Question

The sum and product of mean and variance of a Binomial distribution are $24$ and $128$ respectively then the value of $n$ is

• A. $16$
• B. $32$
• C. $24$
• D. None of these

Answer 1

The correct option is B $32$

Given $np+npq=24$ and $np\left(npq\right)=128$
$np(1+q)=24$ and $n^2{p}^{2}q=128$
$\therefore \frac{n^2{p}^2(1+q{)}^2}{n^2{p}^{2}q}=\frac{24\times 24}{128}$
$\Rightarrow 2q^2-5q+2=0$
$\therefore p=q=1/2$ now $np(1+q)=24$
$\therefore n\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)=24$$\Rightarrow n=\frac{24\times 4}{3}=32$