Chapter 11 Data Analysis and Statistics - Extension - Approximate Binomial Distributions and Test Hypotheses - Practice - Page 765: 15

$0.9759$

$\overline{x}=np=0.09\cdot1221=109.89$ $\sigma=\sqrt{np(1-p)}=\sqrt{1221(0.09)(0.91)}\approx10$ Thus $P(80\leq x\leq130)\approx P(z\leq\frac{130-109.89}{10})-P(z\leq\frac{80-109.89}{10})= P(z\leq2)-P(z\leq-3)=0.9772-0.0013=0.9759$