Question

The standard deviation alone does not measure relative variation. For example, a standard deviation of $1 would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of $1 would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variability is the coefficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula CV = 100(s/ x ). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of 50 lb. There are weights for the two samples.

Sample 1	8.2	7.3	7.4	8.6	7.4
	8.2	8.6	7.5	7.5	7.1
Sample 2	51.8	51.2	51.9	51.6	52.7
	47	50.4	50.3	48.7	48.2

(a) For each of the given samples, calculate the mean and the standard deviation. (Round all intermediate calculations and answers to five decimal places.)

For sample 1
Mean
Standard deviation

For sample 2
Mean
Standard deviation

(b) Compute the coefficient of variation for each sample. (Round all answers to two decimal places.)

CV1
CV2

Answer 1

a).NECESSARY CALCULATION FOR MEAN AND VARIANCE:-

table 1: calculation for mean:-

sample 1()	sample 2()
8.2	51.8
7.3	51.2
7.4	51.9
8.6	51.6
7.4	52.7
8.2	47
8.6	50.4
7.5	50.3
7.5	48.7
7.1	48.2
sum=77.8	sum=503.8

so, we have:-

table 2: calculation of sd:-

sample 1()	sample 2()
8.2	51.8	0.1764	2.0164
7.3	51.2	0.2304	0.6724
7.4	51.9	0.1444	2.3104
8.6	51.6	0.6724	1.4884
7.4	52.7	0.1444	5.3824
8.2	47	0.1764	11.4244
8.6	50.4	0.6724	0.0004
7.5	50.3	0.0784	0.0064
7.5	48.7	0.0784	2.8224
7.1	48.2	0.4624	4.7524
		sum=2.836	sum=30.876

so, we have:-

THE MEAN AND STANDARD DEVIATION TABLE BE:-

	sample 1	sample 2
mean	7.78	50.38
standard deviation	0.56135	1.85221

b). coefficient of variation be:-

	calculation	answer
sample 1	%	CV1 = 7.22 %
sample 2	%	CV2 = 3.68 %

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