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.

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

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by

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

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

That is

So

has a pvalue of 0.7549.

This means that .

We also have that

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

the answer is d on edge

step-by-step explanation: