A Population Has A Mean Of 73.5 And A Standard Deviation Of 2.5 A. Find the mean and standard deviation of x for samples of size 30 B. Find the probability that the mean of a sample of size 30 will be less than 72

Let $X_1, X_2, ..., X_n$ be a random sample from a distribution with mean $\mu$ and variance $\sigma^2$. Then by the Central Limit Theorem if $n$ is sufficiently large, $\bar{X}$ has approximately a normal distribution with $\mu_{\bar{X}}=\mu$ and $\sigma^2_{\bar{X}}=\sigma^2/n.$

The Central Limit Theorem can generally be used if $n>30.$

A.

Given $\mu=73.5, \sigma=2.5, n=30.$

Assume $\bar{X}\sim N(\mu, \sigma^2/n)$

$\mu_{\bar{X}}=\mu=73.5$

$\sigma_{\bar{X}}=\sigma/\sqrt{n}=2.5/\sqrt{30}\approx0.4564$

B.

$P(\bar{X}<72)=P(Z<\dfrac{72-73.5}{2.5/\sqrt{30}})$

$\approx P(Z<-3.2863)\approx0.0005$