Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions - Basic Skills and Concepts: 16

Answer

0.1596.

Work Step by Step

$z_1=\frac{value-mean}{standard \ deviation}=\frac{120-100}{15}=1.333.$ $z_2=\frac{value-mean}{standard \ deviation}=\frac{110-100}{15}=0.667.$ Using the table, the value belonging to 1.333: 0.9082, the value belonging to 0.667: 0.7486. 0.9082-0.7486=0.1596.
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