Answer to Question #243966 in Statistics and Probability for Moses Gato

Question

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm.

a. Find the probability that an individual distance is greater than 218.90 cm.

b. Find the probability that the mean for 25 randomly selected distances is greater than 203.70 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

$\mu=205.5 \\ \sigma = 8.6$

a.

$P(X>218.90) = 1 -P(X<218.90) \\ = 1 -P(Z< \frac{218.90-205.5}{8.6}) \\ = 1 -P(Z< 1.55) \\ = 1 -0.9394 \\ = 0.0606$

b. n=25

$P(\bar{X}>203.70) = 1 -P(Z< \frac{203.70-205.5}{8.6 / \sqrt{25}}) \\ = 1 -P(Z< -1.046) \\ = 1 -0.1477 \\ = 0.8523$

c. Because the population follows a normal distribution and sample mean of normal distribution population always follows a normal distribution.