Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8599 g and a standard deviation of 0.0519 g. A sample of these candies came from a package containing 453 ​candies, and the package label stated that the net weight is 387.0 g.​ (If every package has 453 ​candies, the mean weight of the candies must exceed 387.0/453=0.8542 g for the net contents to weigh at least 387.0​g.)

a. If 1 candy is randomly​ selected, find the probability that it weighs more than0.8542g.The probability is
(Round to four decimal places as​ needed.)

Answer 1

Answer:

The probability is 0.5438

Step-by-step explanation:

Problems of normally distributed samples are 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:

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

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 problem, we have that:

$\mu = 0.8599, \sigma = 0.0519$

a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8542g.

This is 1 subtracted by the pvalue of Z when X = 0.8542. So

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

$Z = \frac{0.8542 - 0.8599}{0.0519}$

$Z = -0.11$

$Z = -0.11$ has a pvalue of 0.4562

1 - 0.4562 = 0.5438

The probability is 0.5438