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

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?

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Lyla 2 weeks 2021-09-15T23:34:34+00:00 1 Answer 0

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    2021-09-15T23:36:15+00:00

    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.

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