In a binomial distribution B(n,p), the sum and the product of the mean and the variance ar…

Question

In a binomial distribution $B(n,p)$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $6(n+p-q)$ is equal to :

Accepted Answer

$$
\begin{aligned}
& \text { Given } \\
& \mathrm{np}+\mathrm{npq}=5 \\
& \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\
& \text { and (np) (npq) }=6 \\
& \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\
& (\mathrm{i})^2 \div(\mathrm{ii}) \\
& \frac{(1+q)^2}{9}=\frac{25}{6} \\
& \Rightarrow 6 \mathrm{q}^2-13 \mathrm{q}+6=0 \\
& \Rightarrow \mathrm{q}=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\
& \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\
& \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\
& \Rightarrow \mathrm{n}=9 \\
& 6(\mathrm{n}+\mathrm{p}-\mathrm{q})=52
\end{aligned}
$$

Solutions · 1