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 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?

Answer 1

To find the percentage of test scores that exceeded 84, we can use z-scores and a z-table or calculator. The approximate percentage is 31.36%. To find the candidate's score at the 98th percentile, we can convert the percentile to a z-score and use the z-score formula. The candidate's score is approximately 97.92.

Step-by-step explanation:

To answer part a of the question, we need to find the percentage of test scores that exceeded 84. Since the distribution of scores is approximately normal, we can use z-scores to find the percentage. To do this, we first need to calculate the z-score for 84, using the formula z = (x - μ) / σ, where x is the given score, μ is the mean, and σ is the standard deviation. Substituting the given values, we have z = (84 - 80) / 8.1 = 0.4938. We can then use a z-table or a calculator to find the percentage of scores beyond this z-score, which is approximately 31.36%.

For part b of the question, we are given that the candidate's score fell at the 98th percentile. This means that the candidate scored higher than approximately 98% of the test takers. To find the corresponding score, we can use the z-score formula again. We need to find the z-score that corresponds to the 98th percentile, which is z = 2.05. Rearranging the formula, we have x = z * σ + μ. Substituting the given values, we have x = 2.05 * 8.1 + 80 = 97.92.

Answer 2

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:

`z=(84-m)/(s) = (84-80)/(8.1)=0.37`

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:

`z=(x-84)/(8.1)=2.06\\ x=2.06*8.1 + 84\\x=100.686`