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Chapter

Seven

McGraw-Hill/Irwin

© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

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Chapter SevenContinuous Probability Continuous Probability DistributionsDistributionsGOALS

When you have completed this chapter, you will be able to:ONE

Understand the difference between discrete and continuous distributions.

TWO

Compute the mean and the standard deviation for a uniform distribution.

THREE

Compute probabilities using the uniform distribution.

FOURList the characteristics of the normal probability distribution.

GoalsGoals

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Chapter Seven continued

GOALSWhen you have completed this chapter, you will be able to:FIVE Define and calculate z values.

SIXDetermine the probability an observation will lie between two points using the standard normal distribution.

SEVENDetermine the probability an observation will be above or below a given value using the standard normal distribution.

EIGHTUse the normal distribution to approximate the binomial probability distribution.

Continuous Probability Continuous Probability DistributionsDistributions

Goals

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Discrete and continuous Discrete and continuous distributionsdistributions

A DiscreteDiscrete distribution is based on random variables which can assume only clearly

separated values.

Discrete distributions studied include:

o Binomial o Hypergeometrico Poisson.

A Continuous Continuous distribution usually

results from measuring something.

Continuous distributions include:

o Uniform o Normal o Others

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The Uniform distributionUniform distribution Is rectangular in shapeIs defined by minimum and maximum valuesHas a mean computed as follows:

a + b 2 =

where a and b are the minimum and maximum values

Has a standard deviation computed as follows:

= (b-a)2

12The uniform distribution

f(x)

x

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Calculates its height as

P(x) = if a < x < b and 0 elsewhere 1(b-a)

Calculates its area as

Area = height* base = *(b-a) 1(b-a)

The uniform distribution

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Suppose the time that you wait on the telephone for a live representative of your phone company to discuss your problem with you is uniformly distributed between 5 and 25 minutes.

What is the mean wait time?

a + b 2 =

= 5+25 2

= 15

What is the standard deviation of the wait time?

=(b-a)2

12

= (25-5)2

12= 5.77

Example 1

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What is the probability of waiting more than ten minutes?

The area from 10 to 25 minutes is

15 minutes. Thus:

P(10 < wait time < 25) = height*base = 1

(25-5) *15 = .75

What is the probability of waiting between 15 and 20 minutes?

The area from 15 to 20 minutes is

5 minutes. Thus:

P(15 < wait time < 20) = height*base = 1

(25-5)*5 = .25

Example 2 continued

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is bell-shapedbell-shaped and has a single peak at the center of the distribution.

Is symmetricalsymmetrical about the mean.

is asymptoticasymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.

Has its mean, mean, , to determine its location and

its standard deviation, standard deviation, , to determine its dispersion.

The NormalNormal probability distribution

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- 5

0 . 4

0 . 3

0 . 2

0 . 1

. 0

x

f(

x

r a l i t r b u i o n : = 0 , = 1

Mean, median, andmode are equal

Theoretically, curve extends to

infinity

a

Characteristics of a Normal Distribution

Normal curve is

symmetrical

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The Standard Normal Probability Distribution

X

z

A z-z-valuevalue is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:

It is also called the

z z distribution.distribution.

The standard standard normalnormal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

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Example 2

X

z

= $2,200 - $2000 $200

= 1.00

The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is

the z-z-valuevalue for a salary of $2,200?

M B A

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EXAMPLE 2 continued

50.1200$

000,2$700,1$

X

zWhat is the z-value for $1,700?

A z-vz-valuealue of 1 indicates that the value of $2,200 is one standard deviation above the

mean of $2,000. A z-vz-valuealue of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

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Areas Under the Normal Curve

Practically all is within three standard deviations of the mean.

+ 3

About 68 percent of the area under the

normal curve is within one standard deviation

of the mean. + 1

About 95 percent is within two standard deviations of the mean.

+ 2

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Example 3

The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water? About 68% of the daily water usage will lie between 15 and 25 gallons (+ 1 ).

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EXAMPLE 4

00.05

2020

X

z

80.05

2024

X

z

What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?

Providence

W arw ick

N ew port

ScituateR es

95

295

RHO DEISLAND

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Example 4 continued

The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881.

We conclude that 28.81 percent of the residents use between 20 and 24 gallons

of water per day.

See the following diagram

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EXAMPLE 4 continued

40.05

2018

X

z

20.15

2026

X

z

What percent of the population use between 18 and 26 gallons per day?

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EXAMPLE 4 continued

We conclude that 54.03 percent of the residents use between 18 and 26 gallons

of water per day.

The area The area associated with a associated with a z-z-value of –0.40 value of –0.40

is .1554.is .1554.

The area The area associated with a associated with a zz-value of 1.20 is -value of 1.20 is

.3849..3849.

Adding these Adding these areas, the result isareas, the result is

.5403..5403.

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EXAMPLE 5

Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?

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EXAMPLE 5 continued

The z-value associated corresponding to 35 percent is about 1.04.

To begin let X be the score that separates an A from a B. If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X.

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EXAMPLE 5 continued

2.772.572)5(04.1725

7204.1

X

X

Those with a score of 77.2 or more earn an A.

We let z equal 1.04 and solve the standard normal equation for X. The result is the score that separates students that earned an A from those that earned a B.

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The Normal Approximation to the Binomial

The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.

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The Normal Approximation continued

Recall for the binomial experiment:

oThere are only two mutually exclusive outcomes (success or failure) on each trial.

oA binomial distribution results from counting the number of successes.

oEach trial is independent.

oThe probability is fixed from trial to trial, and the number of trials n is also fixed.

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Continuity Correction Factor

The value .5 subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).

Continuity Correction FactorContinuity Correction Factor

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For the probability that

fewer than X occur, use the

area below (X-.5).

Continuity Correction Factor

How to Apply the Correction Factor:

For the probability at least X occur, use the

area above (X-.5).

For the probability that more than X

occur, use the area above (X+.5).

For the probability that

X or fewer occur, use the

area below (X+.5).

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EXAMPLE 6

n (. )( )15 200 30

A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras?

This is the mean of a binomial distribution.

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EXAMPLE 6 continued

2 1 30 1 15 255 n ( ) ( )( . ) .

0498.55.25

What is the standard deviation?

What is the variance?

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EXAMPLE 6 continued

88.10498.5

0.305.39

X

z

What is the probability that less than 40 homes in the sample have video cameras?

We use the correction factor (X-.5) for fewer than, so X-.5 is 39.5. The value of z is 1.88.

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From Appendix D the area between 0 and 1.88 on the z scale is .4699.

So the area to the left of 1.88 is .5000 + .4699 = .9699.

The likelihood that less than 40 of the 200 homes have a video camera is about 97%.

EXAMPLE 6 continued

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