Normal Approximation to Binomial Distribution
Let $X$ be a binomially distributed random variable with number of trials $n$ and probability of success $p$.
The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1p)$.
The general rule of thumb to use normal approximation to binomial distribution is that the sample size $n$ is sufficiently large if $np \geq 5$ and $n(1p)\geq 5$.
For sufficiently large $n$, $X\sim N(\mu, \sigma^2)$. That is $Z=\frac{X\mu}{\sigma}=\frac{Xnp}{\sqrt{np(1p)}} \sim N(0,1)$.
Normal Approx. to Binomial Distribution  

No. of Trials ($n$)  
Probability of Success ($p$)  
Select an Option  
Enter the value(s) : 


Results  
Mean ($\mu=np$)  
Standard deviation ($\sqrt{np(1p)}$)  
Required Probability : 
Formula
 $P(X=A)=P(A0.5<X<A+0.5)$
 $P(X<A)=P(X<A0.5)$
 $P(X\leq A)=P(X<A+0.5)$
 $P(A< X\leq B)=P(A0.5<X<B+0.5)$
 $P(A\leq X< B)=P(A0.5<X<B0.5)$
 $P(A\leq X\leq B)=P(A0.5<X<B+0.5)$