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

Coefficient of variation is a relative measure of dispersion. The smaller the coefficient of variation, the better efficient the data is. We can see that the second sample has less coefficient of variation than sample 1. The mean and standard deviations of two samples are shown with all necessary steps and calculations. I hope you can understand this solution.