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 387.0g.)

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.)

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 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 problem, we have that:

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

has a pvalue of 0.4562

1 - 0.4562 = 0.5438

The probability is 0.5438

10000

step-by-step explanation: divide by 2,

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.