Question

Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 46 pounds and standard deviation 3.7 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. (Round your answer to two decimal places.)

Answer 1

Answer:

85.99% of airline passengers incur this fee.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

Mean 46 pounds and standard deviation 3.7 pounds.

This means that $\mu = 46, \sigma = 3.7$

Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

As a proportion, this is 1 subtracted by the pvalue of Z when X = 50. So

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

$Z = \frac{50 - 46}{3.7}$

$Z = 1.08$

$Z = 1.08$ has a pvalue of 0.8599

0.8599*100% = 85.99%

85.99% of airline passengers incur this fee.