The mean height of women in a country (ages 20-29) is 64.5 inches. A random sample of 60 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 inches? Assume $\sigma =2.91$.

Answer :

Given :

The population mean $(\mu )=64.5$

The population standard deviation $(\sigma )=2.91$

The sample size $(n)=60$

Solution :

→ The Probability of Mean Height for the Sample Being Greater Than 65 inches :

$$\therefore P(\overline{x}>65)=P(\frac{\overline{x}–\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{65–64.5}{\frac{2.91}{\sqrt{60}}})$$ $$=P(z>1.33)$$ $$=0.0918$$