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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