Elementary Statistics 2nd Edition
Elementary Statistics 2nd Edition
2nd Edition
ISBN: 9781259724275
Author: William Navidi, Barry Monk
Publisher: McGraw-Hill Education
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Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

    Number of TestsFrequency (in 1000s)
    1  953
    2  423
    3  194
    4    80
    5    29
    6      9
    7      3
    8      1
    Total1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

    Number of TestsFrequency (in 1000s)P (X = x)
    1  9530.563
    2  4230.250
    3  1940.115
    4    800.047
    5    290.017
    6      90.005
    7      30.002
    8      10.001
    Total16921.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution
Check Mark

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

    Number of TestsFrequency (in 1000s)P (X = x)
    1  9530.563
    2  4230.250
    3  1940.115
    4    800.047
    5    290.017
    6      90.005
    7      30.002
    8      10.001
    Total16921.000

Calculation:

  P(Thatstudent took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution
Check Mark

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

  μx=x.p(x)

Calculation:

    Number of TestsFrequency (in 1000s)P (X = x)x.P(x)
    1  9530.5630.56
    2  4230.2500.50
    3  1940.1150.34
    4    800.0470.19
    5    290.0170.09
    6      90.0050.03
    7      30.0020.01
    8      10.0010.00
    Total16921.0001.73

The mean μx of a given distribution is 1.73, because the value of x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution
Check Mark

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

  μx=x.p(x)

Calculation:

    Number of TestsFrequency (in 1000s)P (X = x)x.P(x)x2.P(x)
    1  9530.5630.560.56
    2  4230.2500.501.00
    3  1940.1150.341.03
    4    800.0470.190.76
    5    290.0170.090.43
    6      90.0050.030.19
    7      30.0020.010.09
    8      10.0010.000.04
    Total16921.0001.734.10

  StandardDeviation = x2p(x) [ xp( x )]2=4.10 1.732=1.05

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Chapter 6 Solutions

Elementary Statistics 2nd Edition

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