Question

4.111 Personnel dexterity tests. Personnel tests are designed to test a job applicant’s cognitive and/or physical abilities. The Wonderlic IQ test is an example of the former; the Purdue Pegboard speed test involving the arrangement of pegs on a peg board is an example of the latter. 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 75 and standard deviation 7.5.

a. A particular employer requires job candidates to score at least 80 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 80?

b. The testing service reported to a particular employer that one of its job candidate’s scores fell at the 98th percentile of the distribution (i.e., approximately 98% of the scores were lower than the candidate’s, and only 2%were higher). What was the candidate’s score?

Short answer

a.The percentage of the test scores during the past year exceeding 80 is 25.25%

b. The 98^th percentile score of the student is 90.405

Step-by-step solution

Given information

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 75 and standard deviation 7.5.

Calculating probability

a.

Let x be the random variable to represent the score of the test.

The random variable x follows a normal distribution with mean 75 and standard deviation 7.5

$$\begin{gathered} x~N(75,7.5) \\ P(x>80)=1-P(x\le 80) \\ =1-P\left(\frac{x-75}{7.5}\le \frac{80.75}{7.5}\right) \\ =1-P(z\le 0.666667) \\ =1-747507 \\ =0.252493 \\ \square 0.2525 \end{gathered}$$

So, the percentage of the test scores during the past year exceeding 80 is 25.25%

Calculating percentile score

Let the score of the candidate be p then,

$$\begin{gathered} P(x\le p)=0.98 \\ P\left(\frac{x-75}{7.5}\le \frac{p-75}{7.5}\right)=0.98 \\ P\left(z\le \frac{p-75}{7.5}\right)=0.98 \\ \Phi \left(\frac{p-75}{7.5}\right)=0.98 \\ \frac{p-75}{7.5}={\Phi }^{-1}(0.98) \\ \frac{p-75}{7.5}=2.054 \\ p=75+2.054\times 7.5 \\ p=90.405 \end{gathered}$$

So, the 98^th percentile score of student is 90.405