A) \[\frac{{{n}^{2}}+1}{12}\]
B) \[\frac{\left( n+1 \right)\left( 2n+1 \right)}{6}\]
C) \[\frac{{{n}^{2}}-1}{12}\]
D) None of these
Correct Answer: C
Solution :
[c] Variance\[={{\sigma }^{2}}={{(S.D.)}^{2}}\] \[=\frac{\sum{f.{{d}^{2}}}}{N}-{{\left( \frac{\sum{fd}}{N} \right)}^{2}}=\frac{(n-1)\left( 2n+1 \right)}{6}+{{\left( \frac{n+1}{2} \right)}^{2}}\] \[=\frac{2(2{{n}^{2}}+3n+1)-3({{n}^{2}}+2n+1)}{12}=\frac{{{n}^{2}}-1}{12}\] Hence, option [c] is correct.
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