k 11: Central Limit Question 6, "6.4.16-T, HW score: $83.33 \%, 5$ of 6 points Save The weights of a certain brand of candies are normally distributed with a mean weight of $0.8613 \mathrm{~g}$ and a standard deviation of $0.0524 \mathrm{~g} . \mathrm{A}$ sample of these candies camse from a package containing 466 candios, and the package label stated that the net weight is $397.6 \mathrm{~g}$ (If every package has 466 candies, the mean weight of the candies must exceed $\frac{397.6}{466}=0.8532 \mathrm{~g}$ for the net contents to weigh at least $397.6 \mathrm{~g}$ ) a. If 1 candy is randomly selected, find the probability that it weighs more than $0.8532 \mathrm{~g}$ The probability is (Round to four decimal places as needed.) instructor Clear all Check answer
Solution
Step 1 ：Given that the weights of the candies are normally distributed with a mean (μ) of 0.8613 g and a standard deviation (σ) of 0.0524 g.
Step 2 ：We are asked to find the probability that a randomly selected candy weighs more than 0.8532 g (X).
Step 3 ：We can use the z-score formula to find this probability. The z-score is calculated as \(z = \frac{X - μ}{σ}\).
Step 4 ：Substituting the given values into the formula, we get \(z = \frac{0.8532 - 0.8613}{0.0524} = -0.1546\).
Step 5 ：We can then use a z-table or a statistical function to find the probability that corresponds to this z-score. The probability is approximately 0.5614.
Step 6 ：Thus, the probability that a randomly selected candy weighs more than 0.8532 g is approximately \(\boxed{0.5614}\).
