Explanation
$$ {\textbf{Step 1: Finding mean}} $$
$$ {\text{The first }n\text{ natural numbers are }1,2,3,.....,n.} $$
$$ {\text{Their mean, }}\mathop x\limits^\_ {\text{ = }}\dfrac{{\sum x }}{n}{\text{ }} $$
$$ {\text{ = }}\dfrac{{1 + 2 + 3 + ....... + n}}{n} $$
$$ = \dfrac{{n\left( {n + 1} \right)}}{{2n}} $$
$$ = \dfrac{{n + 1}}{2} $$
$$ {\text{Sum of the square of the first n natural numbers is }}\sum {{x^2}} {\text{ = }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ }} $$
$$ {\textbf{Step 2: Finding standard deviation}} $$
$$ {\text{Thus, the standard deviation }}\sigma {\text{ = }}\sqrt {\dfrac{{\sum {{x^2}} }}{n} - {{\left( {\dfrac{{\sum x }}{n}} \right)}^2}} $$
$$ {\text{ = }}\sqrt {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} $$
$$ {\text{ = }}\sqrt {\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} $$
$$ = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right]} $$
$$ = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2(2n + 1) - 3(n + 1)}}{6}} \right]} $$
$$ = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{4n + 2 - 3n - 3}}{6}} \right]} $$
$$ = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left( {\dfrac{{n - 1}}{6}} \right)} $$
$$ = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $$
$$ {\text{Hence, the S}}{\text{.D}}{\text{. of the first n natural numbers is }}\sqrt {\dfrac{{{n^2} - 1}}{{12}}} $$
$$ {\textbf{ Hence, the correct answer is option B}}{\text{.}} $$
$${\textbf{Step - 1: Finding mean of first n natural numbers}}$$
$${\text{We know that, mean of n observations = }}\dfrac{{{\text{Sum of n observations}}}}{{\text{n}}}$$
$$\therefore {\text{ Mean = }}\dfrac{{{\text{1 + 2 + 3 + }}.....{\text{ + n}}}}{{\text{n}}}$$
$${\text{We know that, sum of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}$$
$$ \Rightarrow {\text{ Mean = }}\dfrac{{\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}}}{{\text{n}}}$$
$$ \Rightarrow {\text{ Mean = }}\dfrac{{{\text{n + 1}}}}{{\text{2}}}$$
$${\textbf{Step - 2: Calculating variance}}$$
$${\text{We know that, variance = }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}^{\text{2}}} }}{{\text{n}}}{\text{ - }}{\left( {{\text{Mean}}} \right)^{\text{2}}}$$
$$\therefore {\text{ Variance = }}\dfrac{{{{\text{1}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ + }}{{\text{3}}^{\text{2}}}{\text{ + }}....{\text{ + }}{{\text{n}}^{\text{2}}}}}{{\text{n}}}{\text{ - }}{\left( {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right)^{\text{2}}}$$
$${\text{Also, sum of squares of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}$$
$$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{{\text{6n}}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$$
$$ \Rightarrow {\text{ Variance = }}\dfrac{{\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$$
$$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{2}}{{\text{n}}^{\text{2}}}{\text{ + 3n + 1}}}}{{\text{6}}}{\text{ - }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ + 2n + 1}}}}{{\text{4}}}$$
$$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{4}}{{\text{n}}^{\text{2}}}{\text{ + 6n + 2 - 3}}{{\text{n}}^{\text{2}}}{\text{ - 6n - 3}}}}{{{\text{12}}}}$$
$$ \Rightarrow {\text{ Variance = }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}$$
$$\mathbf{{\text{Thus, the variance of first n natural numbers is }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}.}$$
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