Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of

8.6 cm

b. Find the probability that the mean for 15 randomly selected distances is greater than 193.70 cm.

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

Solve B & C please!

Answer 1

b) Let X be the overhead reach distances of adult females. Then we know that

X ~ N(195,8.6²) . We have to find the probability that P( > 193.7 ). We will use the result that if X~N() , then ~ N( /n ). Thus ~ N( 195,8.6²/15). Also if ~ N( /n ), then Z = ~ N(0,1).

Hence P( > 193.7 ) = P(Z > (193.7-195) /(8.6/ ) = P(Z>-0.59) = P (Z < 0.59 ) = 0.7224 (using the normal probability tables)

c) We can use the normal distribution even though the sample size does not exceed 30, because the parent population follows normal distribution and the population standard deviation is known. We use t-distribution if the population standard deviation is not known.