Question

Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below. Brittany got a score of 88.388.3; this version has a mean of 63.163.1 and a standard deviation of 1414. Alissa got a score of 236.5236.5; this version has a mean of 219219 and a standard deviation of 2525. Tera got a score of 7.757.75; this version has a mean of 6.66.6 and a standard deviation of 0.50.5. If the company has only one position to fill and prefers to fill it with the applicant who performed best on the aptitude test, which of the applicants should be offered the job?

Answer 1

Answer:

Tera had the higher z-score, so she should be offered the job.

Step-by-step explanation:

Z- score:

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:

Whoever has the higher z-score should get the job.

Brittany:

Scored 88.3, mean 63.1, standard deviation 14. So

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

$Z = \frac{88.3 - 63.1}{14}$

$Z = 1.8$

Alissa:

Scored 236.5, mean 219, standard deviation 25. So

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

$Z = \frac{236.5 - 219}{25}$

$Z = 0.7$

Tera:

Scored 7.75, mean 6.66, standard deviation 0.5. So

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

$Z = \frac{7.75 - 6.66}{0.5}$

$Z = 2.18$

Tera had the higher z-score, so she should be offered the job.