Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) in 2003 told airlines to assume that passengers average 195 pounds in the winter, 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. (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

Answer 1

µ = 195 * 22 = 4290

σ = √ 22 * 35 = 164.16455

X ~ N ( µ = 4290 , σ = 164.16455 )
P ( X > 4500 ) = 1 - P ( X < 4500 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4500 - 4290 ) / 164.16455
Z = 1.2792
P ( ( X - µ ) / σ ) > ( 4500 - 4290 ) / 164.16455 )
P ( Z > 1.2792 )
P ( X > 4500 ) = 1 - P ( Z < 1.2792 )
P ( X > 4500 ) = 1 - 0.8996
P ( X > 4500 ) = 0.1004