Probability

Question

A certain brand of candies have a mean weight of .8587g and standard deviation of .0524. A sample of these candies came from a package containing 442 candies, and the package label stated the netweight is 377.2g. (If every package has 442 candies, the mean weight of the candies must exceed 377.2/442=.8535g for the net contents to weigh at least 377.2g).

a) If one candy is randomly selected, find the Probability that it weighs more than .8535g                         Round to 4 decimals

b) If 442 candies are randomly selected, find the probability that their mean is at least .8538g                   Round to 4 decimals

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 .8538g or greater when 442 candies are selected (IS, Is Not) exceptionally small.

Please show work

Answer 1

The mean weight is 0.8587 g

with a standard deviation ("sigma of the gaussian") of 0.0524.

 

a) The weight of 0.8535 g can be converted to the number of standard deviations below/above the mean using:

 

   z  = (0.8535 - 0.8587) / 0.0524

       = -0.0052/0.0524 ~ -0.0992

 

For the standard normal distribution, the probability that Z is less than 0, written as P(Z<0), is  0.5.  As the value increases the probability P(Z < value) increases also.  These probabilities are given by a normal distribution function or table, e.g.:  http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm

 

In this case we want P(Z>-0.0992). Because the distribution is symmetric, this the same as:  P(Z<+0.0992)

From the table (interpolating between 0.9 = 0.5359 and 0.1 = 0.5398) we get:

    P(Z<0.0992) ~ 0.5395.

 

So the probability the piece weights more than 0.8535 g is 0.5395.

 

b) Adding 442 samples will give a normal distribution that has:

   a mean that is 442 times 0.8587, 

   and a standard deviation that is 0.0524 *√442 .

Dividing by 442 we get "their mean" of 0.8587 g  (as expected)

and "their mean"'s standard deviation of (0.0524 *√442) / 442

   = 0.0524 /√442  ~ 0.0025 

The standard deviation has been reduces by the square root of the number of samples.

 

Now the z value for the 442-sample-mean is:

 

z = (0.8535 - 0.8587) / 0.0025 ~ -2.08

 

Using the table as before, we look up: P(Z<+2.08) = 0.9821

So there is a 98.2 % chance that the mean weight is above the required value.

 

c) I leave it to you to decide if a ~ 2% chance of getting a little bit less than the specified weight is "exceptionally small"  ;-)