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 $σ = 2.91$.

Answer :

Given :

The population mean $(μ) = 64.5$

The population standard deviation $(σ) = 2.91$

The sample size $(n) = 60$

Solution :

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

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