Question

Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below. Demetria got a score of 85.185.1; this version has a mean of 61.1 and a standard deviation of 12. Vincent got a score of 299.2; this version has a mean of 264 and a standard deviation of 22. Tobias got a score of 7.26; this version has a mean of 7.1 and a standard deviation of 0.4.

Required:
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

Due to the higher Z-score, Demetria should be offered the job.

Explanation:

Z-score:

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Whichever applicant had grade with the highest z-score should be offered the job.

Demetria got a score of 85.1; this version has a mean of 61.1 and a standard deviation of 12.

For Demetria, we have
`X = 85.1, \mu = 61.1, \sigma = 12`. So

`Z = (X - \mu)/(\sigma)`

`Z = (85.1 - 61.1)/(12)`

`Z = 2`

Vincent got a score of 299.2; this version has a mean of 264 and a standard deviation of 22.

For Vincent, we have
`X = 299.2, \mu = 264, \sigma = 22`.

`Z = (X - \mu)/(\sigma)`

`Z = (299.2 - 264)/(22)`

`Z = 1.6`

Tobias got a score of 7.26; this version has a mean of 7.1 and a standard deviation of 0.4.

For Tobias, we have
`X = 7.26, \mu = 7.1, \sigma = 0.4`.

`Z = (X - \mu)/(\sigma)`

`Z = (7.26 - 7.1)/(0.4)`

`Z = 0.4`

Due to the higher Z-score, Demetria should be offered the job.