Question

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.

Answer 1

50.40% probability that it weighs more than 0.8544 g.

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.

In this question, we have that:

`\mu = 0.8551, \sigma = 0.0518`

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

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

`Z = (0.8544 - 0.8551)/(0.0518)`

`Z = -0.01`

`Z = -0.01` has a pvalue of 0.4960

1 - 0.4960 = 0.5040

50.40% probability that it weighs more than 0.8544 g.