Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8624 g and a standard deviation of 0.0514 g. A sample of these candies came from a package containing 441 ​candies, and the package label stated that the net weight is 376.7 g.​ (If every package has 441 ​candies, the mean weight of the candies must exceed StartFraction 376.7 Over 441 EndFraction equals0.8541 g for the net contents to weigh at least 376.7 ​g.) 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 441 candies are randomly​ selected, find the probability that their mean weight is at least 0.8541 g. The probability that a sample of 441 candies will have a mean of 0.8541 g or greater is nothing. ​(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? ▼ Yes, No, because the probability of getting a sample mean of 0.8541 g or greater when 441 candies are selected ▼ is is not exceptionally small.

Answer 1

a)probability =0.564

b)probability = 0.9997

c)Yes,

Step by step Explanation:

z-score is needed here, so we need to calculate it by referring to the normal distribution table.

mean weight of 0.8624 g

standard deviation of 0.0514 g

a) If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8541 g. The probability is nothing

Z = (X - μ)/σ

Z = (0.8541 - 0.8624)/0.0514 = -0.1615

The area to the right of this Z-score translates to a probability of 0.564

b) this has almost the same calculation with (a)but with a sample standard deviation

σx = σ/√n =

0.0514/√(441)

=0 .0024

Z = (0.8541 -0 .8624)/.0024 = -3.4583

The area to the right of this Z-score translates to

a probability of 0.9997

c) Yes, because the probability of getting a sample mean of 0.8541or greater when 441 candies are selected is NOT exceptionally small.