completely lost!!!!

Question

A certain brand of candies have a mean weight of 0.8613 g and a standard deviation of 0.0522. A sample of these candies came from a package containing 441 candies, and the package label stated that the net weight is 376.2 g. (If every package has 441 candies, the mean weight of the candies must exceed 376.2/441 = 0.8531 g for the net contents to weigh at least 376.2 g.) 

a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8531 g.
The probability is approximately ?

B If 441 candies are randomly selected, find the probability that their mean weight is at least 0.8531 g.

c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label? because the probability of getting a sample mean of 0.8531 g or greater when 441 candies are selected exceptionally small.

Answer 1

Ryan,

 

In population or samples, the mean is always based on your population mean which in this case is .8613. Your sample point is the desired .8531 mean. The standard deviation needs to factor in the sample size because it grows smaller with larger samples. We call this standard error:

SE=σ/√n=.0522/√441=.0522/21=.00249

 

Now to find the probability we need to calculate a z value and convert that to a probability using the z-table:

z=(x-μ)/SE=(.8531-.8613)/.00249=-.0082/.00249=-3.29

Since the cumulative probability of this is essentially zero (.000495), the probability of finding a candy weighing more than .8531 is 1-.000495=99.95% likely.

 

To find the probability that 441 candies weigh at least .8531, take 99.95% to the 441th power:

.8038

 

As for part c, my best answer here is that with a 99.95% chance the mean weight is being met, the label is probably accurate. Hope that helps!