A) \[\frac{{{n}^{2}}-1}{12}\]
B) \[\frac{{{n}^{2}}-1}{6}\]
C) \[\frac{{{n}^{2}}+1}{6}\]
D) \[\frac{{{n}^{2}}+1}{12}\]
Correct Answer: A
Solution :
Variance \[={{(\text{S}\text{.D}\text{.})}^{2}}\]\[=\frac{1}{n}\Sigma {{x}^{2}}-{{\left( \frac{\Sigma x}{n} \right)}^{2}}\], \[\left( \because \ \ \bar{x}=\frac{\Sigma x}{n} \right)\] \[=\frac{n(n+1)\ (2n+1)}{6n}-{{\left( \frac{n(n+1)}{2n} \right)}^{2}}=\frac{{{n}^{2}}-1}{12}\].
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