Explanation Definition: 1. If a data set contains n data values, the mean of the data set is μ = ∑ x n . This represents the mean of the whole population. 2. The standard deviation of population of sample of n data values is given by σ = ∑ ( x − μ ) 2 n . 3. For a set of data with mean x ¯ and standard deviation s , we define the coefficient of variation, denoted by CV , as C V = s x ¯ ⋅ 100 % . Calculation: From the table given Exercise 22 understand each data value. x 10.3 10.4 18.6 15.9 16.3 14.9 First find the mean of the given data set. To compute the mean, let us find the sum of given data values. The sum is 10.3 + 10.4 + 18.6 + 15.9 + 16.3 + 14.9 = 86.4. Therefore, ∑ x = 86.4. There are 6 data values. Therefore, n = 6. The mean is then ∑ x n = 86.4 6 = 14.4. Next find the population standard deviation. To begin with, find the deviation from the mean for each value. For x = 10.3 the deviation from the mean is x ⋅ μ = 10.3 − 14.4 = − 4.1. For x = 10.4 the deviation from the mean is x ⋅ μ = 10.4 − 14.4 = − 4.0. For x = 18.6 the deviation from the mean is x ⋅ μ = 18.6 − 14.4 = 4.2. For x = 15.9 the deviation from the mean is x ⋅ μ = 15.9 − 14.4 = 1.5. For x = 16.3 the deviation from the mean is x ⋅ μ = 16

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ISBN: 9780134434681

Author: Tom Pirnot

Publisher: PEARSON

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Chapter 14.3, Problem 37E

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The set of data having more relative variability. The average poverty rate in 2013 was 12.08% with a standard deviation of 1.73%. To compare these values with the results in Exercise 22

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Chapter 14.3, Problem 36E

Chapter 14.3, Problem 38E

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The set of data having more relative variability. The average poverty rate in 2013 was 12.08% with a standard deviation of 1.73%. To compare these values with the results in Exercise 22

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The set of data having more relative variability. The average poverty rate in 2013 was 12.08% with a standard deviation of 1.73%. To compare these values with the results in Exercise 22

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Chapter 14 Solutions