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
$np+npq=24$
$\Rightarrow np\left(1+q\right)=24\dots \left(i\right)$
and $np\cdot npq=128$
$\Rightarrow n^2{p}^{2}q=128\dots \left(ii\right)$
Dividing the square of $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{n^2{p}^2\left(1+p^2\right)}{n^2{p}^{2}q}=\frac{24\times 24}{128}$
$\Rightarrow \frac{1+2q+q^2}{q}=\frac{9}{2}$
$\Rightarrow 2+4q+2q^2=9q$
$\Rightarrow 2q^2-5q+2=0$
$\Rightarrow \left(2q-1\right)\left(q-2\right)=0$
$\Rightarrow q=\frac{1}{2}$, $2$ 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.
${\left(\frac{1}{2}+\frac{1}{2}\right)}^{32}$