Assume the weights of Farmer Carl's potatoes are normally distributed with a mean of 7.0 ounces and a standard deviation of 1.3 ounces. He bags his potatoes in groups of 6. You buy a bag and the total weight is 36 ounces. Here we determine how lucky or unlucky you are.

Question

(a) What is the mean potato weight in your bag of 6?

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is less than the mean found in your bag.

Answer 1

In "YOUR BAG" there is no mean. It weighs 36 ounces. If you are talking about each individual potato, 36/6 = 6

(b) Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

Answer 2

A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.2 ounces. A consumer advocacy group wants to determine whether the mean amount is actually less than this. The mean volume of juice for a random sample of 70 bottles was 16.04 ounces with a standard deviation of 0.9 ounces. Find a 95% confidence interval to estimate the true mean amount of juice in these bottles.

A) What is the sample mean?

B) Standard deviation?

C) How many standard deviations do we need to capture 95% estimate?

D) Margin of error?