Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm195 cm and a standard deviation of 8.6 cm8.6 cm. a. Find the probability that an individual distance is greater than 204.30204.30 cm. b. Find the probability that the mean for 2525 randomly selected distances is greater than 192.80 cm.192.80 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

Solution:

Given in the question

Mean =195

Standard deviation =8.6

Solution(A)

P(X>204.30) = 1-P(X<=204.30)

Z = (204.30-195)/8.6 = 1.08

From z table we found p-value

P(X>204.30)= 1-0.8599 = 0.1401

Solution(b)

No. Of sample = 25

P(X>192.80) = 1-P(X<=192.8)

Z = (192.8-195)/8.6/Sqrt(25) = -1.28

From t distribution table

At Df =24, p-value = 0.1064

Solution(C)

Because the original population has a normal distribution, the distribution of sample mean is normal for any sample size.