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Chapter 6Continuous Probability Distributions

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

f (x)f (x)

xx

Uniform

x

f (x) Normal

xx

f (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.

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)f (x)

xx

Uniform

x1x1 x2x2

x

f (x) Normal

x1x1 x2x2

x1x1 x2x2

Exponential

xx

f (x)f (x)

x1x1 x2x2

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.

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

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


� 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


� 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.

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.


� 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


� Probability Table for the Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .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. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .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. . . . . . . . . . .

PepZone

5w-20Motor Oil

P(0 < z < .83)


P(z > .83) = .5 – P(z < .83) = .5- .2967

= .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 = .2967Area = .5 - .2967

= .2033

z

PepZone

5w-20Motor Oil


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

of a stockout to be no more than .04, what should the reorder point be? Assuming that it has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons.

PepZone

5w-20Motor Oil


� Solving for the Reorder Point

PepZone

5w-20Motor Oil

0

Area = .9600

Area = .0400

zz.04



PepZone

5w-20Motor Oil

Step 1: Find the z-value that cuts off an area of .04in 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 . . . . . . . . . . .

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 - .04 = .96)



PepZone

5w-20Motor Oil

Step 2: Convert z.04 to the corresponding value of x.

x = μ + z.05σ = 15 + 1.75(6)

= 25.5 or 26

A reorder point of 26 gallons will place the probabilityof a stockout during leadtime at (slightly less than) .04.



PepZone

5w-20Motor Oil

By raising the reorder point from 20 gallons to 26 gallons on hand, the probability of a stockoutdecreases from about .20 to .04.

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.


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