If a variable X takes values 0, 1, 2,..., n with frequencies ^nC_…

Question

If a variable $X$ takes values $0, 1, 2,..., n$ with frequencies $^nC_0, ^nC_1, ^nC_2 , ... , ^nC_n$, then write variance $X$.

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Answer

$\overline{\text{x}}=\frac{\sum\limits_{\text{i}=0}^{\text{n}}\text{x}_\text{i}\text{f}_\text{i}}{\sum\limits_{\text{i}=0}^{\text{n}}\text{f}_\text{i}}=\frac{0\times^\text{n}\text{C}_\text{o}+1\times^\text{n}\text{C}_1+...+\text{n}\times^\text{n}\text{c}_\text{n}}{^\text{n}\text{C}_\text{o}+^\text{n}\text{C}_1+.....+^\text{n}\text{c}_\text{n}}$
$\Rightarrow\overline{\text{x}}=\frac{\text{n}\times2^{\text{n}-1}}{\frac{2^\text{n}}{\text{n}+1}}$
$=\frac{\text{n}(\text{n}+1)}{2}$
$\therefore\text{Var}(\text{X})=\sigma^2$
$=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=0}\big(\text{x}_\text{i}-\overline{\text{x}}\big)^2$
$=\frac{1}{\text{n}}[(0+1+2+....+\text{n})-\text{n}\overline{\text{x}}]^2$
$\Rightarrow\sigma^2=\frac{1}{\text{n}}\Big[\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}\times\text{n}(\text{n}+1)}{2}\Big]^2$
$=\frac{1}{\text{n}}\Big[\frac{\text{n}(\text{n}+1)}{2}(1-\text{n})\Big]^2$
$=\frac{\text{n}^2}{4\text{n}}(\text{n}+1)^2(\text{n}-1)^2$

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