Answer:
A score of <u>61</u> should be used as the cutoff for the lower 16% and a score of <u>92</u> 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](https://tex.z-dn.net/?f=%5Cmu%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%20X%3D%5Cfrac%7B1%7D%7B30%7D%5Ctimes%202138%3D71.27%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28X-%5Cmu%29%5E%7B2%7D%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B30-1%7D%5Ctimes%203343.8667%7D%3D10.74)
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 <em>z</em>-score corresponding to this probability is, -1.
Compute the value of <em>x</em> as follows:
![z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(-1\times 10.74)\\\\=60.53\\\\\approx 61](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5C%5C%5C%5Cx%3D%5Cmu%2Bz%5Ccdot%20%5Csigma%5C%5C%5C%5C%3D71.27%2B%28-1%5Ctimes%2010.74%29%5C%5C%5C%5C%3D60.53%5C%5C%5C%5C%5Capprox%2061)
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 <em>z</em>-score corresponding to this probability is, 1.96.
Compute the value of <em>x</em> as follows:
![z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(1.96\times 10.74)\\\\=92.3204\\\\\approx 92](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5C%5C%5C%5Cx%3D%5Cmu%2Bz%5Ccdot%20%5Csigma%5C%5C%5C%5C%3D71.27%2B%281.96%5Ctimes%2010.74%29%5C%5C%5C%5C%3D92.3204%5C%5C%5C%5C%5Capprox%2092)
Thus, the complete sentence is:
A score of <u>61</u> should be used as the cutoff for the lower 16% and a score of <u>92</u> should be used as the cutoff for the upper 2.5%.