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)

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