In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 189.7 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 41.4 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 27 passengers.

What is the approximate probability (±0.0001) that the total weight of the passengers exceeds 5511 pounds?

Question

Amina Carrillo

Mathematics

9 July, 00:34

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 189.7 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 41.4 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 27 passengers.

What is the approximate probability (±0.0001) that the total weight of the passengers exceeds 5511 pounds?

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Answers (1)

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Dominik Adams · Answer 1

The only way I can think of to solve this problem is to assume normal distribution.

Since the total weight excess 5511, hence the weight per passenger is at least:

x = 5511 / 27 = 204.1 pounds

Solve for the z score:

z = (x - u) / s

z = (204.1 - 189.7) / 41.4

z = 0.35

From the standard probability tables, the P value using right tailed test is:

P = 0.3632