability distribution
with a mean time between arrivals of with a mean time between arrivals of
3 minutes. Al would like to know the3 minutes. Al would like to know the
probability that the time between two successiveprobability that the time between two successive
arrivals will be 2 minutes or less.arrivals will be 2 minutes or less.
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xx
f(x)f(x)
.1.1
.3.3
.4.4
.2.2
1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10Time Between Successive Arrivals (mins.)Time Between Successive Arrivals (mins.)
Exponential Probability DistributionExponential Probability Distribution
PP((xx << 2) = 1 2) = 1 -- 2.718282.71828--2/32/3 = 1 = 1 -- .5134 = .4866.5134 = .4866
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Exponential Probability DistributionExponential Probability Distribution
A property of the exponential distribution is thatA property of the exponential distribution is thatthe mean, the mean, µµ, and standard deviation, , and standard deviation, σσ, are equal., are equal.
Thus, the standard deviation, Thus, the standard deviation, σσ, and variance, , and variance, σ σ 22, for, forthe time between arrivals at Al’s fullthe time between arrivals at Al’s full--service pump are:service pump are:
σσ = = µµ = 3 minutes= 3 minutes
σσ 2 2 = (3)= (3)2 2 = 9= 9
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Exponential Probability DistributionExponential Probability Distribution
The exponential distribution is skewed to the right.The exponential distribution is skewed to the right.
The skewness measure for the exponential distributionThe skewness measure for the exponential distributionis 2.is 2.
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Relationship between the PoissonRelationship between the Poissonand Exponential Distributionsand Exponential Distributions
The Poisson distributionThe Poisson distributionprovides an appropriate descriptionprovides an appropriate description
of the number of occurrencesof the number of occurrencesper intervalper interval
The exponential distributionThe exponential distributionprovides an appropriate descriptionprovides an appropriate description
of the length of the intervalof the length of the intervalbetween occurrencesbetween occurrences
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End of Chapter 6End of Chapter 6