Answer to Question #150594 in Statistics and Probability for Vladimyr Lubin

Question

The overhead reach distances of adult females are normally distributed with a mean of 195 and a standard deviation of 8.3cm.
a. Find the probability that an individual distance is greater than 204.30 cm.
b. Find the probability that the mean for 15 randomly selected distances is greater than 193.20cm
c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

$\mu=195$

$\sigma=8.3$

a) $P(X>204.30)=P(Z>\frac{204.3-\mu}{\sigma})=P(Z>\frac{204.3-195}{8.3})=P(Z>1.12)=$

$=1-P(Z<1.12)=1-0.8686=0.1314$

b) $n=15$

$P(\bar x>193.20)=P(Z>\frac{193.20-\mu}{\frac{\sigma}{\sqrt n}})=P(Z>\frac{193.20-195}{\frac{8.3}{\sqrt 15}})=P(Z>-0.84)=$

$=1-P(Z<-0.84)=1-0.2005=0.7995$

c) The normal distribution can be used as soon as the original population is normally distributed.