Question

The sum and product of mean and variance of a binomial distribution 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 $np+npq=24$
$np(1+q)=24$
$n^2{p}^2(1+q{)}^2=576$ .....(1)
and $np.npq=128$ ......(2)
Dividing (1) by (2), 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) , $\frac{1}{2}$
$\Rightarrow p=\frac{1}{2}$
So, by (1), we get
$n=32$
So, the distribution is $(\frac{1}{2}+\frac{1}{2}{)}^n$