Q. If the variable takes the values 0, 1, 2, …., $n$ with frequencies proportional to the binomial coefficients $C(n,0),C(n,1),C(n,2),\ldots..,C(n,n)$ respectively, then the variance of the distribution is
A) $n$
B) $\frac{\sqrt{n}}{2}$
C) $\frac{n}{2}$
D) $\frac{n}{4}$
Explanation
(d)
Now,$\mu_{1}^{'} = \frac{\sum_{r = 0}^{n}{{r\ }^{n}C_{r}}}{\sum_{r = 0}^{n}{\ ^{n}C_{r}}} = \frac{n.2^{n - 1}}{2^{2}} = \frac{n}{2}$
$\mu_{2}^{'}$=$\frac{\sum_{r = 0}^{n}{{r^{2}\ }^{n}C_{r}}}{\sum_{r = 0}^{n}{\ ^{n}C_{r}}} = \frac{n(n - 1)}{2^{2}}.2^{n - 2} + \frac{n}{2}$
=$\frac{n(n - 1)}{4}\frac{n}{2}$
$\therefore Variance,$
$\mu_{2}^{'} = \left( \mu_{2}^{'} \right) - \left( \mu_{2}^{'} \right)^{2}$=$\frac{n(n - 1)}{4} + \frac{n}{2} - \frac{n^{2}}{4} = \frac{n}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.