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: Begin by determining the population standard deviation for the Coffee prices. From the given table understand each data value. X 4.74 4.91 4.83 4.99 4.72 4.69 4.79 4.81 4.67 4.61 4.41 4.49 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 4.74 + 4.91 + 4.83 + 4.99 + 4.72 + 4.69 + 4.79 + 4.81 + 4.67 + 4.61 + 4.41 + 4.49 = 56.66. Therefore, ∑ x = 56.66. There are 12 data values. Therefore, n = 12. The mean is then ∑ x n = 56.66 12 = 4.72166666666667. Next find the population standard deviation. To begin with, find the deviation from the mean for each value. For x = 4.74 the deviation from the mean is 0.01833333333333. For x = 4.91 the deviation from the mean is 0.18833333333333. For x = 4.83 the deviation from the mean is 0.10833333333333. For x = 4.99 the deviation from the mean is 0.26833333333333. For x = 4.72 the deviation from the mean is − 0.00166666666667. For x = 4.69 the deviation from the mean is − 0.03166666666667. For x = 4.79 the deviation from the mean is 0.06833333333333. For x = 4.81 the deviation from the mean is 0.08833333333333. For x = 4.67 the deviation from the mean is − 0.05166666666667. For x = 4.61 the deviation from the mean is − 0.11166666666667. For x = 4.41 the deviation from the mean is − 0.31166666666667. For x = 4.49 the deviation from the mean is − 0.23166666666667. Now, square each of the deviation value. For x = 4.74 , ( 0.01833333333333 ) 2 = 0.00033611111111111900. For x = 4.91 , ( 0.18833333333333 ) 2 = 0.03546944444444450000. For x = 4.83 , ( 0.10833333333333 ) 2 = 0.01173611111111110000. For x = 4.99 , ( 0.26833333333333 ) 2 = 0.07200277777777790000. For x = 4.72 , ( − 0.00166666666667 ) 2 = 0.00000277777777777865. For x = 4.69 , ( − 0.03166666666667 ) 2 = 0.00100277777777775000. For x = 4.79 , ( 0.06833333333333 ) 2 = 0.00466944444444445000. For x = 4.81 , ( 0.08833333333333 ) 2 = 0.00780277777777771000. For x = 4.67 , ( − 0.05166666666667 ) 2 = 0.00266944444444445000. For x = 4.61 , ( − 0.11166666666667 ) 2 = 0.01246944444444440000. For x = 4.41 , ( − 0.31166666666667 ) 2 = 0.09713611111111100000. For x = 4.49 , ( − 0.23166666666667 ) 2 = 0.05366944444444440000. Note that each value appears only once. So, their frequency is 1. Hence, the product of the square and the frequency will be same as the square. The sum of these values is 0.29896666666666600000. The population standard deviation is therefore given as follows: σ = ∑ ( x − μ ) 2 n = 0.29896666666666600000 12 = 0.02491388888888890000. Taking square root gives, μ = 0

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Author: Pirnot

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Chapter 14.3, Problem 36E

To determine

To find:

The population standard deviation of each data set given by the following table:

2015 Coffee Prices ($/lb) Gasoline Prices ($/gal)

Jan 4.74 2.11

Feb 4.91 2.25

Mar 4.83 2.48

Apr 4.99 2.49

May 4.72 2.78

Jun 4.69 2.83

Jul 4.79 2.83

Aug 4.81 2.68

Sept 4.67 2.39

Oct 4.61 2.29

Nov 4.41 2.19

Dec 4.49 2.06

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Chapter 14.3, Problem 35E

Chapter 14.3, Problem 37E

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2015	Coffee Prices ($/lb)	Gasoline Prices ($/gal)
Jan	4.74	2.11
Feb	4.91	2.25
Mar	4.83	2.48
Apr	4.99	2.49
May	4.72	2.78
Jun	4.69	2.83
Jul	4.79	2.83
Aug	4.81	2.68
Sept	4.67	2.39
Oct	4.61	2.29
Nov	4.41	2.19
Dec	4.49	2.06

The population standard deviation of each data set given by the following table: 2015 Coffee Prices ($/lb) Gasoline Prices ($/gal) Jan 4.74 2.11 Feb 4.91 2.25 Mar 4.83 2.48 Apr 4.99 2.49 May 4.72 2.78 Jun 4.69 2.83 Jul 4.79 2.83 Aug 4.81 2.68 Sept 4.67 2.39 Oct 4.61 2.29 Nov 4.41 2.19 Dec 4.49 2.06

Solution Summary: The author explains that the prices of Coffee were more stable than those of Gasoline.

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