Answer:
Step-by-step explanation:
FULL QUESTION:
A certain brand of candies have a mean weight of 0.8605 g and a standard deviation of 0.0513. 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 376.7/441 =0.8543 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 0.8543 g.
The probability is ?. (Round to four decimal places as needed.)
b. If 441 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 441 candies will have a mean of 0.8543 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?
Yes/no because the probability of getting a sample mean of 0.8543 g or greater when 441 candies are selected is /is not exceptionally small.
SOLUTION:
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 = (.8543 - .8605)/.0513 = -.1209
The area to the right of this Z-score translates to a probability of 0.5478
Part b) is essentially the same calculation but with a sample standard deviation:
σx = σ/√n = .0513/√(441) = .0024
Z = (.8543 - .8605)/.0024 = -2.5833
The area to the right of this Z-score translates to a probability of 0.9951
c) Yes, because the probability of getting a sample mean of 0.8543 or greater when 441 candies are selected is NOT exceptionally small.
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