Answer
$0.9332$
Work Step by Step
$\overline{x}=np=0.04\cdot460=18.4$
$\sigma=\sqrt{np(1-p)}=\sqrt{460(0.04)(0.96)}\approx4.2$
Thus $P(x\geq12)=1-P(x\leq12)\approx 1-P(z\leq\frac{12-18.4}{4.2})\approx 1-P(z\leq-1.5)=1-0.0668=0.9332$
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: 11
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