Question

If the variable takes the values $0,eb1,2,...n$ with frequencies proportional to binomial coeffcients $C(n,0),C(n,1),C(n,2),....C(n,n)$ respectively, then the variance of the distribution is

• A. $n$
• B. $\frac{\sqrt{n}}{2}$
• C. $\frac{n}{2}$
• D. $\frac{n}{4}$

Answer 1

The correct option is D $\frac{n}{4}$

$\overline{X}=\frac{0.{{}^{n}C}_0+1.{{}^{n}C}_1+2.{{}^{n}C}_2+.....+n.{{}^{n}C}_n}{{{}^{n}C}_0+{{}^{n}C}_1+{{}^{n}C}_2+....+{{}^{n}C}_n}[\because {\sum }^{}{{}^{n}C}_r=2^n,{{}^{n}C}_r=\frac{n}{r}{{}^{n-1}C}_{r-1}]$
$=\frac{1}{2^n}n.2^{n-1}=\frac{n}{2}$
also $\frac{1}{N}\sum f_i{x}_i^2=\frac{1}{2^n}{\sum }_{r=0}^n{r}^2{{}^{n}C}_r=\frac{1}{2^n}\{{\sum }_{r=0}^n\frac{r(r-1)n(n-1)}{r(r-1)}{{}^{n-2}C}_{r-2}+n.2^{n-1}\}$
$=\frac{1}{2^n}\left[n\right(n-1)2^{n-2}+n.2^{n-1}]=\frac{n(n-1)}{4}+\frac{n}{2}=\frac{n}{2}[\frac{n-1}{2}+1]=\frac{n}{2}\left(\frac{n+1}{2}\right)$
$var\left(x\right)=\frac{1}{N}\sum f_i{x}_i^2-{\overline{X}}^2=\frac{n}{2}\left(\frac{n+1}{2}\right)-\frac{n^2}{4}=\frac{n}{4}$