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

Answer 1

Let $X=$ the average cholesterol content of a certain can goods in milligrams

$X\sim N(\mu, \sigma^2/n)$

Then $Z=\dfrac{X-\mu}{\sigma/\sqrt{n}}\sim N(0. 1)$

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

a.  If a single egg is selected, find the probability that the cholesterol content will be more than 220 milligrams.

$n=1$

$$P(X>220)=1-P(X\leq220)$$

$$=1-P(Z\leq\dfrac{220-215}{15/\sqrt{1}})\approx1-P(Z\leq0.33333)$$

$$\approx0.3694$$

b. If a sample of 25 eggs is selected, find the probability that the mean of the sample will be larger than 220 milligrams.

$n=25$

$$P(X>220)=1-P(X\leq220)$$

$$=1-P(Z\leq\dfrac{220-215}{15/\sqrt{25}})\approx1-P(Z\leq1.66667)$$

$$\approx0.0478$$