Answer
$0.8849$
Work Step by Step
$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$
Hence here the z-score: $\frac{90-73}{14.1}\approx1.2$
Then, using the table: $P(x\leq90)\approx P(z\leq1.2)=0.8849$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
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 - Guided Practice for Example 3 - Page 759: 8
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