## Essentials of Statistics (5th Edition)

Published by Pearson

# Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions: 15

0.4972.

#### Work Step by Step

$z_1=\frac{value-mean}{standard \ deviation}=\frac{110-100}{15}=0.67.$ $z_2=\frac{value-mean}{standard \ deviation}=\frac{90-100}{15}=-0.67.$ Using the table, the value belonging to 0.67: 0.7486, the value belonging to -0.67: 0.2514. 0.7486-0.2514=0.4972.

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