CFA Level 1 Exam: Relative Dispersion in Mechanical Aptitude and Finger Dexterity Tests

Relative Dispersion in Mechanical Aptitude and Finger Dexterity Tests

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Question

A large group of inductees was given a mechanical aptitude and a finger dexterity test. The mean score on the mechanical aptitude test was 200, with a standard deviation of 10. The mean and standard deviation for the finger dexterity test were 30 and 6 respectively. What is the relative dispersion in the two groups?

Answers

A. Mechanical 5 percent, finger 20 percent

B. Mechanical 500 percent, finger 200 percent

C. Mechanical 20 percent, finger 10 percent

D. Mechanical 50 percent, finger 200 percent

E. None of these answers

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Explanations

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A. B. C. D. E.

The respective CVs are found from (s*100)/mean. Mechanical: 10*100/200 = 5% and Finger: 6*100/30 = 20%.

To calculate the relative dispersion in the two groups, we need to determine the coefficient of variation (CV) for each group. The coefficient of variation is a measure of relative variability and is calculated by dividing the standard deviation by the mean and multiplying by 100 to express it as a percentage.

For the mechanical aptitude test group: Mean = 200 Standard Deviation = 10

CV (mechanical) = (Standard Deviation / Mean) * 100 = (10 / 200) * 100 = 5%

For the finger dexterity test group: Mean = 30 Standard Deviation = 6

CV (finger) = (Standard Deviation / Mean) * 100 = (6 / 30) * 100 = 20%

Therefore, the relative dispersion in the two groups is Mechanical 5 percent, Finger 20 percent.

The correct answer is option A. Mechanical 5 percent, finger 20 percent.

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