## Personnel tests are designed to test a job applicant’s cognitive and/or physical abilities. A particular dexterity test is administered na

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 72 and standard deviation 8.1. a. A particular employer requires job candidates to score at least 78 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 78? b. The testing service reported to a particular employer that one of its job candidate’s scores fell at the 95th percentile of the distribution (i.e., approximately 95% of the scores were lower than the candidate’s, and only 5% were higher). What was the candidate’s score?

## Answers ( )

Answer:(a) 22.96% of the test scores during the past year exceeded 78.(b) The candidate’s score was 85.32.Step-by-step explanation:We are given that 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 72 and standard deviation 8.1.

Let X = distribution of test scoresSO, X ~ Normal()

The z score probability distribution for normal distribution is given by;Z = ~ N(0,1)

where, = population mean score = 72

= standard deviation = 8.1

(a) Now, percentage of the test scores during the past year which exceeded 78 is given by = P(X > 78)P(X > 78) = P( > ) = P(Z > 0.74) = 1 – P(Z < 0.74)

= 1 – 0.7704 =

0.2296The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.

Therefore, 22.96% of the test scores during the past year exceeded 78.(b)Now, we given that the testing service reported to a particular employer that one of its job candidate’s scores fell at the 95th percentile of the distribution and we have to find the candidate’s score, that means;P(X > x) = 0.05 {where x is the required candidate score}

P( > ) = 0.05

P(Z > ) = 0.05

Now, in the z table the critical value of x which represents the top 5% area is given as 1.645, i.e;x = 72 + 13.32 =

85.32Hence, the candidate’s score was 85.32.