A study of the scores on an in-plant course in management principles and the years of service of the employees enrolled in the course resulted in these statistics:
Mean test score was 200 with a standard deviation of 40
Mean number of years of service was 20 years with a standard deviation of 2 years
In comparing the relative dispersion of the two distributions, what are the coefficients of variation (CV)?
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A. B. C. D. E.Explanation
The respective CVs are found from (s*100)/mean. Test: 40*100/200 = 20% and Service: 2*100/20 = 10%.
To calculate the coefficient of variation (CV), you need to divide the standard deviation of each distribution by its respective mean and then multiply the result by 100 to express it as a percentage.
For the test scores: Mean test score = 200 Standard deviation of test scores = 40
CV for test scores = (Standard deviation of test scores / Mean test score) * 100 = (40 / 200) * 100 = 20%
For the years of service: Mean years of service = 20 years Standard deviation of years of service = 2 years
CV for years of service = (Standard deviation of years of service / Mean years of service) * 100 = (2 / 20) * 100 = 10%
Therefore, the coefficient of variation for the test scores is 20% and for the years of service is 10%.
Among the answer choices provided, the closest match is D. Test 20%, service 10%, which correctly represents the coefficients of variation for the two distributions. Hence, the correct answer is D.