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 $\sigma^2=V(X)=np(1-p)$.

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 $n(1-p)\geq 5$.

For sufficiently large $n$, $X\sim N(\mu, \sigma^2)$. That is $Z=\frac{X-\mu}{\sigma}=\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1)$.

Normal Approx. to Binomial Distribution
No. of Trials ($n$)
Probability of Success ($p$)
Select an Option
Enter the value(s) :

Mean ($\mu=np$)
Standard deviation ($\sqrt{np(1-p)}$)
Required Probability :


  • $P(X=A)=P(A-0.5<X<A+0.5)$
  • $P(X<A)=P(X<A-0.5)$
  • $P(X\leq A)=P(X<A+0.5)$
  • $P(A< X\leq B)=P(A-0.5<X<B+0.5)$
  • $P(A\leq X< B)=P(A-0.5<X<B-0.5)$
  • $P(A\leq X\leq B)=P(A-0.5<X<B+0.5)$