Binomial Applet

Binomial Applet

http://stattrek.com/Tables/Binomial.aspx

Chapter 6

Continuous Probability Distributions

Discrete vs. Continuous Random Variables

1. For discrete random variables we find the probability the variable will take on a specific value. For continuous random variable we find the probability the variable will fall in some range.

2. For discrete random variables we find the probability associated with some value using a probability function. For a continuous random variable we find the area under a curve between two points.

Probability Density Curves

1. The area under the curve is equal to 12. The area under the curve between two values of

the variable measures the probability the value will fall between those values

3. A probability density curve is defined by a probability density function

Uniform Probability Distribution

The probability of the variable falling into some interval is the same for all equally-sized intervals.

Uniform Probability Density Function

0)(

,

)(

xfbxandaxfor

bdaandbcagivenbxaforab

cddxcf

Mean and Variance of a Uniform Distribution

12

)(

2)(

2abxVar

baxE

Uniform Distribution, ExampleAssume that a delivery truck takes between 200 minutes and 250 minutes to complete its delivery route. Assume that the amount of time taken is uniformly distributed.

What is the probability the truck will take between 210 and 230 minutes?

Uniform Distribution, ExampleWhat is the probability the truck will take between 210 and 230 minutes?

(230 – 210)/(250 – 200) = 20/50 = 0.4

Uniform Distribution, Example

What is the probability the truck will take more than 240 minutes?

Uniform Distribution, ExampleWhat is the probability the truck will take more than 240 minutes?

(250 – 240)/(250 – 200) = 10/50 = 0.2

Uniform Distribution, Example

What is the probability the truck will take less than 220 minutes?

Uniform Distribution, ExampleWhat is the probability the truck will take less than 220 minutes?

(220 – 200)/(250 – 200) = 20/50 = 0.4

Uniform Distribution, Example

What is the mean and variance of this distribution?

E(x) = (250 + 200)/2 = 225Var(x) = (250 – 200)2/12 = 2500/12 = 208.33

Importance of the Normal Distribution

•Many variables have a distribution similar to that of the normal distribution•The means of samples are normally distributed (given samples of 30 or more)

Characteristics of the Normal Distribution

• Symmetrical and bell-shaped• The tails of the distribution asymptotically converge on the x axis• The location of the distribution on the x axis is determined by the mean of the distribution• The shape of the distribution is determined by the variance• The total area under the curve is equal to 1, the areas to either side of the mean equal 0.5• Moving out a given number of standard deviations will always capture the same share of the distribution

Normal Distributions

Wikipedia

Normal Probability Distribution

xm – 3s m – 1s

m – 2sm + 1s

m + 2sm + 3s

m

68.26%

95.44%

99.72%

Standard Normal Distribution

m = 0=1s

To convert a normal distribution to standard normal:z = (x - m)/s

Characteristics of the Normal Distribution

http://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html

Normal Probability Density Function

2

2

2

2

1)(

x

exf

Standard Normal Table

Standard Normal Table

Standard Normal Table

Normal Distribution, Exercises

Find: P(z < 2)

P(z < 2) = .9772

Normal Distribution, Exercises

Find: P(z < -2)

P(z < -2) = .0228

Normal Distribution, Exercises

Find: P(0 < z < 2)

P(0 < z < 2) = .9772 - .5 = .4772OrP(0 < z < 2) = .5 - .0228 = .4772

Normal Distribution, Exercises

Find: P(-1 < z < 0)

P(-1 < z < 0) = P(0 < z < 1) = .8413 - .5 = .3413OrP(-1 < z < 0) = .5 - .1587 = .3413

Normal Distribution, Exercises

Find: P(z > -1)

P(z > -1) = P(z < 1) = .8413OrP(z > -1) = 1 - .1587 = .8413

Normal Distribution, Exercises

Find: P(z > 2)

P(z > 2) = 1 - .9772 = .0228OrP(z > 2) = P(z < -2) = .0228

Normal Distribution, Exercises

Find: P(1 < z < 1.5)

P(1 < z < 1.5) = .9332 - .8413 = 0.0919OrP(1 < z < 1.5) = P(-1.5 < z < -1) = .1587 - .0668 = .0919

Normal Distribution, Exercises

Find: P(-1.5 < z < -0.5)

P(-1.5 < z < -0.5) = P(0.5 > z > 1.5) = .9332 - .6915 = .2417OrP(-1.5 < z < -0.5) = .3085 -.0668 = .2417

Normal Distribution, Exercises

Find: P(-1 < z < 2)

P(-1 < z < 2) = .9772 - .1587 = 0.8185

Standard Normal Probability Distribution, Example

Pep Zone sells auto parts and supplies includinga popular multi-grade motor oil. When thestock of this oil drops to 20 gallons, areplenishment order is placed. Pep

Zone

5w-20Motor Oil

The store manager is concerned that sales are being lost due to stockouts while waiting for an order.It has been determined that demandduring replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know theprobability of a stockout, P(x > 20).

Standard Normal Probability Distribution, Example

PepZone

5w-20Motor Oil

Standard Normal Probability Distribution, Example

1. Solve for the z score:Z = (x – m)/s= (20 – 15)/6

= .832. Look up value in standard normal table

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-20Motor Oil

P(z < .83)

Standard Normal Probability Distribution, Example

Standard Normal Probability Distribution, Example

3. Compute the area under the curve to the right of 0.83

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

= .2033

If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be?

PepZone5w-20Motor Oil

Standard Normal Probability Distribution, Example

Solving for the Reorder Point

PepZone5w-20Motor Oil

0

Area = .9500

Area = .0500

zz.05

Standard Normal Probability Distribution, Example

PepZone5w-20Motor Oil

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 thecomplement of thetail area (1 - .05 = .95)

Standard Normal Probability Distribution, Example

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

PepZone5w-20Motor Oil

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

Standard Normal Probability Distribution, Example

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

x = m+z.05s = 15 + 1.645(6) = 24.87 or 25

PepZone5w-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.

Standard Normal Probability Distribution, Example

Practice Homework

P. 241, #12, 20P. 246, #30