A study of business faculty at state supported institutions in Ohio revealed that the mean salary for nine months is $52,000 and the standard deviation of the sample is $3,000. The study also showed that the faculty had been employed an average (mean) of 15 years with a standard deviation of 4 years. How does the relative dispersion in the distribution of salaries compare with that of the lengths of service?
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A. B. C. D. E.A
The coefficient of variation = (s*100)/mean. Salary: 3,000*100/52,000 = 5.7%. Service: 4*100/15 = 26.7%.
To compare the relative dispersion in the distribution of salaries and the lengths of service, we can use the coefficient of variation (CV). The coefficient of variation is a measure of relative variability and is calculated as the ratio of the standard deviation to the mean, expressed as a percentage.
For salaries: Mean salary = $52,000 Standard deviation of salaries = $3,000
CV of salaries = (Standard deviation of salaries / Mean salary) * 100 = ($3,000 / $52,000) * 100 ≈ 5.77%
For lengths of service: Mean length of service = 15 years Standard deviation of lengths of service = 4 years
CV of lengths of service = (Standard deviation of lengths of service / Mean length of service) * 100 = (4 years / 15 years) * 100 ≈ 26.67%
Therefore, the relative dispersion in the distribution of salaries is approximately 5.77%, while the relative dispersion in the distribution of lengths of service is approximately 26.67%.
Comparing these values to the given answer choices, we find that the closest match is option A, which states that salaries have a relative dispersion of about 6% and lengths of service have a relative dispersion of about 27%. So the correct answer is A. Salaries about 6%, service about 27%.