Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions - Basic Skills and Concepts - Page 268: 24

Answer

a) $0.01$ % b) $0.01$ % c) No d) $70.1$ inches

Work Step by Step

a) We find the $z$-score: $\frac{51.6−69.5}{2.4}=-7.46$ This is less than $−3.5$, so only about $0.01$ % percent of men fit. b) We find the $z$-score: $\frac{51.6−63.8}{2.4}=-4.69$ This is less than $−3.5$, so only about $0.01$ % percent of women fit. c) It does not, because most of the passengers cannot enter the aircraft without bending. d) We know that the $z$-score is $0.25$. Hence: $=(0.25)(2.4)+69.5=70.1$ inches
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