Problem 16BSC

Loading M&M PackagesM&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518 g (based on Data Set 20 in Appendix B). The M&M candies used in Data Set 20 came from a package containing 465 candies, and the package label stated that the net weight is 396.9 g. (If every package has 465 candies, the mean weight of the candies must exceed 396.9/465 = 0.8535 g for the net contents to weigh at least 396.9 g.)

a.If 1 M&M plain candy is randomly selected, find the probability that it weighs more than 0.8535 g.

b.If 465 M&M plain candies are randomly selected, find the probability that their mean weight is at least 0.8535 g.

c.Given these results, does it seem that the Mars Company is providing M&M consumers with the amount claimed on the label?

Problem 16BSC

Answer:

Step1 of 3:

We have M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518 g (based on Data Set 20 in Appendix B). The M&M candies used in Data Set 20 came from a package containing 465 candies, and the package label stated that the net weight is 396.9 g. (If every package has 465 candies, the mean weight of the candies must exceed 396.9/465 = 0.8535 g for the net contents to weigh at least 396.9 g.)

That is 0.8565 and 0.0518.

Step2 of 3:

We need to find,

a.If 1 M&M plain candy is randomly selected, find the probability that it weighs more than 0.8535 g.

b.If 465 M&M plain candies are randomly selected, find the probability that their mean weight is at least 0.8535 g.

c.Given these results, does it seem that the Mars Company is providing M&M consumers with the amount claimed on the label?

Step 3 of 3:

a).

If 1 M&M plain candy is randomly selected,the probability that it weighs more than 0.8535 g is given by

Consider the z statistics when x = 0.8535

P(x 0.8535) = 1- P(Z )

= 1 - P()

= 1 - P()

= 1 - P(-0.0579)

= 1 - 0.4761

= 0.5239

Probability of -0.0579 is calculated by using standard normal table(area under normal curve).In area under normal curve we have to see row -0.05 under column 0.07

P(-0.0579) = 0.4761.

Therefore, the probability that it weighs more than 0.8535 g is 0.4761.

b).

If 465 M&M plain candies are randomly selected, the probability that their mean weight is at least 0.8535 g is given by

Consider the z statistics when x = 0.8535

P(x 0.8535) = 1- P(Z )

= 1 - P()

= 1 - P()

= 1 - P(-1.25)

= 1 - 0.1056

= 0.8944

Probability of -1.25 is calculated by using standard normal table(area under normal curve).In area under normal curve we have to see row -1.2 under column 0.05

P(-1.25) = 0.1056.

Therefore,the probability that their mean weight is at least 0.8535 g is 0.1056.

c).

Whether the bags contain the labeled amount cannot be answered from the one bag examined. If each bag is filled with exactly 465 candies, then the company needs to make some adjustments – because then the probability a bag contains less than the claimed weight is 1 – 0.8944 = 0.1056 [and since 0.1056 > 0.05, it would not be unusual to get a bag with less than the claimed weight]. In practice, however, the bags are filled by volume or weight – and not by candy count

Instead of filling each bag with exactly 465 M&Ms, the company probably fills the bags so that the weight is as stated. In any event, the company appears to be doing a good job of filling the bags.