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

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{50 - 47.88}{3.09}$

$Z = 0.69$

$Z = 0.69$ 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.