Continuous Probability Distributions

1 1

Continuous Probability Continuous Probability DistributionsDistributions

2 2

Continuous Probability DistributionsContinuous Probability Distributions

The Uniform DistributionThe Uniform Distribution

aa bb

The Normal DistributionThe Normal Distribution

The Exponential DistributionThe Exponential Distribution

3 3


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

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

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

4 4


The probability of the continuous The probability of the continuous random variable assuming a specific random variable assuming a specific value is 0. value is 0.

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

5 5

aa bb


aa bb

aa bbx1x2x1

x1

P(x1 ≤ x≤ x2) P(x≤ x1)

P(x≥ x1)

P(x≥ x1)= 1- P(x<x1)

6 6

The Uniform Probability DistributionThe Uniform Probability Distribution

Uniform Probability Density FunctionUniform Probability Density Function

f f ((xx) = 1/() = 1/(bb - - aa) for ) for aa << xx << bb

= 0 elsewhere= 0 elsewhere

wherewhere

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

The probability of the continuous random variable The probability of the continuous random variable assuming a specific value is 0. assuming a specific value is 0.

P(x=xP(x=x11) = 0) = 0

7 7

Example: Slater's BuffetExample: Slater's Buffet

Slater customers are charged for the amount of saladSlater customers are charged for the amount of salad

they take. Sampling suggests that the amount of saladthey take. Sampling suggests that the amount of salad

taken is uniformly distributed between 5 ounces and 15taken is uniformly distributed between 5 ounces and 15

ounces.ounces.

Probability Density Function Probability Density Function

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

= 0 elsewhere= 0 elsewhere

wherewhere

xx = salad plate filling weight = salad plate filling weight

8 8

Example: Slater's BuffetExample: Slater's Buffet

What is the probability that a customer will takeWhat is the probability that a customer will take

between 12 and 15 ounces of salad?between 12 and 15 ounces of salad?

F (x )F (x )

x x55 1010 15151212

1/101/10

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

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

9 9


x x

f (x )f (x )

55 15151212

1/101/10P(8<x < 12) = ?P(8<x < 12) = ?

88

P(8<x < 12) = (1/10)(12-8) = .4P(8<x < 12) = (1/10)(12-8) = .4

10 10


x x

f (x )f (x )

55 15151212

1/101/10P(0<x < 12) = ?P(0<x < 12) = ?

P(0<x < 12) = P(5<x < 12)== (1/10)(12-5) = .7P(0<x < 12) = P(5<x < 12)== (1/10)(12-5) = .7

11 11


Uniform Probability Density FunctionUniform Probability Density Function

f f ((xx) = 1/() = 1/(bb - - aa) for ) for aa << xx << bb

= 0 elsewhere= 0 elsewhere Expected Value of Expected Value of xx

E(E(xx) = () = (aa + + bb)/2)/2 Variance of Variance of xx

Var(Var(xx) = () = (bb - - aa))22/12/12

wherewhere

aa = smallest value the variable can = smallest value the variable can assumeassume

bb = largest value the variable can assume = largest value the variable can assume

12 12

Normal DistributionNormal Distribution

13 13

Before Starting Normal DistributionBefore Starting Normal Distribution

x x

f (x )f (x )

55 15151212

1/101/10

P(8<x < 12) = ?P(8<x < 12) = ?

P( < 8) = ?P( < 8) = ?

x x

f (x )f (x )

55 15151212

1/101/10

P( < 12) = ?P( < 12) = ?

P(8<x < 12) = .7-.3 = .4P(8<x < 12) = .7-.3 = .4

14 14

The Normal Probability DistributionThe Normal Probability Distribution

Graph of the Normal Probability Density Graph of the Normal Probability Density FunctionFunction

xx

f f ((x x ))

15 15

The Normal CurveThe Normal Curve

The shape of the normal curve is often The shape of the normal curve is often illustrated as a bell-shaped curve. illustrated as a bell-shaped curve.

The highest point on the normal curve is at the The highest point on the normal curve is at the mean of the distribution.mean of the distribution.

The normal curve is symmetric.The normal curve is symmetric.

The standard deviation determines the width The standard deviation determines the width of the curve.of the curve.

16 16

The Normal CurveThe Normal Curve

The total area under the curve the same as The total area under the curve the same as any other probability distribution is 1.any other probability distribution is 1.

The probability of the normal random variable The probability of the normal random variable assuming a specific value the same as any assuming a specific value the same as any other continuous probability distribution is 0. other continuous probability distribution is 0.

Probabilities for the normal random variable Probabilities for the normal random variable are given by areas under the curve.are given by areas under the curve.

17 17

The Normal Probability Density FunctionThe Normal Probability Density Function

f x e x( ) ( ) / 12

2 2 2

f x e x( ) ( ) / 1

2

2 2 2

wherewhere

= mean= mean

= standard deviation= standard deviation

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

18 18

The Standard Normal Probability Density The Standard Normal Probability Density FunctionFunction

wherewhere

= 0= 0

= 1= 1

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

19 19

The table will give this probability

Given any Given any positive value for positive value for zz, the table , the table will give us the following probability will give us the following probability

Given positive z

The probability that we find using The probability that we find using the table is the probability of having the table is the probability of having a standard normal variable between a standard normal variable between

0 and the given positive z.0 and the given positive z.

20 20

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

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549

.7 .2580 .2612 .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

Given Given zz find the probability find the probability

21 21

Given this probability between 0 and .5

Given any probability between 0 Given any probability between 0 and .5,and .5,, the table will give us the , the table will give us the

following positive following positive zz value value

The table will give us this positive z

22 22

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

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549

.7 .2580 .2612 .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

Given the probability find Given the probability find zz find find

23 23

10%

40%

What is the What is the zz value where probability of a value where probability of a standard normal variable to be greater than z standard normal variable to be greater than z

is .1 is .1

24 24

Standard Normal Probability Standard Normal Probability DistributionDistribution

(Z Distribution)(Z Distribution)

25 25

Standard Normal Probability DistributionStandard Normal Probability Distribution

A random variable that has a normal distribution A random variable that has a normal distribution with a mean of zero and a standard deviation of with a mean of zero and a standard deviation of one is said to have a one is said to have a standard normal standard normal probability distributionprobability distribution..

The letter The letter z z is commonly used to designate this is commonly used to designate this normal random variable.normal random variable.

The following expression convert any Normal The following expression convert any Normal Distribution into the Standard Normal Distribution Distribution into the Standard Normal Distribution

zx

zx

26 26

Example: Pep ZoneExample: Pep Zone

Pep Zone sells auto parts and supplies including Pep Zone sells auto parts and supplies including multi-grade motor oil. When the stock of this oil multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is drops to 20 gallons, a replenishment order is placed.placed.

The store manager is concerned that sales are being The 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 leadtime demand is It has been determined that leadtime demand is normally distributed with a mean of 15 gallons and normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. a standard deviation of 6 gallons.

In Summary; we have a In Summary; we have a NN (15, 6): A normal (15, 6): A normal randomrandom

variable with mean of 15 and std of 6. variable with mean of 15 and std of 6.

The manager would like to know the probability of a The manager would like to know the probability of a stockout, P(stockout, P(xx > 20). > 20).

27 27

zz = ( = (xx - - )/)/

= (20 - 15)/6= (20 - 15)/6

= .83= .83

Standard Normal DistributionStandard Normal Distribution

00 .83.83

Area = .5Area = .5zz

28 28

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

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549

.7 .2580 .2612 .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


29 29

The Probability of Demand Exceeding 20The Probability of Demand Exceeding 20

00 .83.83

Area = .2967Area = .2967

Area = .5Area = .5

Area = .2033Area = .2033

zz

The Standard Normal table shows an area The Standard Normal table shows an area of .2967 for the region between the of .2967 for the region between the zz = 0 line = 0 line and the and the z z = .83 line above. The shaded tail = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a area is .5 - .2967 = .2033. The probability of a stockout is .2033.stockout is .2033.

30 30

If the manager of Pep Zone wants the probability of aIf the manager of Pep Zone wants the probability of a

stockout to be no more than .05, what should thestockout to be no more than .05, what should the

reorder point be?reorder point be?

Let Let zz.05.05 represent the represent the zz value cutting the tail area value cutting the tail area of .05. of .05.

Area = .05Area = .05

Area = .5 Area = .5 Area = .45 Area = .45

00 zz.05.05


31 31

Using the Standard Normal Probability TableUsing the Standard Normal Probability Table

We now look-up the .4500 area in the We now look-up the .4500 area in the Standard Normal Probability table to find the Standard Normal Probability table to find the corresponding corresponding zz.05.05 value. value. zz.05.05 = 1.645 is a = 1.645 is a reasonable estimate. reasonable estimate. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 .

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

.

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 .


32 32

The corresponding value of The corresponding value of xx is given by is given by

xx = = + + zz.05.05

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

= 24.87= 24.87

A reorder point of 24.87 gallons will place theA reorder point of 24.87 gallons will place the

probability of a stockout during leadtime probability of a stockout during leadtime at .05. at .05.

Perhaps Pep Zone should set the reorder point Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05.at 25 gallons to keep the probability under .05.


33 33

7171 6666 6161 6565 54549393

6060 8686 7070 7070 73737373

5555 6363 5656 6262 76765454

8282 7979 7676 6868 53535858

8585 8080 5656 6161 61616464

6565 6262 9090 6969 76767979

7777 5454 6464 7474 65656565

6161 5656 6363 8080 56567171

7979 8484

A firm has assumed that the distribution of the aptitude test of people applying for a job in this firm is normal. The following sample is available.

Example: Aptitude Test Example: Aptitude Test

34 34

We first need to estimate mean and standard We first need to estimate mean and standard deviationdeviation

Example: Mean and Standard Deviation Example: Mean and Standard Deviation

41.1049

04.5310

1

)(

42.6850

3421

2

n

xxs

n

xx i

35 35

What test mark has the property of having 10% What test mark has the property of having 10% of test marks being less than or equal to itof test marks being less than or equal to it

To answer this question, we should first answer To answer this question, we should first answer the followingthe following

What is the standard normal value (What is the standard normal value (zz value), such value), such that 10% of z values are less than or equal to that 10% of z values are less than or equal to it?it?

10%

zz Values Values

36 36

We need to use standard Normal distribution in We need to use standard Normal distribution in Table 1.Table 1.

10%

10%

zz Values Values

37 37

10%

40%

zz Values Values

38 38

40%

z = 1.28

10%

z = - 1.28

zz Values Values

39 39

The standard normal value (The standard normal value (zz value), such that 10% of z value), such that 10% of z values are less than or equal to it is z = -1.28values are less than or equal to it is z = -1.28

The normal value of test marks such that 10% of random The normal value of test marks such that 10% of random variables are less than it is 55.1. variables are less than it is 55.1.

zσ

μx

28.1x

10.41

68.42

1.5542.68)28.1(41.10x

zz Values and Values and xx Values Values

To transform this standard normal value to a similar value To transform this standard normal value to a similar value in our example, we use the following relationshipin our example, we use the following relationship

40 40

Following the same procedure, we could find z values for Following the same procedure, we could find z values for cases where 20%, 30%, 40%, …of random variables are cases where 20%, 30%, 40%, …of random variables are less than these values. Following the same procedure, less than these values. Following the same procedure, we could transform we could transform zz values into values into x x values.values.

Lower 10%Lower 10% -1.28-1.28 55.155.1

Lower 20%Lower 20% -.84-.84 59.6859.68

Lower 30%Lower 30% -.52-.52 63.0163.01

Lower 40%Lower 40% -.25-.25 65.8265.82

Lower 50%Lower 50% 00 68.4268.42

Lower 60%Lower 60% .25.25 71.0271.02

zz Values and Values and xx Values Values

zx

41 41

Victor Computers manufactures and sells aVictor Computers manufactures and sells a

general purpose microcomputer. As part of a study to general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to evaluate sales personnel, management wants to determine if the annual sales volume (number of units determine if the annual sales volume (number of units sold by a salesperson) follows a normal probability sold by a salesperson) follows a normal probability distribution.distribution.

A simple random sample of 30 of the salespeople A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below.was taken and their numbers of units sold are below.

33 43 44 45 52 52 56 58 63 6433 43 44 45 52 52 56 58 63 64

64 65 66 68 70 72 73 73 74 7564 65 66 68 70 72 73 73 74 75

83 84 85 86 91 92 94 98 102 10583 84 85 86 91 92 94 98 102 105

(mean = 71, standard deviation = 18.54)(mean = 71, standard deviation = 18.54)

Partition this Normal distribution into 6 equal probability Partition this Normal distribution into 6 equal probability partsparts

Example : Victor ComputersExample : Victor Computers

42 42

Areas = 1.00/6 = .1667

Areas = 1.00/6 = .1667

717153.0253.0263.0363.03 78.9778.97

88.98 = 71 + .97(18.54)88.98 = 71 + .97(18.54)

Example : Victor ComputersExample : Victor Computers

43 43

The End The End

44 44

Normal Approximation Normal Approximation of Binomial Probabilitiesof Binomial Probabilities

When the number of trials, When the number of trials, nn, becomes large, evaluating , becomes large, evaluating the binomial probability function by hand or with a the binomial probability function by hand or with a calculator is difficult.calculator is difficult.

The normal probability distribution provides an easy-to-The normal probability distribution provides an easy-to-use approximation of binomial probabilities where use approximation of binomial probabilities where nn > > 2020, , np np >> 5 5, and , and nn(1 - (1 - pp)) >> 5. 5.

Set Set = = npnp

Add and subtract a Add and subtract a continuity correction factorcontinuity correction factor because a because a continuous distribution is being used to approximate a continuous distribution is being used to approximate a discrete distribution. For example, discrete distribution. For example,

PP((xx = 10) is approximated by = 10) is approximated by PP(9.5 (9.5 << xx << 10.5). 10.5).

np p( )1 np p( )1

45 45

The Exponential Probability DistributionThe Exponential Probability Distribution

Exponential Probability Density FunctionExponential Probability Density Function

for for xx >> 0, 0, > 0 > 0

wherewhere = mean = mean

ee = 2.71828 = 2.71828 Cumulative Exponential Distribution FunctionCumulative Exponential Distribution Function

wherewhere xx00 = some specific value of = some specific value of xx

f x e x( ) / 1

f x e x( ) / 1

/0

01)( xexxP

46 46

The time between arrivals of cars at Al’s The time between arrivals of cars at Al’s Carwash Carwash

follows an exponential probability distribution follows an exponential probability distribution with awith a

mean time between arrivals of 3 minutes. Al mean time between arrivals of 3 minutes. Al would likewould like

to know the probability that the time between to know the probability that the time between twotwo

successive arrivals will be 2 minutes or less.successive arrivals will be 2 minutes or less.

PP((xx << 2) = 1 - 2.71828 2) = 1 - 2.71828-2/3-2/3 = 1 - .5134 = 1 - .5134 = .4866= .4866

Example: Al’s CarwashExample: Al’s Carwash

47 47

Example: Al’s CarwashExample: Al’s Carwash

Graph of the Probability Density FunctionGraph of the Probability Density Function

xx

F (x )F (x )

.1.1

.3.3

.4.4

.2.2

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10

P(x < 2) = area = .4866P(x < 2) = area = .4866

48 48

The EndThe End

Documents

Continuous Probability Distributions