Krzys’ Ostaszewski: http://www.math.ilstu.edu/krzysio/
Author of the “Been There Done That!” manual for Course P/1
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Exercise for March 24, 2007
A new super-size passenger airplane has 400 seats in coach class and 100 seats in the
business class, with these being the only two classes available. The weight of the luggage
(all luggage pieces combined) of an individual coach passenger follows a continuous
probability distribution with mean of 100 pounds and the standard deviation of 10
pounds, while the weight of the luggage (all luggage pieces combined) of a business class
passenger follows a continuous probability distribution with mean of 80 pounds and the
standard deviation of 5 pounds. The weights of individual passengers luggages are
independent. You are given that the 95-th percentile of the standard normal distribution is
1.644853627. Approximate the smallest possible number w such that the probability that
the total weight of the entire luggage on the plane is less than w with probability 0.95,
assuming that the airplane is full.
A. 48339
B. 52500
C. 55401
D. 544531
E. 33356286
Solution.
Let us write X for the total weight of all of coach passengers luggage. Then X is the sum
of weights of individual passengers luggages, and since these are independent and
identically distributed, and there are 400 coach passengers, X is approximately normal.
We also know that E ( X ) = 400 !100 = 40000 and Var ( X ) = 400 !100 = 40000. Let us
also write Y for the total weight of all of business passengers luggage. Then Y is the sum
of weights of individual passengers luggages, and since these are independent and
identically distributed, and there are 100 business passengers, Y is approximately normal.
We know that E (Y ) = 100 ! 80 = 8000 and Var (Y ) = 100 ! 25 = 2500. The total weight of
all luggage is the sum of X and Y. Furthermore, X and Y are independent, and since a sum
of two independent normal random variables is normal, X + Y is approximately normal,
with
E ( X + Y ) = 40000 + 8000 = 48000
and
Var ( X + Y ) = 40000 + 2500 = 42500.
If we write ! for the CDF of the standard normal distibution, then from the table given
for Course P, ! (1.644 ) = 0.9495, while ! (1.645 ) = 0.9505. Using linear interpolation,
we approximate the 95-th percentile of the standard normal distribution as 1.6445. The
95-th percentile of X + Y is therefore approximately equal to
48000 + 1.6445 ! 42500 " 48339.
Answer A.
© Copyright 2007 by Krzysztof Ostaszewski.
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