Question

Many firms use on-the-job training to teach their employees new software. Suppose you work in the personnel department of a firm that just finished training a group of its employees in new software, 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 73 and 4, respectively, and the distribution of scores is mound-shaped and symmetric. Suppose the trainee in question received a score of 68. Compute the trainee's z-score.

Answer 1

Answer:

The trainee's z-score is -1.25.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 73, \sigma = 4$

Suppose the trainee in question received a score of 68. Compute the trainee's z-score.

This is Z when X = 68. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{68 - 73}{4}$

$Z = -1.25$

The trainee's z-score is -1.25.