Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below. Norma got a score of 84.5; this version has a mean of 67.5 and a standard deviation of 10. Rebecca got a score of 306.5; this version has a mean of 293 and a standard deviation of 27. Morgan got a score of 8.38; this version has a mean of 7.1 and a standard deviation of 0.8. 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

Norma should be offered the job.

Step-by-step explanation:

For Norma;

Population mean; μ = 67.5

Sample mean; x¯ = 84.5

standard deviation; σ = 10

Formula for the test statistic is;

z = (x¯ - μ)/σ

z = (84.5 - 67.5)/10

z = 1.7

For Rebecca;

Population mean; μ = 293

Sample mean; x¯ = 306.5

standard deviation; σ = 27

Formula for the test statistic is;

z = (x¯ - μ)/σ

z = (306.5 - 293)/27

z = 0.5

For Morgan;

Population mean; μ = 7.1

Sample mean; x¯ = 8.38

standard deviation; σ = 0.8

Formula for the test statistic is;

z = (x¯ - μ)/σ

z = (8.38 - 7.1)/0.8

z = 1.6

The person with highest z-score is Norma and as such he should get the job.

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