Question

Part 2: Free Response 11. The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces. Company managers do not want the weight of a chocolate bar to fall below 8 ounces, for fear that consumers will complain. (a) Find the probability that the weight of a randomly selected candy bar is less than 8 ounces.

Answer 1

0.1587 = 15.87% probability that the weight of a randomly selected candy bar is less than 8 ounces.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces.

This means that
`\mu = 8.1, \sigma = 0.1`.

(a) Find the probability that the weight of a randomly selected candy bar is less than 8 ounces

This is the pvalue of Z when X = 8. So

`Z = (X - \mu)/(\sigma)`

`Z = (8 - 8.1)/(0.1)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

0.1587 = 15.87% probability that the weight of a randomly selected candy bar is less than 8 ounces.