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    Chapter

    Seven

    McGraw-Hill/Irwin 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.

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    Chapter Seven

    Continuous Probability Distributions

    GOALSWhen 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.

    FOUR

    List the characteristics of the normal probability distribution.

    Goals

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    Chapter Seven continued

    GOALSWhen you have completed this chapter, you will be able to:

    FIVE

    Define and calculatezvalues.

    SIX

    Determine the probability an observation will lie between two points

    using the standard normal distribution.

    SEVEN

    Determine the probability an observation will be above or below a

    given value using the standard normal distribution.

    Continuous Probability Distributions

    Goals

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    Discrete and continuous

    distributions

    A Discretedistributionis based on random

    variables which canassume only clearly

    separated values.

    Discrete distributions

    studied include:

    o Binomial

    o Poisson.

    A Continuousdistribution usually

    results from measuringsomething.

    Continuous distributions

    include:

    o Uniform

    o Normalo Others

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    The Uniform distribution

    Is rectangular in shape

    Is defined by minimum and

    maximum values

    Has a mean computed as

    follows:

    a + b

    2m =

    where aand bare

    the minimum and

    maximum values

    Has a standard deviation

    computed as follows:

    s = (b-a)2

    12

    The uniform distribution

    P(x)

    xa b

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    Calculates its height as

    P(x) = ifa< x < band 0 elsewhere1

    (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 liverepresentative 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

    2m==

    5+25

    2= 15

    What is the standarddeviation of the wait time?

    s = (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 15to 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 symmetrical about the mean.

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

    Has its mean, m, to determine its location andits standard deviation, s, to determine itsdispersion.

    The Normal probability distribution

    is bell-shaped and has a single peak at the

    center of the 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 : m = 0 , s2 = 1

    Mean, median, and

    mode are equal

    Theoretically,

    curve extends to

    infinity

    a

    Characteristics of a Normal Distribution

    Normal

    curve is

    symmetrical

    7-10

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    The Standard Normal

    Probability Distribution

    s

    m-

    Xz

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

    It is also called the

    zdistribution.

    The standardnormal distribution

    is a normal distributionwith a mean of 0 and astandard deviation of 1.

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    Example 2

    s

    m-

    X

    z

    = $2,200 - $2000$200

    = 1.00

    The bi-monthly

    starting salaries of

    recent MBAgraduates follows

    the normal

    distribution with a

    mean of $2,000 anda standard deviation

    of $200. What is

    the z-value for asalary of $2,200?

    MBA

    3

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    EXAMPLE 2 continued

    50.1200$

    000,2$700,1$-

    -

    -

    s

    mXz

    What is the

    z-value for$1,700?

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

    mean of $2,000. A z-value of1.50 indicatesthat $1,700 is 1.5 standard deviation below themean of $2000.

    7 14

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    Areas Under the Normal

    Curve

    Practically all is within three standard

    deviations of the mean.

    m + 3s

    About 68 percent of

    the area under the

    normal curve is withinone standard deviation

    of the mean.

    m + 1s

    About 95 percent is within two standard

    deviations of the mean.

    m + 2s

    7 15

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    Example 3

    The daily water usage per

    person in Providence,

    Rhode Island is normallydistributed with a mean of

    20 gallons and a standard

    deviation of 5 gallons.

    About 68 percent of thoseliving in New Providence

    will use how many gallons

    of water?

    About 68% of the daily

    water usage will lie between

    15 and 25 gallons (+ 1s ).

    7 16RHODE

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    EXAMPLE 4

    00.05

    2020

    -

    -

    s

    mXz

    80.05

    2024

    -

    -

    s

    mXz

    What is the probability that

    a person from Providence

    selected at random will usebetween 20 and 24 gallons

    per day?

    Providence

    Warwick

    Newport

    ScituateRes

    95

    295

    RHODE

    ISLAND

    7 17

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    Example 4 continued

    The area under a normal

    curve between az-value of0 and az-value of 0.80 is

    0.2881.

    We conclude that 28.81

    percent of the residents use

    between 20 and 24 gallonsof water per day.

    See the following diagram

    7 18

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    7 19

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    EXAMPLE 4 continued

    40.052018

    -

    -

    -

    s

    mXz

    20.15

    2026

    -

    -

    s

    mXz

    What percent of

    the population use

    between 18 and 26

    gallons per day?

    7 20

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    EXAMPLE 4 continued

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

    of water per day.

    The area

    associated with a

    z-value of

    0.40 is.1554.

    The area

    associated with a

    z-value of 1.20 is.3849.

    Adding these

    areas, the result is

    .5403.

    7 21

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    EXAMPLE 5

    Professor Mann has

    determined that the scores

    in his statistics course areapproximately 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?

    7-22

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    EXAMPLE 5 continued

    Thez-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 ormore earn anA.

    We letzequal 1.04 and

    solve the standard normal

    equation forX. The resultis the score that separates

    students that earned anA

    from those that earned aB.