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 242: 5.31a

Answer

$p(10\leq x\leq12)=.3830$

Work Step by Step

$z=\frac{x-\mu}{\sigma}=\frac{10-11}{2}=-.5$ $z=\frac{x-\mu}{\sigma}=\frac{12-11}{2}=.5$ $p(10\leq x\leq12)=p(-.5\leq z\leq.5)=p(-.5\leq z\leq0)+p(0\lt z\leq.5)=2p(0\lt z\leq.5)=2(.1915)=.3830$ According to Table IV, page 773: $p(0\lt z\leq.5)=.1915$
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