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 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?

Answer 1

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 scores

SO, X ~ Normal($\mu=72, \sigma^{2} =8.1^{2}$)

The z score probability distribution for normal distribution is given by;

                             Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean score = 72

            $\sigma$ = 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( $\frac{X-\mu}{\sigma}$ > $\frac{78-72}{8.1}$ ) = P(Z > 0.74) = 1 - P(Z < 0.74)

                                                       = 1 - 0.7704 = 0.2296

The 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( $\frac{X-\mu}{\sigma}$ > $\frac{x-72}{8.1}$ ) = 0.05

             P(Z >  $\frac{x-72}{8.1}$ ) = 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;

                        $\frac{x-72}{8.1} = 1.645$

                      ${x-72} = 1.645 \times 8.1$

                            x = 72 + 13.32 = 85.32

Hence, the​ candidate's score was 85.32.