Question

The sum and product of mean and variance of a binomial distribution are are $24$ and $128$ respectively. The binomial distribution is:

• A. ${(\frac{1}{2}+\frac{1}{2})}^{32}$
• B. ${(\frac{3}{10}+\frac{7}{10})}^{32}$
• C. ${(\frac{1}{50}+\frac{49}{50})}^{32}$
• D. ${(\frac{1}{3}+\frac{2}{3})}^{32}$

Answer 1

The correct option is A ${(\frac{1}{2}+\frac{1}{2})}^{32}$

Let the binomial distribution be $(p+q{)}^n$
We have, mean $=np$, variance $=npq$
Given $np+npq=24$
$\Rightarrow np(1+q)=24$
$\Rightarrow n^2{p}^2(1+q{)}^2=576\cdots (1)$
and $np.npq=128\cdots \left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$, we get $\frac{(1+q{)}^2}{q}=\frac{9}{2}$
​$\Rightarrow 2q^2+4q+2=9q\Rightarrow 2q^2-5q+2=0$
​$\Rightarrow q=2$ (not possible) or $q=\frac{1}{2}$
$\Rightarrow p=\frac{1}{2}$
So, by $\left(1\right)$, we get $n=32$
So, the distribution is ${(\frac{1}{2}+\frac{1}{2})}^{32}$