Chapter 11 Data Analysis and Statistics - Extension - Approximate Binomial Distributions and Test Hypotheses - Practice - Page 765: 12

$0.4589$

$\overline{x}=np=0.04\cdot460=18.4$ $\sigma=\sqrt{np(1-p)}=\sqrt{460(0.04)(0.96)}\approx4.2$ Thus $P(6\leq x\leq18)=\approx P(\frac{6-18.4}{4.2}\leq z\leq\frac{18-18.4}{4.2})\approx P(z\leq-0.1)-P(z\leq-3)=0.4602-0.0013=0.4589$