Question

The average cholesterol content is 180 m deviation of 12 milligrams assume that the variable is normally distributed give a sample of 16 pounds is selected what is the probability that the mean of the sample with the greater than 172 milligrams

Answer 1

$$P(X>172)=1-P(Z\le\dfrac{172-180}{12/\sqrt{16}})$$

$$=1-P(Z\le-2.6667)\approx0.99617$$