Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8622 g and a standard deviation of 0.0523 9. A sample of these candies came from a package containing 445 candies, and the package label stated that the net weight is 380 2 9. Of every package has 445 candies, the mean weight of the candies must exceed 245 -0.8543 g for the net content to weigh at least 38029.) 1 candy is randomly selected, find the probability that it weighs more than 0.8543 9. The probabilis (Round to four decimal places as needed.) b. 11 445 candies we randomly selected, find the probability that their mean weight is at least 0.85439 The probably that a sample of 445 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? because the probability of getting a sample mean of 0.8543 g or greater when 445 candies are selected exceptionally small

Answer 1

a)

µ = 0.8622, σ = 0.0523

P(X > 0.8543) =

= P( (X-µ)/σ > (0.8543-0.8622)/0.0523)

= P(z > -0.1511)

= 1 - P(z < -0.1511)

Using excel function:

= 1 - NORM.S.DIST(-0.1511, 1)

= 0.5600

b)

µ = 0.8622, σ = 0.0523, n = 445

P(X̅ > 0.8543) =

= P( (X̅-μ)/(σ/√n) > (0.8543-0.8622)/(0.0523/√445) )

= P(z > -3.1864)

= 1 - P(z < -3.1864)

Using excel function:

= 1 - NORM.S.DIST(-3.1864, 1)

= 0.9993

c)

Yes because the probability of getting a sample mean of 0.8543g or greater when 445 candies are selected is not exceptionally small.