Question

The mean and variance of a binomial distribution are $12$ and $3$ respectively. Find the probability distribution.

Answer 1

Let the parameters of the distribution be $n,p$.

Give the mean is $12\Rightarrow np=12$

Variance is $3\Rightarrow npq=3$

So, $q=\frac{3}{12}=\frac{1}{4}$

$P=1-q\Rightarrow P=1-\frac{1}{4}=\frac{3}{4}$

$np=12\Rightarrow \frac{3}{4}n=12\Rightarrow n=16$

$\therefore$ The binomial distribution is ,

$P(x=r)={}^{16}{C}_r{\left(\frac{3}{4}\right)}^r{\left(\frac{1}{4}\right)}^{16-r}$ for $r=0,1,2,....16$

$={}^{16}{C}_r.\frac{3^r}{4^{16}}.$ for $r=0,1,2,...,16$