Question

The average cholesterol content of a certain can goods is 215 milligrams and the cholesterol deviation is 15 milligrams.assume the variable is normally distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 miligrams?

Answer 1

Let $X=$ the mean of the sample: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=215\ mg, \sigma=15\ mg, n=25$

$$P(X>220)=1-P(X\leq 220)$$

$$=1-P(Z\leq \dfrac{220-215}{15/\sqrt{25}})\approx1-P(Z\leq 1.6667)$$

$$\approx0.04779$$