Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) in 2003 told airlines to assume that passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 22 passengers. What is the approximate probability that the total weight of the passengers exceeds 4500 pounds? Use the four-step process to guide your work.

Answer 1

Answer:

There is a 33.72% probability that the total weight of the passengers exceeds 4500 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:

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

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.

In this problem, we have that:

There are 22 passengers. Passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. This means that $\mu = 22*190 = 4180, \sigma = 22*35 = 770$.

What is the approximate probability that the total weight of the passengers exceeds 4500 pounds?

This probability is 1 subtracted by the pvalue of Z when $X = 4500$. So

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

$Z = \frac{4500 - 4180}{770}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

This means that there is a 1-0.6628 = 0.3372 = 33.72% probability that the total weight of the passengers exceeds 4500 pounds.