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 80 and standard deviation 8.1. a. A particular employer requires job candidates to score at least 84 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 84? 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?

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mariacsinning mariacsinning · Answer 1 · 2020-03-27T03:23:28+01:00

Answer:

a.35.57%

b. 100.686

Step-by-step explanation:

To find the percentage of the test scores that exceeded 84, we need to standardize 84 as:

[tex]z=\frac{84-m}{s} = \frac{84-80}{8.1}=0.37[/tex]

Where m is the mean of the scores and s is the standard deviation. Now, using the standard normal table, we can know the following probability:

P(Z > 0.37) = 0.3557

So, the percentage of the test scores that exceeded 84 were 35.57%

Now, to find the candidate's score that fell at the 98th percentile of the distribution we need to find th z-value that satisfies:

P(Z < z) = 0.98

So, using the standard normal table we have:

z = 2.06

Then, the candidate's score x is 100.686 and it is calculated as:

[tex]z=\frac{x-84}{8.1}=2.06\\ x=2.06*8.1 + 84\\x=100.686[/tex]