Krzys’ Ostaszewski: http://www.krzysio.net
Author of the BTDT Manual (the “Been There Done That!” manual) for Course P/1
http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic)
Instructor for online P/1 seminar: http://smartURL.it/onlineactuary
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If you have questions about these exercises, please send them by e-mail to:
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Dr. Ostaszewski’s online exercise posted March 19, 2011
The average height of an adult American male is 176 cm, with a standard deviation of 6
cm, and for an American female the average is 163 cm, with a standard deviation of 5
cm. Assuming that the heights of each group are normally distributed, and that heights of
individual people are independent of each other, what is the probability that a randomly
selected American male is taller than a randomly selected American female?
A. 0.70
B. 0.75
C. 0.85
D. 0.90
E. 0.95
Solution.
Let Y be the random height of an American male, and X be the random height of an
American female. We are looking for the probability Pr Y > X . Note that because of
independence, Y − X is normally distributed. We have
E Y − X = E Y − E X = 176 − 163 = 13,
(
and
)
(
( )
)
(
)
( )
( )
( )
Var Y − X = Var Y + Var X = 62 + 52 = 61.
Therefore, if we write Z for a standard normal random variable, and Φ for the
cumulative distribution function of a standard normal random variable, we obtain
⎛ Y − X − 13 0 − 13 ⎞
⎛
13 ⎞
Pr Y > X = Pr Y − X > 0 = Pr ⎜
>
⎟ = Pr ⎜ Z > −
⎟=
⎝
61
61 ⎠
61 ⎠
⎝
(
)
(
(
)
)
⎛
⎛
⎛ 13 ⎞
⎛ 13 ⎞ ⎞
⎛ 13 ⎞
13 ⎞
= 1− Φ⎜ −
= 1 − ⎜1 − Φ ⎜
≈
= 1 − Pr ⎜ Z ≤ −
⎟ = Φ⎜
⎟
⎟
⎟
⎝
⎝
⎝ 61 ⎠ ⎠
⎝ 61 ⎟⎠
⎝
61 ⎠
61 ⎠
⎛ 13 ⎞
≈ Φ 1.66 ≈ 0.95.
≈ Φ⎜
⎝ 61 ⎟⎠
(
)
Answer E.
© Copyright 2011 by Krzysztof Ostaszewski.
All rights reserved. Reproduction in whole or in part without express written
permission from the author is strictly prohibited.