M&M plain candies have weights that are normally distrbuted with a mean weight of 0.8565 gram and a standard deviation of 0.0518 gram. A random sample of 100 M&M candies is obtained from a package containing 465 candies: the package label states that the net weight is 396.9 grams. If every package has 465 candies, the mean weight of the candies must exceed 396.9/465=0.8535 for the net contents to weigh at least 3969.9 grams.

Question

M&M plain candies have weights that are normally distrbuted with a mean weight of 0.8565 gram and a standard deviation of 0.0518 gram. A random sample of 100 M&M candies is obtained from a package containing 465 candies: the package label states that the net weight is 396.9 grams. If every package has 465 candies, the mean weight of the candies must exceed 396.9/465=0.8535 for the net contents to weigh at least 3969.9 grams.
A. If 1 M&M plain cand is randomly selected, how likely is it that it weighs more than 0.8535 gram?
B. If 465 M&M plain candies are randomly selected, how likely is it that their mean weight is at least 0.8535 gram?

PsyDAG · Answer

A. Z = (score-mean)/SD (for distribution of scores)

B. Z = (mean1 - mean2)/standard error (SE) of difference between means (for distribution of means)

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since 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 proportions related to these Z scores.