Question

1. Many firms use on-the-job training to teach their employees computer programming. Suppose you work in the personnel department of a firm that just finished training a group of its employees to program, and you have been requested to review the performance of one of the trainees on the final test that was given to all trainees. The mean and standard deviation of the test scores are 72 and 5, respectively, and the distribution of scores is bell-shaped and symmetric. Suppose the trainee in question received a score of 68. Compute the trainee's z-score.

a. z = -0.88

b. z = -0.80

c. z = 0.8

d. z = 0.88

2. The cholesterol levels (in milligrams per deciliter) of 30 adults are listed below. Find the interquartile range for the cholesterol level of the 30 adults.

154 16 165 165 10 11 12 10 14 185 18 19 10 12 15 18 18 200 200 200 25 205 211 215 220 220 225 238 255 265

a. 211

b. 30

c. 180

d. 31

Answer 1

1).$\mu=72,\sigma=5, X = 68$

the trainee's z-score is :-

$z=\left ( \frac{X-\mu}{\sigma} \right )=\left ( \frac{68-72}{5} \right )=\mathbf{0.80(b)}$

2). sample size (n) = 30

first quartile be:-

$=\frac{1}{4}*(n+1)^{th}obs=\frac{1}{4}*(30+1)^{th }obs=7.75^{th}obs\approx 8 ^{th}obs=180$

third quartile be:-

$=\frac{3}{4}*(n+1)^{th}obs=\frac{3}{4}*(30+1)^{th }obs=23.25^{th}obs\approx 23^{rd}obs=211$

the interquartile range be:-

= third quartile - first quartile

$=211-180$

$=\mathbf{31(d)}$

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