Question

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?

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