Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 182 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 30 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 21 passengers. What is the approximate probability that the total weight of the passengers exceeds 4222 pounds? (Round your answer to four decimal places.)

Answer 1

Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

$W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}$

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by $\phi$ . The values of $\phi$ can be found in the attached file.

$P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018$

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.  

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