Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8549 g and a standard deviation of 0.052 g. A sample of these candies came from a package containing 452 ​candies, and the package label stated that the net weight is 386.1g.​ (If every package has 452 ​candies, the mean weight of the candies must exceed StartFraction 386.1 Over 452 EndFraction =0.8541 g for the net contents to weigh at least 386.1g.) a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8541 g. The probability is nothing. ​(Round to four decimal places as​ needed.) b. If 452 candies are randomly​ selected, find the probability that their mean weight is at least 0.8541 g. The probability that a sample of 452 candies will have a mean of 0.8541 g or greater is ________. ​(Round to four decimal places as​ needed.) c. Given these​ results, does it seem that the candy company is providing consumers with the amount claimed on the​ label? ▼ No, Yes, because the probability of getting a sample mean of 0.8541 g or greater when 452 candies are selected ▼ is not is exceptionally small.

Answer 1

Answer)

As the data is normally distributed, we can use standard normal z table to estimate the probability.

Mean = 0.8549

S.d = 0.052

A)

P(x>0.8541)

Z score = (x-mean)/s.d

Z = (0.8541-0.8549)/0.052

Z = -0.02

From z table, p(z>-0.02) = 0.508

B)

P(x>0.8541)

For sample

Z = (x-mean)/(s.d/√n)

Z = (0.8541-0.8549)/(0.052/√452)

Z = -0.33

From z table, p(z>-0.33) = 0.6293

Yes because the probability of getting a sample mean of 0.8541 or greater when sample 452 candies are selected is not exceptionally small