1/71 Statistics Continuous Probability Distributions

1/71

Statistics

Continuous Probability Distributions


Uniform Probability Distribution Normal Probability Distribution Normal Approximation of Binomial

Probabilities Exponential Probability Distribution


f (x)f (x)

x x

Uniform

x

f (x) Normal

xx

f (x)f (x) Exponential

STATISTICS in PRACTICE Procter & Gamble (P&G) produces

and markets such products as

detergents, disposable diapers,

bar soaps, and paper towels. The Industrial Chemicals Division

of P&G is a supplier of fatty alcohols derived from natural substances such as coconut oil and from petroleum- based derivatives.

STATISTICS in PRACTICE

The division wanted to know the economic

risks and opportunities of expanding its

fatty-alcohol production facilities, so it

called in P&G’s experts in probabilistic

decision and risk analysis to help.


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.


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 f(x) between x1 and x2.

f(x) 0≧ 1)(

dxxf


f (x)f (x)

x x

Uniform

x1 x1 x2 x2

x

f (x) Normal

x1 x1 x2 x2 x1 x1 x2 x2

Exponential

xx

f (x)f (x)

x1

x1

x2 x2


The expected value of a continuous random variable x is:

The variance of a continuous random variable x is:

dxxxfxE )()(

dxxfxxVar )()()( 22


If the random variable x has the density function f(x), the probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be

2

1

)()( 21

x

x

dxxfxxxP


Cumulative probability function: If the random variable x has the density

function f(x), the cumulative distribution function for x x≦ 2 is:

2

)()()( 22

x

dxxfxxPxF

Uniform Probability Distribution

where: a = smallest value the variable can assume

b = largest value the variable can assume

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

A random variable is uniformly distributed whenever the probability is proportional to the interval’s length.

The uniform probability density function is:

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

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


Expected Value of x

Variance of x


Example:

Random variable x = the flight time of an

airplane traveling from Chicago to New

York. Suppose the flight time can be any

value in the interval from 120 minutes to 140

minutes.


Assume every 1-minute interval being equally likely, x is said to have a uniform

probability distribution and the probability density function is

elsewhere

xforxf

140120

0

20/1)(


Example: Uniform Probability Density Function for Flight Time


What is the probability that the flight time is between 120 and 130 minutes? That is, what is ?

Area provides Probability of Flight Time Between 120 and 130 Minutes

)130120( xP


Applying these formulas to the uniform distribution for flight times from Chicago to New York, we obtain

and σ=5.77 minutes.

33.3312

)120140()(

1302

)140120()(

2

xVar

xE


Example: Slater's Buffet Slater customers are charged

for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces

and 15 ounces.

Uniform Probability Density Function

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

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

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

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

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


Uniform Probability Distributionfor Salad Plate Filling Weight

f(x)f(x)

x x

55 1010 1515

1/101/10

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


f(x)f(x)

x x

55 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 customer

will take between 12 and 15 ounces of salad?

1212


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

Heightsof people


It has been used in a wide variety of applications:

Scientific measurements

Scientific measurements

Amountsof rainfall

Amountsof rainfall


It has been used in a wide variety of applications:

Test scores

Test scores

Normal Probability Distribution Normal Probability Density Function

2 2( ) / 21( )

2xf x e

μ = mean, σ = standard deviation, π = 3.14159

e = 2.71828

where:

The distribution is symmetric; its skewness measure is zero.

Normal Probability Distribution Characteristics

x

The entire family of normal probability distributions is defined by its mean μ and its standard deviation σ .


Standard Deviation s

Mean mx

The highest point on the normal curve is at the mean, 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

s = 15

s = 25

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

x

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


.5 .5

x


of values of a normal random variable are within of its mean.

68.26%+/- 1 standard deviation


95.44%+/- 2 standard deviations


99.72%+/- 3 standard deviations

Characteristics


x

m – 3s m – 1sm – 2s

m + 1sm + 2s

m + 3sm

68.26%

95.44%99.72%

Characteristics

Standard Normal Probability Distribution

A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution.

s = 1

0

z

The letter z is used to designate the standard normal random variable.



Areas, or probabilities, for The Standard Normal Distribution



Example: What is the probability that the z value for the standard normal random variable will be between .00 and 1.00?



Example: = ? Table 6.1 to show that the probability of a z value

between z = .00 and z = 1.00 is .3413 the normal distribution is symmetric, therefore,

=

= .3413 + .3413 = .6826

)00.100.1( zP

)00.100(.)00.00.1( zPzP)00.100.1( zP


Example: = ? Table 6.1 to show that the probability of a z

value between z = .00 and z = 1.00 is .3413 the normal distribution is symmetric,

therefore,

=

= .3413 + .3413 = .6826

)00.100.1( zP

)00.100(.)00.00.1( zPzP)00.100.1( zP



Example: = ? =

the normal distribution is symmetric and

=

= .1915 + .5000 = .6915.

)50.( zP

)50.( zP )00.(.)00.50.( zPzP

)50.00(.)00.50.( zPzP)50.( zP



Example: = ? Probability of a z value between z = 0.00 and

z = 1.00 is .3413, and Probability of a z value between z = 0.00 and z =1.58 is .4429 .

Hence, Probability of a z value between z = 1.00 and z = 1.58 is .4429 — .3413 = .1016.

)58.100.1( zP


Example: = ? Probability of a z value between z = 0.00

and z = 1.00 is .3413, and Probability of a z value between z = 0.00 and z =1.58 is.4429.

Hence, Probability of a z value between z = 1.00 and z = 1.58 is .4429 — .3413 = .1016

)58.100.1( zP


Example: find a z value such that the probability of obtaining a larger z value is .10.


An area of approximately .4000 (actually .3997) will be between the mean and z 1.28.* In terms of the question originally asked, the probability is approximately .10 that the z value will be larger than 1.28.

Converting to the Standard Normal Distribution

Standard Normal ProbabilityDistribution

zx

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


Standard Normal Density Function

2 / 21( )

2zf x e

z = (x – m)/s = 3.14159e = 2.71828

where:


Example: Pep Zone

Pep Zone sells auto parts and supplies including a popular multi-grade

motor oil. When the stock of this

oil drops to 20 gallons, a replenishment order is placed.

PepZone5w-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 demand during replenishment

lead-time is normally distributed

with a mean of 15 gallons and a

standard deviation of 6 gallons.



Example: Pep Zone

The manager would like to know the

probability of a stockout, P(x > 20).



Example: Pep Zone

z = (x - )/ = (20 - 15)/6 = .83

Solving for the Stockout Probability

Step 1: Convert x to the standard normal distribution.


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

see next slide


Cumulative Probability Table for the Standard Normal Distributionz .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

. . . . . . . . . . .


P(z < .83)


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

= .2033


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


Probability of a stockout P(x >

20)



0 .83

Area = .7967Area = 1 - .7967

= .2033

z




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



Solving for the Reorder Point


0

Area = .9500

Area = .0500

zz.05




Step 1: Find the z-value that cuts off an area of .05 in the right tail of the standard normal 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 of the tail area (1 - .05 = .95)




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

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

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

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




By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about .20 to .05. This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer’s desire to make a purchase.


Normal Approximation of Binomial Probabilities

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

The normal probability distribution provides an easy-to-use approximation of binomial probabilities where n > 20, np > 5,

and n(1 - p) > 5.


Add and subtract 0.5 (a continuity correction

factor) because a continuous distribution is

being used to approximate a discrete

distribution. For example,

P(x = 10) is approximated by P(9.5 < x < 10.5).

Set

(1 )np p

= np


Example: To find the binomial probability of 12 successes in 100 trials and p = .1.

n= 100, p = .1. np = 10 > 5 and

n(1 - p) = (100)(.9) =90 > 5. Mean, np = (100)(.1) = 10 > 5, Variance,

np(1 - p) = (100)(.1)(.9) =9. Compute the area under the corresponding normal curve between 11.5 and 12.5.


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 betweenvehicle arrivalsat a toll booth

Time requiredto completea questionnaire

Time requiredto completea questionnaire

Distance betweenmajor defectsin a highway

Distance betweenmajor defectsin a highway

SLOW

Density Function


where: = mean

e = 2.71828

0 0,xfor ,1

)( /

xexffor x > 0, > 0

Cumulative Probabilities


P x x e x( ) / 0 1 o

where:

x0 = some specific value of x

Computing Probabilities for the Exponential Distribution

Example: The Schips loading dock example.

x = loading time and μ=15, which gives us

What is the probability that loading a truck will take between 6 minutes and 18 minutes?

15/0

01)( xexxP

Computing Probabilities for the Exponential Distribution

Since, and The probability that loading a truck will

take between 6 minutes and 18 minutes is equal to

.6988 -- .3297 = .3691.

3297.1)6( 15/6 exP

6988.1)18( 15/18 exP


Example: Al’s Full-Service Pump

The time between arrivals of

cars at Al’s full-service gas pump

follows an exponential probability

distribution with a mean time between arrivals of 3 minutes. Al would like to know theprobability that the time between two successive arrivals will be 2 minutes or less.

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

Time Between Successive Arrivals (mins.)


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


A property of the exponential distribution is that the mean, m, and standard deviation, s, are equal.

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

s = m = 3 minutes

s 2 = (3)2 = 9


The exponential distribution is skewed to the right.

The skewness measure for the exponential distribution is 2.

Relationship between the Poisson and Exponential Distributions

The Poisson distributionprovides an appropriate descriptionof the number of occurrencesper interval

The Poisson distributionprovides an appropriate descriptionof the number of occurrencesper interval

The exponential distributionprovides an appropriate descriptionof the length of the intervalbetween occurrences

The exponential distributionprovides an appropriate descriptionof the length of the intervalbetween occurrences

Documents

1/71 Statistics Continuous Probability Distributions