Central Limit Theorem 6.0

Question

A population of values has a normal distribution with  and . You intend to draw a random sample of size .

Find the probability that a sample of size  is randomly selected with a mean between 103.4 and 118.5.

P(103.4 < M < 118.5) = 

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Standardize the normal random variables to find the z-scores

z = (x - µ) / (σ / √n)

z = (103.4 - 100.3) / (68.6 / √120)

z = 0.495

z = (x - µ) / (σ / √n)

z = (118.5 - 100.3) / 68.6 / √120

z = 2.91

Since z tables are cumulative, if we subtract the probabilities that correspond to these z-scores, it will give us the area in between to the two P(103.4 < Z < 118.5)

With a z-score of 0.495, the probability is .6897. With a z-score of 2.91, the probability is .9982.

.9982 - .6897 = .3085 or 30.85%