If a variable x takes values 0, 1, 2, ....,n with frequencies proportional to binomial coefficients ${\mathrm{n}}_{{\mathrm{C}}_0}, {\mathrm{n}}_{{\mathrm{C}}_1}, {\mathrm{n}}_{{\mathrm{C}}_2},......, {\mathrm{n}}_{{\mathrm{C}}_{\mathrm{n}}},$ then the variance of x is

$\begin{array}{rcl} & & {\mathrm{x}}_{\mathrm{i}} 0 1 2 ....... \mathrm{n}\\ & & {\mathrm{f}}_{\mathrm{i}} {}^{\mathrm{n}}{\mathrm{c}}_0 {}^{\mathrm{n}}{\mathrm{c}}_1 {}^{\mathrm{n}}{\mathrm{c}}_2......{}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{n}}\\ \overline{\mathrm{x}}& =& \frac{\sum {\mathrm{f}}_{\mathrm{i}} {\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ & =& \frac{{\displaystyle \frac{\mathrm{n}}{2}} 2^{\mathrm{n}}}{2^{\mathrm{n}}}=\frac{\mathrm{n}}{2}\end{array}$

Variance $=\frac{1}{\mathrm{N}} \sum {\mathrm{f}}_{\mathrm{i}} {{\mathrm{x}}_{\mathrm{i}}}^2-{\left(\overline{\mathrm{x}}\right)}^2$

$\begin{array}{rcl}\sum {\mathrm{f}}_{\mathrm{i}} {{\mathrm{x}}_{\mathrm{i}}}^2& =& \sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{r}}^2 {}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \left(\mathrm{r}\left(\mathrm{r}-1\right){}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}+\mathrm{r}·{}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}\right)\\ & =& 2(\mathrm{n}-1) \sum _{\mathrm{r}=2}^{\mathrm{n}} {}^{\mathrm{n}-2}{\mathrm{c}}_{\mathrm{r}-2}+\mathrm{n} \sum _{\mathrm{r}=1}^{\mathrm{n}} {}^{\mathrm{n}-1}{\mathrm{c}}_{\mathrm{r}-1}\\ & =& \mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-1}+\mathrm{n}·2^{\mathrm{n}-1}\end{array}$

Variance

             $$\begin{gathered} =\frac{\sum {\mathrm{f}}_{\mathrm{i}} {{\mathrm{x}}_{\mathrm{i}}}^2}{\mathrm{N}}-{\left(\overline{\mathrm{x}}\right)}^2 \\ =\frac{{\displaystyle \frac{\mathrm{n}(\mathrm{n}-1)}{4}} 2^{\mathrm{n}}+\mathrm{n} {\displaystyle \frac{2^{\mathrm{n}}}{2}}}{2^{\mathrm{n}}}-\frac{{\mathrm{n}}^2}{4} \\ =\frac{\mathrm{n} (\mathrm{n}+1)}{4}-\frac{{\mathrm{n}}^2}{4} \\ =\frac{\mathrm{n}}{4} \end{gathered}$$