Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 11 - Probability and Calculus - Chapter Review - Review Exercises - Page 602: 35

Answer

34.31%
1568133340

Work Step by Step

From the normal curve table, the area to the left of $z=-1.17$ is $-0.1210$ From the normal curve table, the area to the left of $z=-0.09$ is $-0.4641$ Therefore, the area to the right is $P(-1.17\lt Z \lt -0.09)=-0.4641-(-0.1210)=-0.3431$ The proportion is $-0.3431$ or $34.31\%$
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