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

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

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.

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

This means that

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

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

x = 3 ± √(5+y)