Answer
$p(x\gt120)=.0329$
Work Step by Step
$z_0=\frac{x-\mu}{\sigma}=\frac{120-105.3}{8.0}=1.8375$
According to table IV, page 773:
$p(0\lt z\lt 1.84)=.0948$
$p(x\gt120)=p(z\gt1.8375)=p(z\gt0)-p(0\lt z\lt1.84)=.5-.4671=.0329$
Statistics (12th Edition)
by McClave, James T.; Sincich, Terry T.
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: 5.38a
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