© 2005 Thomson/South-Western Slide 1

1

1Slide© 2005 Thomson/South-Western

Slides Prepared byJOHN S. LOUCKS

St. Edward’s University


Chapter 6Continuous Probability Distributions

� Uniform Probability Distribution� Normal Probability Distribution� Exponential Probability Distribution

f (x)

x

Uniform

x

f (x) Normal

x

f (x) Exponential


Continuous Probability Distributions

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

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

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

2


Continuous Probability Distributions

� The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2.

f (x)

x

Uniform

x1 x2

x

f (x) Normal

x1 x2

x1 x2

Exponential

x

f (x)

x1 x2


Uniform Probability Distribution

where: a = smallest value the variable can assumeb = largest value the variable can assume

f (x) = 1/(b – a) for a < x < b= 0 elsewhere

� A random variable is uniformly distributedwhenever the probability is proportional to the interval’s length.

� The uniform probability density function is:


Var(x) = (b - a)2/12

E(x) = (a + b)/2


� Expected Value of x

� Variance of x

3



� Example: Slater's BuffetSlater customers are charged

for the amount of salad they take. Sampling suggests that theamount of salad taken is uniformly distributedbetween 5 ounces and 15 ounces.


� Uniform Probability Density Function

f(x) = 1/10 for 5 < x < 15= 0 elsewhere

where:x = salad plate filling weight



� Expected Value of x

� Variance of x

E(x) = (a + b)/2= (5 + 15)/2= 10

Var(x) = (b - a)2/12= (15 – 5)2/12= 8.33


4


� Uniform Probability Distributionfor Salad Plate Filling Weight

f(x)

x5 10 15

1/10

Salad Weight (oz.)



f(x)

x5 10 15

1/10

Salad Weight (oz.)

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

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

12



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.

5


Heightsof people


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

Scientificmeasurements


Amountsof rainfall


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

Testscores



� Normal Probability Density Function

2 2( ) /21( )2

xf x e µ σ

σ π− −=

2 2( ) /21( )2

xf x e µ σ

σ π− −=

µ = meanσ = standard deviationπ = 3.14159e = 2.71828

where:

6


The distribution is symmetric; its skewnessmeasure is zero.


� Characteristics

x


The entire family of normal probabilitydistributions is defined by its mean µ and itsstandard deviation σ .


� Characteristics

Standard Deviation σ

Mean µx


The highest point on the normal curve is at themean, which is also the median and mode.


� Characteristics

x

7



� Characteristics

-10 0 20

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

x



� Characteristics

σ = 15

σ = 25

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

x


Probabilities for the normal random variable aregiven by areas under the curve. The total areaunder the curve is 1 (.5 to the left of the mean and.5 to the right).


� Characteristics

.5 .5

x

8



� Characteristics

of values of a normal random variableare within of its mean.68.26%

+/- 1 standard deviation


+/- 2 standard deviations


+/- 3 standard deviations



� Characteristics

xµ – 3σ µ – 1σ

µ – 2σµ + 1σ

µ + 2σµ + 3σµ

68.26%95.44%99.72%


Standard Normal Probability Distribution

A random variable having a normal distributionwith a mean of 0 and a standard deviation of 1 issaid to have a standard normal probabilitydistribution.

9


σ = 1

0z

The letter z is used to designate the standardnormal random variable.



� Converting to the Standard Normal Distribution


z x=

− µσ

z x=

− µσ

We can think of z as a measure of the number ofstandard deviations x is from µ.



� Standard Normal Density Function

π−=

2 /21( )2

zf x eπ

−=2 /21( )

2zf x e

z = (x – µ)/σπ = 3.14159e = 2.71828

where:

10



� Example: Pep ZonePep Zone sells auto parts and supplies including

a popular multi-grade motor oil. When thestock of this oil drops to 20 gallons, areplenishment order is placed.

PepZone5w-20

Motor Oil


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

The manager would like to know theprobability of a stockout, P(x > 20).


PepZone5w-20

Motor Oil

� Example: Pep Zone


z = (x - µ)/σ= (20 - 15)/6= .83

� Solving for the Stockout Probability

Step 1: Convert x to the standard normal distribution.

PepZone

5w-20Motor Oil

Step 2: Find the area under the standard normalcurve to the left of z = .83.

see next slide


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� Cumulative Probability Table for the 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

P(z < .83)



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

= .2033


Step 3: Compute the area under the standard normalcurve to the right of z = .83.

PepZone

5w-20Motor Oil

Probabilityof a stockout P(x > 20)




0 .83

Area = .7967Area = 1 - .7967

= .2033

z

PepZone

5w-20Motor Oil


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� Standard Normal Probability DistributionIf the manager of Pep Zone wants the probability

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

PepZone

5w-20Motor Oil



� Solving for the Reorder Point

PepZone

5w-20Motor Oil

0

Area = .9500

Area = .0500

zz.05




PepZone

5w-20Motor Oil

Step 1: Find the z-value that cuts off an area of .05in the right tail of the standard normaldistribution.

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

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .94411.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .95451.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .96331.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .97061.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 . . . . . . . . . . .

We look up the complement of the tail area (1 - .05 = .95)


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PepZone

5w-20Motor Oil

Step 2: Convert z.05 to the corresponding value of x.

x = µ + z.05σ = 15 + 1.645(6)

= 24.87 or 25

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




PepZone

5w-20Motor Oil

By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockoutdecreases from about .20 to .05.

This is a significant decrease in the chance that PepZone will be out of stock and unable to meet acustomer’s desire to make a purchase.



Normal Approximation of Binomial Probabilities

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

The normal probability distribution provides aneasy-to-use approximation of binomial probabilitieswhere n > 20, np > 5, and n(1 - p) > 5.

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Normal Approximation of Binomial Probabilities

� Set µ = np

σ = −(1 )np pσ = −(1 )np p

� Add and subtract 0.5 (a continuity correction factor)because a continuous distribution is being used toapproximate a discrete distribution. For example, P(x = 10) is approximated by P(9.5 < x < 10.5).


Exponential Probability Distribution

� The exponential probability distribution is useful in describing the time it takes to complete a task.

� The exponential random variables can be used to describe:

Time betweenvehicle arrivalsat a toll booth

Time requiredto complete

a questionnaire

Distance betweenmajor defectsin a highway

SLOW


� Density Function


where: µ = meane = 2.71828

f x e x( ) /= −1µ

µf x e x( ) /= −1µ

µ for x > 0, µ > 0

15


� Cumulative Probabilities


P x x e x( ) /≤ = − −0 1 o µP x x e x( ) /≤ = − −0 1 o µ

where:x0 = some specific value of x



� Example: Al’s Full-Service PumpThe time between arrivals of cars

at Al’s full-service gas pump followsan exponential probability distributionwith a mean time between arrivals of 3 minutes. Al would like to know theprobability that the time between two successivearrivals will be 2 minutes or less.


x

f(x)

.1

.3

.4

.2

1 2 3 4 5 6 7 8 9 10Time Between Successive Arrivals (mins.)


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

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A property of the exponential distribution is thatthe mean, µ, and standard deviation, σ, are equal.

Thus, the standard deviation, σ, and variance, σ 2, forthe time between arrivals at Al’s full-service pump are:

σ = µ = 3 minutes

σ 2 = (3)2 = 9



The exponential distribution is skewed to the right.

The skewness measure for the exponential distributionis 2.


Relationship between the Poissonand Exponential Distributions

The Poisson distributionprovides an appropriate description

of the number of occurrencesper interval

The exponential distributionprovides an appropriate description

of the length of the intervalbetween occurrences

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End of Chapter 6

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© 2005 Thomson/South-Western Slide 1