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 - Applying the Concepts - Basic - Page 242: 5.39b

Answer

$p(48\lt x\lt52)=.2434$

Work Step by Step

$z=\frac{x-\mu}{\sigma}=\frac{48-50}{6.4}=-.31$ $z=\frac{x-\mu}{\sigma}=\frac{52-50}{6.4}=.31$ According to table IV, page 773: $p(0\lt z\lt .31)=.1217$ $p(48\lt x\lt52)=p(-.31\lt z\lt.31)=2p(0\lt z\lt .31)=2(.1217)=.2434$
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