Question

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

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

Answer 1

Answer: (A)

$\frac{n}{4}$

Here, $\mu 1^{\prime }=\frac{{\sum }_0^{n}r.{}^n{C}_r}{{\sum }_0^n{}^n{C}_r}=\frac{{\sum }_0^{n}r.\frac{n}{r}{.}^{n-1}{C}_{r-1}}{{\sum }_0^n{}^n{C}_r}$

$=$$\frac{{\sum }_{r=1}^n{}^{n-1}{C}_{r-1}}{{\sum }_{r=0}^n{}^n{C}_r}$ $=\frac{n{.2}^{n-1}}{2^n}=\frac{n}{2}$

and $\mu 2^{\prime }=\frac{{\sum }_0^n{r}^2.{}^n{C}_r}{{\sum }_{r=0}^n{}^n{C}_r}$

$=\frac{1}{2^n}{\sum }_0^n{\{\left(r\right(r-1\left)\right)+r\}}^n{C}_r$ $=\frac{1}{2^n}\sum _0^{n}r{(r-1)}^n{C}_r+\frac{n}{2}$ $=\frac{1}{2^n}\sum _0^{n}r(r-1)\frac{n(n-1)}{r(r-1)}{}^{n-2}{C}_{r-2}+\frac{n}{2}$ $=\frac{n(n-1)}{2^n}{.2}^{n-2}+\frac{n}{2}$ $=\frac{n(n-1)}{4}+\frac{n}{2}$

$\therefore$ Variance ${\sigma }^2=\mu 2^{\prime }-{\left(\mu 1^{\prime }\right)}^2$ $=\frac{n(n-1)}{4}+\frac{n}{2}-{\left(\frac{n}{2}\right)}^2=\frac{n}{4}$