Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 11 Data Analysis and Statistics - 11.3 Use Normal Distributions - 11.3 Exercises - Skill Practice - Page 760: 20

Answer

$0.9893$

Work Step by Step

$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$ Hence here the z-score is: $\frac{80-64}{7}\approx 2.3$ Then, using the table: $P(x\leq 80)\approx P(z\leq 2.3)=0.9893$
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