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: 5.82d

Answer

$P(x\lt950)=.0573$

Work Step by Step

$P(x\lt950)=P(x\leq949)=P(z\leq\frac{(949+.5)-1000}{31.46})=P(z\leq-1.61)=P(z\geq1.61)=P(z\geq0)-P(0\leq z\leq1.61)=.5-.4463=.0573$
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