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 79 and standard deviation 6.6.

a. A particular employer requires job candidates to score at least 83 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 83?

b. The testing service reported to a particular employer that one of its job​ candidate's scores fell at the

90th percentile of the distribution​ (i.e., approximately 90​% of the scores were lower than the​ candidate's, and only

10​% were​ higher). What was the​ candidate's score?

PLEASE ANSWER QUESTION B.

b. The​ candidate's score was_______________​(Round to the nearest whole number as​ needed.)

Answer 1

Given that, mean $(\mu)$ =79 and standard deviation $(\sigma)$ = 6.6

a) We want to find, P(X > 83)

$P(X > 83) = P(\frac {X -\mu}{\sigma}> \frac {83-79}{6.6}) = P(Z > 0.61) = 0.2709$

In percentage: 0.2709 * 100% = 27.09%

Therefore, approximately 27.09% of the test scores during the past year exceeded 83

b) We want to find, the value of x such that, P(X < x) = 0.90

$P(X < x) = 0.90$

$= >P(\frac {X -\mu}{\sigma}< \frac {x-79}{6.6}) =0.90$

$= >P(Z < \frac {x-79}{6.6}) =0.90$

$= > \frac {x-79}{6.6}=Z_{0.90}$

$= > \frac {x-79}{6.6}= 1.28$

$= > x =( 1.28*6.6) &plus; 79$

$= > x = 8.448 &plus; 79$

$= > x = 87.448$

$= > x \approx 87$

Therefore, the candidate's score was 87

Note: If you want score to be next whole number then it would be 88