Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

Answer:

There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.

Step-by-step explanation:

Problems dealing with normally distributed samples can be addressed using the z-score formula.

For a set with the mean and a standard deviation , the z-score for a measure X is calculated by

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.

For this case

Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus

What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?

That translates to

Thus

has a p-value of 0.7549.

Additionally, we have that

There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.