What is the probability of this question?

Question

The weight of a certain brand of candies are normally distributed with a mean weight of 0.8 538G and a standard deviation of 0.0 519G. A sample of these candies came from a package containing 453 candies in the package label stated that the net weight is 386.5 G.(If every package has 453 candies, the main weight of the candies must exceed 386.5

——— = 0.8532 g for the net continents the way 386.5 g.

453

a.) If 1 candy is randomly selected, find the probability that it weighs more than 0.8532 G.

The probability is:

Answer 1

the z value for the data is z=(score - mean)/sr. dev. = (.8532-.8538)/.0519 = -.01156

using the TI 84, normalcdf (-0.01156,100, 0, 1 ) = prob z > -0.01156 = .5046

Answer 2

How you solve this depends on your available technology. Here's an old-school, low-tech version.

The weight of a single candy comes from a Normal distribution with Mean = 0.8538 and StdDev = 0.0518. You can first find the probability that a single random candy weighs less than 0.8532 as follows:

1) Convert to an equivalent z-score:

1. z = (ActualValue - Mean)/StdDev = (.8532-.8538)/.0519 = -0.01156...

2) Use a z-score/Standard Normal table to find the probability of having a z-score less than this. If you are using a table, round the above z-score to -0.01; the corresponding value in the table is .4960.

Since the probability that a candy weighs less than the given amount is .4960, the probability that a candy weighs more than the given amount is 1 - .4960 = .5040.

Note that it's much easier, quicker, and more accurate to solve this with technology. If you can specify what you have available, you might get a more useful answer.