The weights of a certain brand of candies are normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g. A sample of these candies came from a package containing 466 candies, and the package label stated that the net weight is 397.9 g. (If every package has 466 candies, the mean weight of the candies must exceed 397.9466 Required:
If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g. The probability is:

The probability is 0.508 = 50.8%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g.

This means that

If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g.

This is 1 subtracted by the pvalue of Z when X = 0.8535. So

has a pvalue of 0.492

1 - 0.492 = 0.508

The probability is 0.508 = 50.8%.

what? ?

what type of question is thiss

step-by-step explanation:

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles