Question

Personnel tests are designed to test a job​ applicant's cognitive​ and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8.a. A particular employer requires job candidates to score at least 83 on the dexterity test. Approximately what percentatge of the test scores during the past year exceeded 83?

Answer 1

26.11% of the test scores during the past year exceeded 83.

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

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:

It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that
`\mu = 78, \sigma = 7.8`.

Approximately what percentatge of the test scores during the past year exceeded 83?

This is 1 subtracted by the pvalue of Z when
`X = 83`. So:

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

`Z = (83 - 78)/(7.8)`

`Z = 0.64`

`Z = 0.64` has a pvalue of 0.7389.

This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.