Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.0511 g. A sample of these candies came from a package containing 469 ​candies, and the package label stated that the net weight is 400.3 g.​ (If every package has 469 ​candies, the mean weight of the candies must exceed StartFraction 400.3 Over 469 EndFraction equals0.8536 g for the net contents to weigh at least 400.3 ​g.)

Required:
a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8536 g.
b. If 441 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?

Answer 1

Answer:

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

$\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535$

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002$

$P(X>0.8536)=P(z>0.002)=0.4992$

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288$

$P(X_s>0.8543)=P(z>0.3288)=0.37115$

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89$

$P(X_s>0.8556)=P(z>0.89)=0.18673$

The probability is P=0.187, so the claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.