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

• A. 50
• B. 53
• C. 52
• D. 51

Answer 1

Answer: (C)

$ n p+n p q=5, n p \cdot n p q=6$
$ n p(1+q)=5, n^2 p^2 q=6$
$ n^2 p^2(1+q)^2=25, n^2 p^2 q=6 $
$ \frac{6}{q}(1+q)^2=25 $
$6 q^2+12 q+6=25 q$
$ 6 q^2-13 q+6=0 $
$ 6 q^2-9 q-4 q+6=0 $
$(3 q-2)(2 q-3)=0 $
$ q=\frac{2}{3}, \frac{3}{2}, q=\frac{2}{3} \text { is accepted }$
$ p=\frac{1}{3} \Rightarrow n \cdot \frac{1}{3}+n \cdot \frac{1}{3} \cdot \frac{2}{3}=5 $
$ \frac{3 n+2 n}{9}=5$
$ n=9$
So $6(n+p-q)=6\left(9+\frac{1}{3}-\frac{2}{3}\right)=52$