Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8614 g and a standard deviation of 0.0516 g. A sample of these candies came from a package containing 441 ​candies, and the package label stated that the net weight is 376.4 g.​ (If every package has 441 ​candies, the mean weight of the candies must exceed StartFraction 376.4 Over 441 EndFraction equals0.8536 g for the net contents to weigh at least 376.4 ​g.) a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8536 g.

Answer 1

Answer:

There is a 55.97% probability that the randomly selected candy weighs more than 0.8596g.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8614 g and a standard deviation of 0.0516 g. This means that $\mu = 0.8614, \sigma = 0.0516$.

a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8536 g.

This is 1 subtracted by the pvalue of Z when $X = 0.8536$. So:

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

$Z = \frac{0.8536 - 0.8614}{0.0516}$

$Z = -0.15$

$Z = -0.15$ has a pvalue of 0.4403. This means that there is a 1-0.4403 = 0.5597 = 55.97% probability that the randomly selected candy weighs more than 0.8596g.