Explanation Definition: 1. If a data set contains n data values, the mean x ¯ of the data set is x ¯ = ∑ x n . 2. If x is a data value in a set whose mean is x ¯ , then x − x ¯ is called x’s deviation from the mean. 2. We denote the standard deviation of a sample of n data values by s = ∑ ( x − x ¯ ) 2 n − 1 . 4. We calculate the standard deviation, s, of a sample that is given as a frequency distribution as s = ∑ ( x − x ¯ ) 2 ⋅ f n − 1 . Here f is the frequency of data value x , and n = ∑ f is the number of data values in the distribution. Calculation: We will consider here two examples to prove that it is not true that the more the numbers there are in a distribution, the larger the standard deviation. Suppose we have the following data. x f 2 2 3 4 4 2 5 0 6 4 Here there are total 12 numbers or 12 data values. First find the mean of the given data set. To compute the mean, let us find all products x ⋅ f of the given values and their respective frequencies. For x = 2 and f = 2 , the product x ⋅ f = 2 ⋅ 2 = 4. For x = 3 and f = 4 , the product x ⋅ f = 3 ⋅ 4 = 12. For x = 4 and f = 2 , the product x ⋅ f = 4 ⋅ 2 = 8. For x = 5 and f = 0 , the product x ⋅ f = 5 ⋅ 0 = 0. For x = 6 and f = 4 , the product x ⋅ f = 6 ⋅ 4 = 24. The sum of these products is 4 + 12 + 8 + 0 + 24 = 48. Therefore, ∑ ( x ⋅ f ) = 48. The sum of the frequencies is 2 + 4 + 2 + 0 + 4 = 12. Therefore, ∑ f = 12. The mean is then ∑ ( x ⋅ f ) ∑ f = 48 12 = 4. Next find the standard deviation. To begin with, find the deviation from the mean for each value. For x = 2 the deviation from the mean is x − x ¯ = 2 − 4 = − 2. For x = 3 the deviation from the mean is x − x ¯ = 3 − 4 = − 1. For x = 4 the deviation from the mean is x − x ¯ = 4 − 4 = 0. For x = 5 the deviation from the mean is x − x ¯ = 5 − 4 = 1. For x = 6 the deviation from the mean is x − x ¯ = 6 − 4 = 2. Now, square each of the deviation value. For x = 2 , ( − 2 ) 2 = 4. For x = 3 , ( − 1 ) 2 = 1. For x = 4 , ( 0 ) 2 = 0. For x = 5 , ( 1 ) 2 = 1. For x = 6 , ( 2 ) 2 = 4. Now multiply each deviation squared value by the frequency of the corresponding data value. For x = 2 with frequency 2, 4 ⋅ 2 = 8. For x = 3 with frequency 4, 1 ⋅ 4 = 4. For x = 4 , with frequency 2, 0 ⋅ 2 = 0. For x = 5 , with frequency 0, 1 ⋅ 0 = 0. For x = 6 , with frequency 4, 4 ⋅ 4 = 16. The sum of these values is 8 + 4 + 0 + 0 + 16 = 28

6th Edition

ISBN: 9780134434681

Author: Tom Pirnot

Publisher: PEARSON

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Chapter 14.3, Problem 42E

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Whether the following statement is true or false:

The more the numbers there are in a distribution, the larger the standard deviation.

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Chapter 14.3, Problem 41E

Chapter 14.3, Problem 43E

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Chapter 14 Solutions

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Whether the following statement is true or false: The more the numbers there are in a distribution, the larger the standard deviation.

Whether the following statement is true or false: The more the numbers there are in a distribution, the larger the standard deviation.

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Chapter 14 Solutions

Whether the following statement is true or false:

The more the numbers there are in a distribution, the larger the standard deviation.