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

Explanation

Given: Mean` =9 ` and variance` =6`

`therefore {np}=9 ldots(1) `

`{npq}=6 ldots(2)`

Dividing `eq (2)` by eq` (1)`, we get `q=frac{2}{3} and p=1-q=frac{1}{3}`

As `n p=9, ` substituting the value of `p`, we get

`frac{{n}}{3}=9 text { or } {n}=27 `

`{P}({X}={r})=^{27} C_{r}(frac{1}{3})^{r}(frac{2}{3})^{27-r}, r=0,1,2 ldots 27`