The weights of a certain brand of candies are normally distributed with a mean weight of .8551 g and a standard deviation of 0.0518 g. A sample of these candies came from a package containing 467 candies, and the package label stated that the net weight is 399 g. (If every package has 467467 candies, the mean weight of the candies must exceed 399.0/467 = .8544 g for the net contents to weigh at least 399 g.)If 1 candy is randomly selected, find the probability that it weighs more than 0.8544 g.

50.40% probability that it weighs more than 0.8544 g.

Step-by-step explanation:

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

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

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.

In this question, we have that:

If 1 candy is randomly selected, find the probability that it weighs more than 0.8544 g.

This is 1 subtracted by the pvalue of Z when X = 0.8544. So

has a pvalue of 0.4960

1 - 0.4960 = 0.5040

50.40% probability that it weighs more than 0.8544 g.

f(x) = y

f(-3) = -5

step-by-step explanation: