In response to the increasing weight of airline passengers, the Federal Aviation

Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 190 pounds in the summer, including clothing and carry-on baggage. Passengers vary, though. A reasonable standard deviation is 35 pounds. Suppose a commuter plane carries 19 passengers. (a) What is the approximate probability that the total weight of the passengers exceeds 4000 pounds? (b) What is the approximate probability that the average passenger weight exceeds 215 pounds?

Explanation / Answer

Let X = weight of a passenger. Then, we assume X ~ N(µ, 2),

Given that it can be assumed that passengers’ average weight is 190 pounds and standard deviation is 35 pounds, µ = 190 and = 35.

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel ……………………………..(5)

Part (a)

Approximate probability that total weight of 19 passengers exceeds 4000 pounds

= P(average weight exceeds 210.526 pounds)

= P(Xbar > 210.526) = P[Z > {(19)(210.526 - 190)/35}] [vide (2) and (4) of Back-up Theory]

= P(Z > 2.271) = 1 – P(Z < 2.271 ) = 1 – 0.9884 = 0.0116 ANSWER

Part (b)

Approximate probability the average weight of 19 passengers exceeds 215 pounds

= P(Xbar > 215) = P[Z > {(19)(215 - 190)/35}] [vide (3) and (4) of Back-up Theory]

= P(Z > 2.766) = 1 – P(Z < 2.766 ) = 1 – 0.9972 = 0.0028 ANSWER

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