Question

ACTIVITY:
The average cholesterol content of a certain canned good is 215 mg, and the standard
deviation is 15 mg. Assume that the variable is normally distributed. If a sample of 25 canned goods
are selected, what is the probability that the mean of the sample will be larger than 220 mg

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 that,

$X= 220\\\mu=215\\\sigma = 15\\n=25$

$z=\dfrac{X-\mu}{\sigma/\sqrt n}=\dfrac{220-215}{15/\sqrt{25}}=\dfrac{25}{15}=1.67$

Thus, probability that mean of sample is greater than 220 mg is

$P(z>220)=P(z>1.67)$

So, from standard normal distribution table:

$P(z>1.67)=0.0475$

Hence, The probability that mean of sample is greater than 220 mg is 0.0475