Respuesta :
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:
[tex]\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[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(-1\times 10.74)\\\\=60.53\\\\\approx 61[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma}\\\\x=\mu+z\cdot \sigma\\\\=71.27+(1.96\times 10.74)\\\\=92.3204\\\\\approx 92[/tex]
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%.
