Question

The weights of a certain brand of candies are normally distributed with a mean weight of0.8612g and a standard deviation of 0.0514g. A sample of these candies came from a package containing 452 candies, and the package label stated that the net weight is 385.9g.​ (If every package has452candies, the mean weight of the candies must exceed 385.9 Over 452 =0.8538g for the net contents to weigh at least 385.9

​g.)a. If 1 candy is randomly​ selected, find the probability that it weighs more than

0.8538

g.The probability is

​(Round to four decimal places as​ needed.)

b. If 452candies are randomly​ selected, find the probability that their mean weight is at least 0.8538g.The probability that a sample of

452candies will have a mean of 0.8538g 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

0.8538

g or greater when

452

candies are selected

▼

is not

is

exceptionall small

Answer 1

For part a) we need to calculate a z-score and refer to the normal distribution tables.

Remember that not all distribution tables read the same way, but they should show you by a diagram whether the percentage given is for the area to the left of the z or to the right.

Z = (X - μ)/σ      or       Z = (.8538 - .8612)/.0514 = -0.14

P ( X>0.8538 )=P ( Z>−0.14 )

Use the standard normal table to conclude that:

P (Z>−0.14)=0.5557

Part b) is essentially the same calculation but with a sample standard deviation:

σx = σ/√n = .0514/√(452) = .0024

Z = (.8538 - .8612)/.0024 = -3.08

P ( X>0.8538 )=P ( Z>−3.08 )

Use the standard normal table to conclude that:

P (Z>−3.08)=0.999

The area to the right of this Z-score translates to a probability of 0.999

c)  Yes, because the probability of getting a sample mean of 0.8538 or greater when 441 candies are selected is NOT exceptionally small.