Question

Suppose it is known that the mean and the standard deviation for a given population are 65 and 8, respectively. What is the probability that a sample of 100 drawn from this population will have a mean greater than 66?

Answer 1

$P(\bar X >66)=P(Z>\frac{66-65}{\frac{8}{\sqrt{100}}})=P(Z>1.25)=1-P(Z<1.25)=0.1056.$