Question

If a variable $x$ takes values $0,1,2,\dots ,n$ with frequencies proportional to the binomial coefficients ${}^n{C}_0,{}^n{C}_1,{}^n{C}_2,\dots$ ${}^n{C}_n$, then var(X) is

• A. $\frac{n^2-1}{12}$
• B. $\frac{n}{2}$
• C. $\frac{n}{4}$
• D. None of these

Answer 1

Answer: (C)

We have
$\bar{X}=\frac{0^n{C}_0+1^n{C}_1+2^n{C}_2+\cdots +n^n{C}_n}{{}^n{C}_0+{}^n{C}_1+{}^n{C}_2+\cdots +{}^n{C}_n}$
$=\frac{\sum _{r=0}^n{r}^n{C}_r}{\sum _{r=0}^n{}^n{C}_r}$
$=\frac{1}{2^n}\sum _{r=1}^{n}r\frac{n}{r}{}^{n-1}{C}_{r-1}\left[\because \sum _{r=0}^n{}^n{C}_r=2^n;{}^n{C}_r=\frac{n}{r}{}^{n-1}{C}_{r-1}\right]$
$=\frac{n}{2^n}\sum _{r=1}^n{}^{n-1}{C}_{r-1}=\frac{n}{2^n}{2}^{n-1}=\frac{n}{2}\left[\because \sum _{r=1}^n{}^{n-1}{C}_{r-1}=2^{n-1}\right]$
and $\frac{1}{N}\sum f_i{x}_i^2=\frac{1}{2^n}\sum r^{2n}{C}_r=\frac{1}{2^n}\sum _{r=0}^n[r(r-1)+r{]}^n{C}_r$
$=\frac{1}{2^n}\left\{\sum _{r=0}^{n}r-(r-1){}^n{C}_r+\sum _{r=0}^{n}r{}^n{C}_r\right\}$
$=\frac{1}{2^n}\left\{\sum _{r=2}^{n}r(r-1)\frac{n}{r}\frac{n-1}{r-1}{}^{n-2}{C}_{r-2}+\sum _{r=1}^{n}r\frac{n}{r}{}^{n-1}{C}_{r-1}\right\}$
$=\frac{1}{2^n}\left\{n(n-1)2^{n-2}+n{2}^{n-1}\right\}=\frac{n(n-1)}{4}+\frac{n}{2}$
$\therefore$ Var(X) $=\frac{1}{N}\sum f_i{x}_i^2-{\bar{X}}^2=\frac{n(n-1)}{4}+\frac{n}{2}-\frac{n^2}{4}=\frac{n}{4}$