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