Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions: 16

Answer

0.1613.

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: 0.9099, the value belonging to 0.667: 0.7486. Hence: 0.9099-0.7486=0.1613.
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