Explanation
\textbf{Step - 1: Finding the mean}
\text{The } n \text{ natural numbers are } 1, 2, 3,..., n
\text{The Sum of } n \text{ natural number is } \dfrac{{n(n+1)}}{{2}}
\text{Mean} = \dfrac{ \dfrac{{n(n+1)}}{{2}}}{n} = \dfrac{{n+1}}{{2}}
\textbf{Step - 2: Finding the Variance}
\text{Variance} = \dfrac{{\sum(x_i)^2}}{{n}} - (\text{Mean})^2
\sum(x_i)^2 = \dfrac{{1^2 +2^2 + ...+ n^2}}{{n}}
\text{Since } 1^2 +2^2 + ...+ n^2 = \dfrac{{n(n+1)(2n+1)}}{{6}}
\dfrac{{\sum(x_i)^2}}{{n}} = \dfrac{{n(n+1)(2n+1)}}{{6n}}
\text{Variance} = \dfrac{{n(n+1)(2n+1)}}{{6n}} - (\dfrac{{n+1}}{{2}})^2
=\dfrac{{(n+1)}}{{2}}\times(\dfrac{{2n+1}}{{3}}-\dfrac{{n+1}}{{2}})
= \dfrac{{(n+1)}}{{2}}(\dfrac{{4n+2-3n-3}}{{6}})
=\dfrac{{n+1}}{{2}}\times\dfrac{{n-1}}{{6}}
= \dfrac{{n^2 - 1}}{{12}}
{\textbf{Hence, the correct answer is Option B}.}
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