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
$np + npq = 15$
$\Rightarrow np\left( 1 + q\right) = 15\quad\ldots\left(i\right)$
and $\left(np \right)^{2} npq = 117$
$\Rightarrow n^{2}p^{2}\left(1+q^{2} \right)= 117\quad\ldots\left(ii\right)$
Dividing the square of $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{n^{2}p^{2}\left(1+p^{2}\right)}{n^{2}p^{2}\left(1+q^{2}\right)} = \frac{15\times15}{117}$
$\Rightarrow \frac{1+2q+q^{2}}{1+q^{2}} = \frac{25}{13}$
$\Rightarrow 6q^{2} - 13q + 6 = 0$
$\Rightarrow \left(3q - 2\right)\left(2q - 3\right) = 0$
$\Rightarrow q = \frac{2}{3}$, $\frac{3}{2}$ but $q \ne \frac{3}{2}\quad\left(\because 0 \le q \le 1\right)$
$\therefore q = \frac{2}{3}$
$\Rightarrow p = 1 - q = 1 -\frac{2}{3} = \frac{1}{3}$
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.
$\left(\frac{2}{3}+\frac{1}{3}\right)^{27}$