The weights of a certain brand of candies are normally distributed with a mean weight of 0.8547 g and a standard deviation of 0.0511 g. a sample of these candies came from a package containing 463 candies, and the package label stated that the net weight is 395.2 g. (if every package has 463 candies, the mean weight of the candies must exceed startfraction 395.2 over 463 endfraction 395.2 463equals=0.8535 g for the net contents to weigh at least 395.2 g.) a. if 1 candy is randomly selected, find the probability that it weighs more than 0.8535 g. the probability is .(round to four decimal places as needed.)

There is a 69.15% probability that it weighs more than 0.8535 g.

Step-by-step explanation:

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

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

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, we have that

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8547 g, so .

We have a sample of 463 candies, so we have to find the standard deviation of this sample to use in the place of in the Z score formula. We can do this by the following formula:

Find the probability that it weighs more than 0.8535

This is 1 subtracted by the pvalue of Z when

So

has a pvalue of 0.3085.

This means that there is a 1-0.3085 = 0.6915 = 69.15% probability that it weighs more than 0.8535 g.

i would say the third one.

step-by-step explanation: liner would mean in a line and the only one that doesn't look like it's even thinking about forming a line would be the third one