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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 8.1}{0.1}$

$Z = -1$

$Z = -1$ 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.