Question

If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

Answer 1

Given: Mean = 9 and variance = 6

$$\begin{gathered} \therefore \mathit{}np\mathit{}=9...\left(1\right) \\ \mathit{}npq=6...\left(2\right) \\ \mathrm{Dividing}\mathrm{eq}\left(2\right)\mathrm{by}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathit{}q\mathit{}=\frac{2}{3}\mathrm{and}\mathit{}p=1-q=\frac{1}{3} \\ \mathrm{As}\mathit{}np=9,\mathrm{substituting}\text{the}\mathrm{value}\mathrm{of}p,\mathrm{we}\text{get} \\ \frac{n}{3}=9\mathrm{or}\mathit{}n=27 \\ P(X=r)={}^{27}C_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{27-r},r=0,1,2....27 \end{gathered}$$