Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321755936
ISBN 13: 978-0-32175-593-3

Chapter 5 - Continuous Random Variables - Exercises 5.73 - 5.92 - Applying the Concepts - Basic - Page 256: 5.81d

Answer

$P(x\geq20)=.1762$

Work Step by Step

$P(x\geq20)=1-P(x\leq19)=1-P(z\leq\frac{(19+.5)-16.25}{3.49})=1-P(z\leq.93)=P(z\geq0)-P(0\leq z\leq.93)=.5-.3238=.1762$
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