Binomial Distribution

In a Bernoulli trial the outcome can be either "success" or "failure". If $$p$$ denotes the probability of success then the probability of failure is $$1-p$$. When doing $$n$$ trials the probability of getting exactly $$k$$ successes is given by

$P\;(X=x)=\binom{n}{k}p^k(1-p)^{n-k}$

where $$X$$ is a binomially distributed random variable. We use the notation: $$X\sim B(n,p)$$.

Mean and variance for binomial distribution

The mean of one trial is given by $$\mu = 1\cdot p+0\cdot (1-p)=p$$. The mean of $$n$$ independent trials is hence $$\mu = np$$.

The variance of one trial is given by

$\sigma^2=(1-\mu)^2\cdot p + (0-\mu)^2\cdot (1-p)= (1-p)^2\cdot p + p^2\cdot (1-p)=p\cdot (1-p)$

The variance of $$n$$ independent trials is hence $$\sigma^2=np(1-p)$$.

$\mu = np \hspace{1cm} \sigma^2=np(1-p)$

Binomial distribution in GeoGebra

By using the tool "Probability Calculator" , you can find all probabilities for a binomial distribution.

Binomial and normal distribution

Using the mean $$\mu = np$$ and the variance $$\sigma^2=np(1-p)$$ of the binomial distribution as parameters in the function describing the normal pdf

$f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{\left( - \dfrac{(x-\mu)^2}{2\sigma^2}\right)}$

we get a continuous approximation of the binomial distribution. When $$n$$ is large and when $$p$$ is not too close to 0 or 1, the normal distribution is a very good approximation of the binomial distribution.

by Malin Christersson under a Creative Commons Attribution-Noncommercial-Share Alike 2.5 Sweden License

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