Explanation Definition: 1. While computing the mean of a frequency distribution, write all products x ⋅ f of the values and their frequencies in a new column of the table. Represent the sum of these products by ∑ ( x ⋅ f ) . Denote the sum of the frequencies by ∑ f . The mean is then ∑ ( x ⋅ f ) ∑ f . 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. 3. 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: Observe the figure carefully. Notice that the grades depend upon the mean and the standard deviation. To determine the grade we will first determine the mean and standard deviation. From the given diagram understand the frequency of each data value. Average value x Frequency f 72 2 71 2 73 2 70 1 79 1 65 1 74 1 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 = 72 and f = 2 , the product x ⋅ f = 72 ⋅ 2 = 144. For x = 71 and f = 2 , the product x ⋅ f = 71 ⋅ 2 = 142. For x = 73 and f = 2 , the product x ⋅ f = 73 ⋅ 2 = 146. For x = 70 and f = 1 , the product x ⋅ f = 70 ⋅ 1 = 70. For x = 79 and f = 1 , the product x ⋅ f = 79 ⋅ 1 = 79. For x = 65 and f = 1 , the product x ⋅ f = 65 ⋅ 1 = 65. For x = 74 and f = 1 , the product x ⋅ f = 74 ⋅ 1 = 74. The sum of these products is 144 + 142 + 146 + 70 + 79 + 65 + 74 = 720. Therefore, ∑ ( x ⋅ f ) = 720. The sum of the frequencies is 2 + 2 + 2 + 1 + 1 + 1 + 1 = 10. Therefore, ∑ f = 10. The mean is then ∑ ( x ⋅ f ) ∑ f = 720 10 = 72. Next find the standard deviation. To begin with, find the deviation from the mean for each value. For x = 72 the deviation from the mean is x ⋅ x ¯ = 72 − 72 = 0. For x = 71 the deviation from the mean is x ⋅ x ¯ = 71 − 72 = − 1. For x = 73 the deviation from the mean is x ⋅ x ¯ = 73 − 72 = 1. For x = 70 the deviation from the mean is x ⋅ x ¯ = 70 − 72 = − 2

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ISBN: 9780134506470

Author: Pirnot

Publisher: PEARSON

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Set Theory

Every field in the present world involves a set. The theory of sets was developed by German mathematician Georg Cantor while working on the “problems of trigonometric series”.

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Chapter 14.3, Problem 28E

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The grade of the person earning the 76 using the given diagram, if the averages in the class are 72, 71, 73, 70, 71, 79, 65, 73, 74, and 72.

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Chapter 14.3, Problem 27E

Chapter 14.3, Problem 29E

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The grade of the person earning the 76 using the given diagram, if the averages in the class are 72, 71, 73, 70, 71, 79, 65, 73, 74, and 72.

Solution Summary: The author explains that the grade of the person earning the 76 among the averages is B.

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To find: The grade of the person earning the 76 using the given diagram, if the averages in the class are 72, 71, 73, 70, 71, 79, 65, 73, 74, and 72.

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Chapter 14 Solutions

To find:

The grade of the person earning the 76 using the given diagram, if the averages in the class are 72, 71, 73, 70, 71, 79, 65, 73, 74, and 72.