Answer to Question #281202 in Statistics and Probability for Sammy

Question

The overhead read distances of adult females are normally distributed with a mean of 195 cm any standard deviation of 8.6 cm. Find the probability that the individual distance is greater than 204.30 cm. Find the probability that the mean of 15 randomly selected distances is greater than 193.50 cm. Why can’t 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.