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: 12



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 $P(6\leq x\leq18)=\approx P(\frac{6-18.4}{4.2}\leq z\leq\frac{18-18.4}{4.2})\approx P(z\leq-0.1)-P(z\leq-3)=0.4602-0.0013=0.4589$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.