Question

If the variable takes values $0,1,2,3,...,n$ with frequencies, proportional to ${}^n{C}_0,{}^n{C}_1,{}^n{C}_2,...,{}^n{C}_n$ respectively, then variance is

• A. $\frac{n}{4}$
• B. $\frac{n}{3}$
• C. $\frac{2n}{5}$
• D. None of these

Answer 1

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

$x_i=0,1,2,...,n$
$f_i={}^n{C}_0,{}^n{C}_1,{}^n{C}_2,...,{}^n{C}_n$ respectively.
Variance $=\frac{\sum x_i^2{f}_i}{\sum f_i}-{\left(\frac{\sum x_i{f}_i}{\sum f_i}\right)}^2$
Now we can solve it one by one
$\left(\frac{\sum x_i{f}_i}{\sum f_i}\right)=\left(\frac{\sum _{r=0}^{n}r{}^n{C}_r}{\sum _{r=0}^n{}^n{C}_r}\right)=\frac{1}{2^n}\left(\sum _{r=0}^{n}r{}^n{C}_r\right)$
$=\frac{1}{2^n}\left(\sum _{r=0}^{n}r\frac{n}{r}{}^{n-1}{C}_{r-1}\right)=\frac{n}{2^n}\left(\sum _{r=0}^n{}^{n-1}{C}_{r-1}\right)$
$=\frac{n}{2^n}\left(2^{n-1}\right)=\frac{n}{2}$

Now,
$\left(\frac{\sum x_i^2{f}_i}{\sum f_i}\right)=\left(\frac{\sum _{r=0}^n{r}^2{(}^n{C}_r)}{\sum _{r=0}^n{}^n{C}_r}\right)=\frac{1}{2^n}\left(\sum _{r=0}^n{r}^2{(}^n{C}_r)\right)$
$=\frac{1}{2^n}\left(\sum _{r=0}^n(r^2+r-r){}^n{C}_r\right)$
$=\frac{1}{2^n}\left(\sum _{r=0}^n\left(r\right(r-1\left)\right){}^n{C}_r\right)+\frac{1}{2^n}\left(\sum _{r=0}^n\left(r\right){}^n{C}_r\right)$
$=\frac{n^2}{4}-\frac{n}{4}+\frac{n}{2}$

So, Variance $=\frac{\sum x_i^2{f}_i}{\sum f_i}-{\left(\frac{\sum x_i{f}_i}{\sum f_i}\right)}^2$
$=\frac{n^2}{4}-\frac{n}{4}+\frac{n}{2}-{\left(\frac{n}{2}\right)}^2=\frac{n}{4}$