Introduction to Probability and Statistics
Introduction to Probability and Statistics
14th Edition
ISBN: 9781133103752
Author: Mendenhall, William
Publisher: Cengage Learning
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Chapter 2.5, Problem 2.34E

a.

To determine

To find:The relative frequency histogram for the data.

a.

Expert Solution
Check Mark

Answer to Problem 2.34E

The relative frequency histogram for the datais shown in Figure-1.

Explanation of Solution

Given information: The data is,

    Washington0Van Buren4Buchanan0
    Adams5W.H. Harrison10Lincoln4
    Jefferson6Tyler*15A. Johnson5
    Madison0Polk0Grant4
    Monroe2Taylor6Hayes8
    JO. Adams4Fillmore*2Garfield7
    Jackson0Pierce3Arthur3
    Cleveland5Coolidge2Nixon2
    B. Harrison*3Hoover2Ford4
    McKinley2F.D.Roosevelt6Carter4
    T. Roosevelt*6Truman1Reagan*4
    Taft3Eisenhower2G.H.W. Bush6
    Wilson*3Kennedy3Clinton1
    Harding0L.B.Johmson2G.W. Bush2
    Obama2

Calculation:

The histogram is shown below.

  Introduction to Probability and Statistics, Chapter 2.5, Problem 2.34E

Figure-1

Thus, the relative frequency histogram for the data is shown in Figure-1.

b.

To determine

To find: The mean and standard deviation.

b.

Expert Solution
Check Mark

Answer to Problem 2.34E

The mean and standard deviation is 3.5582 and 2.9139 .

Explanation of Solution

Given information: The data is,

    Washington0Van Buren4Buchanan0
    Adams5W.H. Harrison10Lincoln4
    Jefferson6Tyler*15A. Johnson5
    Madison0Polk0Grant4
    Monroe2Taylor6Hayes8
    JO. Adams4Fillmore*2Garfield7
    Jackson0Pierce3Arthur3
    Cleveland5Coolidge2Nixon2
    B. Harrison*3Hoover2Ford4
    McKinley2F.D.Roosevelt6Carter4
    T. Roosevelt*6Truman1Reagan*4
    Taft3Eisenhower2G.H.W. Bush6
    Wilson*3Kennedy3Clinton1
    Harding0L.B.Johmson2G.W. Bush2
    Obama2

Calculation:

The mean is,

  x¯=xn=15343=3.5582

The variance is,

  s2=1n1[ x i 2 ( x ) 2n]=1431[901 ( 153 ) 243]=8.4906

The standard deviation is,

  s=8.4906=2.9139

Thus, the mean and standard deviation is 3.5582 and 2.9139 .

c.

To determine

To find: The percentage falling into the given intervals.

c.

Expert Solution
Check Mark

Answer to Problem 2.34E

The percentage falling into the given intervalsis 95% and 99.7% .

Explanation of Solution

Given information: The data is,

    Washington0Van Buren4Buchanan0
    Adams5W.H. Harrison10Lincoln4
    Jefferson6Tyler*15A. Johnson5
    Madison0Polk0Grant4
    Monroe2Taylor6Hayes8
    JO. Adams4Fillmore*2Garfield7
    Jackson0Pierce3Arthur3
    Cleveland5Coolidge2Nixon2
    B. Harrison*3Hoover2Ford4
    McKinley2F.D.Roosevelt6Carter4
    T. Roosevelt*6Truman1Reagan*4
    Taft3Eisenhower2G.H.W. Bush6
    Wilson*3Kennedy3Clinton1
    Harding0L.B.Johmson2G.W. Bush2
    Obama2

Calculation:

The interval is,

  (x¯±s)=3.5582±2.9139=(0.6443,6.4721)

The measurement falling into the intervals are 33.

The interval is,

  (x¯±2s)=3.5582±2×2.9139=(2.2696,9.386)

The measurement falling into the intervals are 41.

The interval is,

  (x¯±3s)=3.5582±3×2.9139=(5.1835,12.2999)

The measurement falling into the intervals are 42.

According to the Empirical formula the percentage of values falling into the intervals are 95% and 99.7%

Thus, the percentage falling into the given intervals is 95% and 99.7% .

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Chapter 2 Solutions

Introduction to Probability and Statistics

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