Many firms use on-the-job training to teach their employees computer programming, Suppose you work in the personnel department o
f a firm that just finished training a group of its employees to program, and you have been requested to review the performance of one of the trainees on the final test that was given to all trainees. The mean and standard deviation of the test scores are 80 and 5, respectively. Assuming nothing is known about the distribution, what percentage of test-takers scored better than the trainee who scored 90? Answer 1 Point lKeypadTables At least 75% At least 25% At most 25% Approximately all None of the above
The fraction of the distribution that can exceed n standard deviations from the mean is 1/n². For a standard deviation of 5 and a mean of 80, a score of 90 is (90-80)/5 = 2 standard deviations from the mean. Scores better than that can be at most (1/2)² = 1/4 = 25% of the total number of scores.
