Question

The difference between mean and variance of a binomial distribution is $1$ and the difference of their squares is $11$. Find the distribution.

Answer 1

Let the binomial distribution be ${(q+p)}^n$.
here, mean $=np$ and variance $=npq$
Now, $np-npq=1$
$np(1-q)=\mathrm{1.........}\left(i\right)$
and ${\left(np\right)}^2-{\left(npq\right)}^2=11$
${\left(np\right)}^2(1-q^2)=\mathrm{11.......}\left(ii\right)$
Dividing equation $\left(ii\right)$ by the square of equation $\left(i\right)$, we get
$\frac{{\left(np\right)}^2(1-q^2}{{\left(np\right)}^2{(1-q)}^2}=\frac{11}{1}$
$\frac{(1+q)(1-q)}{(1-q)(1-q)}=11\Rightarrow$ $1+q=11=11-11q$
$\Rightarrow$ $q=\frac{10}{12}=\frac{5}{6}$
$\therefore$ $p=1-q=1-\frac{5}{6}=\frac{1}{6}$
Putting the values of $p$ and $Q$ in $\left(i\right)$, we get
$n\frac{1}{6}(1-\frac{5}{6})=1$
$n=36$
Hence, the binomial distribution is ${(\frac{5}{6}+\frac{1}{6})}^{36}$.