Question

many firms use on-the-job training to teach their employees computer programming. Suppose you work in the personnel department of 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 stardard deviation of the test scores are 80 and 5, respectively. Assume nothing is known about the distribution, what percentage of test-takers scored better than the trainee who scored 90

Answer 1

To find the percentage of test-takers who scored better than the trainee who scored 90, calculate the z-score and find the area under the normal distribution curve to the right of that z-score.

Step-by-step explanation:

To find the percentage of test-takers who scored better than the trainee who scored 90, we need to calculate the z-score of the trainee's score and then find the area under the normal distribution curve to the right of that z-score. The formula to calculate the z-score is z = (x - mean) / standard deviation. Plugging in the values given, we get z = (90 - 80) / 5 = 2.

To find the area to the right of this z-score, we can use a standard normal distribution table or a calculator. Looking up the z-score of 2 in the table, we find that the area to the left of it is approximately 0.9772. Therefore, the percentage of test-takers who scored better than the trainee who scored 90 is approximately (1 - 0.9772) * 100 = 2.28%.

Answer 2

At most 25%

Step-by-step explanation:

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.