Answer
$0.1587$
Work Step by Step
$\overline{x}=np=0.09\cdot1221=109.89$
$\sigma=\sqrt{np(1-p)}=\sqrt{1221(0.09)(0.91)}\approx10$
Thus $P(x\leq100)\approx P(z\leq\frac{100-109.89}{10})\approx P(z\leq-1)=0.1587$
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 - Extension - Approximate Binomial Distributions and Test Hypotheses - Practice - Page 765: 14
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