Answer to Question #201314 in Statistics and Probability for lemy karl ayyo

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

Now, Given

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

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

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

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

So, from standard normal distribution table:

$P(Z>1.67)=0.0475$

Answer:

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