Question

If the sum and the product of the mean and variance of a binomial distribution are $24$ and $128$ respectively, then find the distribution.

• A. $\left(\frac{1}{4}+\frac{3}{4}\right)^{32}$
• B. $\left(\frac{1}{2}+\frac{1}{2}\right)^{30}$
• C. $\left(\frac{1}{2}+\frac{1}{2}\right)^{32}$
• D. $\left(\frac{1}{4}+\frac{3}{4}\right)^{30}$

Answer 1

Answer: (C)

According to question, we have
$np + npq = 24$
$\Rightarrow np\left( 1 + q\right) = 24\quad\ldots\left(i\right)$
and $np \cdot npq = 128$
$\Rightarrow n^{2}p^{2}q = 128\quad\ldots\left(ii\right)$
Dividing the square of $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{n^{2}p^{2}\left(1+p^{2}\right)}{n^{2}p^{2}q} = \frac{24\times24}{128}$
$\Rightarrow \frac{1+2q+q^{2}}{q} = \frac{9}{2}$
$\Rightarrow 2 + 4q + 2q^{2} = 9q$
$\Rightarrow 2q^{2}- 5 q + 2 = 0$
$\Rightarrow \left(2q - 1\right) \left(q - 2\right) = 0$
$\Rightarrow q = \frac{1}{2}$, $2$ but $q \ne 2\left(\because 0 \le q \le 1\right)$
$\Rightarrow q = \frac{1}{2}$
$\therefore p = 1 - q = 1 -\frac{1}{2} = \frac{1}{2}$
From $\left(i\right)$, we get
$n\cdot\frac{1}{2}\left(1+\frac{1}{2}\right) = 24$
$\Rightarrow n\cdot\frac{3}{4} = 24$
$\Rightarrow n = 32$
Hence, the binomial distribution is $\left(q+p\right)^{n}$ i.e.
$\left(\frac{1}{2}+\frac{1}{2}\right)^{32}$