Question

Consider a normally distributed population with a mean of 4 and a variance of 16. If a random sample of size 16 is drawn, what is the probability that the sample mean is at most 4?

Answer 1

Solution:

Given, $\mu=4,\sigma^2=16,n=16$

$X\sim(\mu,\sigma)$

Then, $\bar X\sim(\mu,\sigma/\sqrt n)$

$P(\bar X\le4)=P(z\le \dfrac{4-4}{4/4}) \\=P(z\le 0) \\=0.5$