Algebra 2 (1st Edition)

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: 13



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 $P(x\geq140)=1-P(x\leq140)\approx1- P(z\leq\frac{130-109.89}{10})\approx 1-P(z\leq2)=1-0.9772=0.0228$
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