Question

A human resource company performs employment screening for large companies in southern California.​ Recently, the company began working with a new client and formulated a new test just for this company. Thirty people were given the test. The data below reflect the scores of those 30 people. The human resource company has in the past issued a rejection letter with no interview to the lower​ 16% taking the test. They also send the upper​ 2.5% directly to the company without an interview. Everyone else is interviewed. Based on the data and assuming a​ bell-shaped distribution, what scores should be used for the two​ cutoffs?
79 79 7151 65 71 65 90 60 62 78 76 88 69 65
91 72 56 61 83 61 78 68 66 52 79 88 74 71 69
A score of ________should be used as the cutoff for the lower 16% and a score of________ (Round to the nearest integer as needed.)

Answer 1

Answer:

A score of 61 should be used as the cutoff for the lower 16% and a score of 92 should be used as the cutoff for the upper 2.5%.

Step-by-step explanation:

The data set for the scores of the 30 people are:

S = {79, 79, 71, 51, 65, 71, 65, 90, 60, 62, 78, 76, 88, 69, 65, 91, 72, 56, 61, 83, 61, 78, 68, 66, 52, 79, 88, 74, 71, 69}

Compute the mean and standard deviation:

$\mu=\frac{1}{n}\sum X=\frac{1}{30}\times 2138=71.27\\\\\sigma=\sqrt{\frac{1}{n-1}\sum (X-\mu)^{2}}=\sqrt{\frac{1}{30-1}\times 3343.8667}=10.74$

It is provided that in the past the company issued a rejection letter with no interview to the lower​ 16% taking the test.

That is the probability of no interview is, P (X > x) = 0.16.

That is, P (Z > z) = 0.16.

The z-score corresponding to this probability is, -1.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(-1\times 10.74)\\\\=60.53\\\\\approx 61$

It is also provided that the upper​ 2.5% directly to the company without an interview.

That is the probability of no interview is, P (X < x) = 0.025.

That is, P (Z < z) = 0.025.

The z-score corresponding to this probability is, 1.96.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(1.96\times 10.74)\\\\=92.3204\\\\\approx 92$

Thus, the complete sentence is:

A score of 61 should be used as the cutoff for the lower 16% and a score of 92 should be used as the cutoff for the upper 2.5%.