Question

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds. a) What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds? Give your answer to four decimal places.

Answer 1

There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.

Explanation:

Problems of normally distributed samples can be 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 = (X - \mu)/(\sigma)`

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, so
`\mu = 47.88, \sigma = 3.09`

What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds?

That is
`P(X > 50)`

So

`Z = (X - \mu)/(\sigma)`

`Z = (50 - 47.88)/(3.09)`

`Z = 0.69`

`Z = 0.69` has a pvalue of 0.7549.

This means that
`P(X \leq 50) = 0.7549`.

We also have that

`P(X \leq 50) + P(X > 50) = 1`

`P(X > 50) = 1 - 0.7549 = 0.2451`

There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.