Normal Approximation to the Binomial

$ P(X=4)=b(4;15,0.4)=0.1268$

$\displaystyle z_1=\frac{3.5-6}{1.897}=-1.32,~and~z_2=\frac{4.5-6}{1.897}=-0.79
$

$\displaystyle P(X=4) \approx P(3.5 < X <4.5)=P(-1.32 <Z < -0.79)
$

$\displaystyle =P(Z<-0.79)-P(Z<-1.32)
$

$\displaystyle =0.2148-0.0934=0.1214
$

Figure 2: Normal approximation of $ b(x; 15,0.4)$ and $ \sum_{x=7}^9 b(x; 15,0.4)$.
\includegraphics[scale=1]{figures/06-22}

Cem Ozdogan 2010-05-03