Statistics (12th Edition)

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

Chapter 5 - Continuous Random Variables - Exercises 5.20 - 5.52 - Learning the Mechanics - Page 241: 5.26g

Answer

$p(z\geq-2.33)=.9901$

Work Step by Step

According to Table IV, page 773: $p(0\lt z\leq2.33)=.4901$ $p(z\geq-2.33)=p(-2.33\leq z\lt0)+p(z\geq0)=p(0\lt z\leq2.33)+p(z\geq0)=.4901+.5=.9901$
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