In response to the increasing weight of airline passengers t

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 190.1 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 36.5 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 (±0.0001) that the total weight of the passengers exceeds 4572 pounds?

Solution

This is like asking the probability of the mean exceeding 4572/22 = 207.8181818.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    207.8181818      
u = mean =    190.1      
n = sample size =    22      
s = standard deviation =    36.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.276866828      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.276866828   ) =    0.01139709 [ANSWER]