Question

If the sum of the mean and variance of a binomial distribution is $15$ and the sum of their squares is $117$, then find the distribution.

• A. ${\left(\frac{2}{3}+\frac{1}{3}\right)}^{25}$
• B. ${\left(\frac{1}{2}+\frac{1}{2}\right)}^{25}$
• C. ${\left(\frac{1}{2}+\frac{1}{2}\right)}^{27}$
• D. ${\left(\frac{2}{3}+\frac{1}{3}\right)}^{27}$

Answer 1

Answer: (D)

According to question, we have
$np+npq=15$
$\Rightarrow np\left(1+q\right)=15\dots \left(i\right)$
and ${\left(np\right)}^{2}npq=117$
$\Rightarrow n^2{p}^2\left(1+q^2\right)=117\dots \left(ii\right)$
Dividing the square of $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{n^2{p}^2\left(1+p^2\right)}{n^2{p}^2\left(1+q^2\right)}=\frac{15\times 15}{117}$
$\Rightarrow \frac{1+2q+q^2}{1+q^2}=\frac{25}{13}$
$\Rightarrow 6q^2-13q+6=0$
$\Rightarrow \left(3q-2\right)\left(2q-3\right)=0$
$\Rightarrow q=\frac{2}{3}$, $\frac{3}{2}$ but $q\ne \frac{3}{2}\left(\because 0\le q\le 1\right)$
$\therefore q=\frac{2}{3}$
$\Rightarrow p=1-q=1-\frac{2}{3}=\frac{1}{3}$
From $\left(i\right)$, we get
$n\cdot \frac{1}{3}\left(1+\frac{1}{2}\right)=15$
$\Rightarrow n\cdot \frac{1}{3}.\frac{5}{3}=15$
$\Rightarrow n=27$
Hence, the binomial distribution is ${\left(q+p\right)}^n$ i.e.
${\left(\frac{2}{3}+\frac{1}{3}\right)}^{27}$