Question

Suppose that the weights of airline passenger bags are normally distributed with a mean of 49.02 pounds and a standard deviation of 3.83 pounds. a) What is the probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds

Answer 1

0.601 = 60.1% probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds

Explanation:

Normal Probability Distribution:

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

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

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

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.

Mean of 49.02 pounds and a standard deviation of 3.83 pounds.

This means that
`\mu = 49.02, \sigma = 3.83`

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

This is the p-value of Z when X = 50. So

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

`Z = (50 - 49.02)/(3.83)`

`Z = 0.256`

`Z = 0.256` has a p-value of 0.601.

0.601 = 60.1% probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds