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 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 70?
a. Approximentaly 75%
b. Approximentaly all
c. At most 75%
d. At least 75%
e. None of the above

d. At least 75%

Step-by-step explanation:

Since nothing is known about the distribution, we use the Chebyshev Theorem.

Chebyshev Theorem

The Chebyshev Theorem can also be applied to non-normal distribution. It states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

An in general terms, the percentage of measures within k standard deviations of the mean is given by .

In this problem, we have that:

Mean = 80, standard deviation = 5.

What percentage of test-takers scored better than the trainee who scored 70?

70 = 80 - 2*5

So 70 is two standard deviations below the mean.

Due to the Chebyshev Theorem, we know that at least 75% of the measures are within 2 standard deviations of the mean, between 70 and 90, which means that this percentage is at least 75%, and the correct answer is given by option d.

could you provide figure.

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement