Question

If the variable takes the values $0,1,2,...,n$ with frequencies proportional to the binomial coefficients $C\left(n,0\right),C\left(n,1\right),C\left(n,2\right),...,C\left(n,n\right)$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

$\frac{n}{4}$

Explanation for the correct option :

Step 1. Find the total number of observations.

The frequencies of the variable is proportional to the binomial coefficients $C\left(n,0\right),C\left(n,1\right),C\left(n,2\right),...,C\left(n,n\right)$.

So the total number of observations is:

$\begin{array}{rcl}N& =& {}^{n}C_0+{}^{n}C_1+{}^{n}C_2+...+{}^{n}C_n\\ & =& \sum _{i=0}^n{}^{n}C_i\left[\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}{\mathbf{2}}^{\mathbf{n}}\right]\\ & =& 2^n\end{array}$

Step 2. Find the mean.

The mean of variables $0,1,2,...,n$ with frequencies proportional to the binomial coefficients $C\left(n,0\right),C\left(n,1\right),C\left(n,2\right),...,C\left(n,n\right)$ is given as:

$\begin{array}{rcl}\overline{X}& =& \frac{0{}^{n}C_0+1{}^{n}C_1+2{}^{n}C_2+...+n{}^{n}C_n}{N}\\ & =& \frac{{\displaystyle \sum _{i=0}^n}i{}^{n}C_i}{2^n}\mathbf{}\mathbf{[}\mathbf{N}\mathbf{=}{\mathbf{2}}^{\mathbf{n}}\mathbf{]}\\ & =& \frac{\sum _{i=1}^{n}i·{\displaystyle \frac{n}{i}}{}^{n-1}C_{i-1}}{2^n}\mathbf{[}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{r}}{}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{]}\\ & =& \frac{n\sum _{i=1}^n{}^{n-1}C_{i-1}}{2^n}\\ & =& \frac{n\times 2^{n-1}}{2^n}\\ & =& n\times 2^{n-1-n}\\ & =& n\times 2^{-1}\\ & =& \frac{n}{2}\end{array}$

Step 3. Find the value of $\frac{1}{N}\sum _{i=0}^n{f}_i{x_i}^2$.

Here the variable are $0,1,2,...,n$ and its frequency us given as $C\left(n,0\right),C\left(n,1\right),C\left(n,2\right),...,C\left(n,n\right)$ so the term can be evaluated as:

$\begin{array}{rcl}\frac{1}{N}\sum _{i=0}^n{f}_i{x_i}^2& =& \frac{1}{2^n}\sum _{i=0}^n{}^{n}C_i{\times i}^2\\ & =& \frac{1}{2^n}\sum _{i=0}^n\left[i(i-1)+i\right]{}^{n}C_i\\ & =& \frac{1}{2^n}\left[\sum _{i=0}^{n}i(i-1){}^{n}C_i+\sum _{i=0}^{n}i{}^{n}C_i\right]\\ & =& \frac{1}{2^n}\left[\sum _{i=2}^{n}i(i-1)\times \frac{n}{i}\times \frac{\left(n-1\right)}{\left(i-1\right)}{}^{n-2}C_{i-2}+\sum _{i=1}^{n}i\times \frac{n}{i}{}^{n-1}C_{i-1}\right]\mathbf{[}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{r}}{}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{]}\\ & =& \frac{1}{2^n}\left[n(n-1)\sum _{i=2}^n{}^{n-2}C_{i-2}+n\sum _{i=1}^n{}^{n-1}C_{i-1}\right]\\ & =& \frac{1}{2^n}\left[n(n-1)\times 2^{n-2}+n\times 2^{n-1}\right]\mathbf{}\mathbf{}\mathbf{[}\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}{\mathbf{2}}^{\mathbf{n}}\mathbf{]}\\ & =& n(n-1)\times 2^{n-2-n}+n\times 2^{n-1-n}\\ & =& n(n-1)\times 2^{-2}+n\times 2^{-1}\\ & =& \frac{n(n-1)}{4}+\frac{n}{2}\end{array}$

Step 4. Find the variance of the distribution.

The variance of the distribution is given as:

$\mathrm{Variance}=\begin{array}{l}\frac{1}{N}\end{array}\begin{array}{l}\sum _{i=0}^n\end{array}f$

Substitute $\begin{array}{rcl}\frac{1}{N}\sum _{i=0}^n{f}_i{x_i}^2& =& \frac{n(n-1)}{4}+\frac{n}{2}\end{array}$ and $\overline{X}=\frac{n}{2}$.

$\begin{array}{rcl}\mathrm{Variance}& =& \frac{n(n-1)}{4}+\frac{n}{2}-{\left(\frac{n}{2}\right)}^2\\ & =& \frac{n(n-1)+2n-n^2}{4}\\ & =& \frac{n^2-n+2n-n^2}{4}\\ & =& \frac{n}{4}\end{array}$

Hence, the correct option is D.