CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 2 - MCQExams.com

$$  {\textbf{Step 1: Finding mean}} $$

               $$  {\text{The first }n\text{ natural numbers are }1,2,3,.....,n.} $$

               $$  {\text{Their mean, }}\mathop x\limits^\_ {\text{ = }}\dfrac{{\sum x }}{n}{\text{ }} $$

               $$  {\text{ = }}\dfrac{{1 + 2 + 3 + ....... + n}}{n} $$

               $$   = \dfrac{{n\left( {n + 1} \right)}}{{2n}} $$

               $$   = \dfrac{{n + 1}}{2} $$

               $$  {\text{Sum of the square of the first n natural numbers is }}\sum {{x^2}} {\text{ = }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ }} $$

$$  {\textbf{Step 2: Finding standard deviation}} $$

               $$  {\text{Thus, the standard deviation }}\sigma {\text{ = }}\sqrt {\dfrac{{\sum {{x^2}} }}{n} - {{\left( {\dfrac{{\sum x }}{n}} \right)}^2}}  $$

               $$  {\text{ = }}\sqrt {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}}  $$

               $$  {\text{ =  }}\sqrt {\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}}  $$

               $$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right]}  $$

               $$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2(2n + 1) - 3(n + 1)}}{6}} \right]}  $$

               $$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{4n + 2 - 3n - 3}}{6}} \right]}  $$

               $$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left( {\dfrac{{n - 1}}{6}} \right)}  $$

               $$   = \sqrt {\dfrac{{{n^2} - 1}}{{12}}}  $$

               $$  {\text{Hence, the S}}{\text{.D}}{\text{. of the first n natural numbers is }}\sqrt {\dfrac{{{n^2} - 1}}{{12}}}  $$

$$  {\textbf{ Hence, the correct answer is option B}}{\text{.}} $$

$${\textbf{Step  - 1: Finding mean of first n natural numbers}}$$

                    $${\text{We know that, mean of n observations  =  }}\dfrac{{{\text{Sum of n observations}}}}{{\text{n}}}$$

                    $$\therefore {\text{ Mean  =  }}\dfrac{{{\text{1  +  2  +  3  +  }}.....{\text{  +  n}}}}{{\text{n}}}$$

                    $${\text{We know that, sum of first n natural numbers  =  }}\dfrac{{{\text{n}}\left( {{\text{n  +  1}}} \right)}}{{\text{2}}}$$

                    $$ \Rightarrow {\text{ Mean  =  }}\dfrac{{\dfrac{{{\text{n}}\left( {{\text{n  +  1}}} \right)}}{{\text{2}}}}}{{\text{n}}}$$

                    $$ \Rightarrow {\text{ Mean  =  }}\dfrac{{{\text{n  +  1}}}}{{\text{2}}}$$

$${\textbf{Step  - 2: Calculating variance}}$$

                    $${\text{We know that, variance  =  }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}^{\text{2}}} }}{{\text{n}}}{\text{  -  }}{\left( {{\text{Mean}}} \right)^{\text{2}}}$$

                    $$\therefore {\text{ Variance  =  }}\dfrac{{{{\text{1}}^{\text{2}}}{\text{  +  }}{{\text{2}}^{\text{2}}}{\text{  +  }}{{\text{3}}^{\text{2}}}{\text{  +  }}....{\text{ +   }}{{\text{n}}^{\text{2}}}}}{{\text{n}}}{\text{  -  }}{\left( {\dfrac{{{\text{n  +  1}}}}{{\text{2}}}} \right)^{\text{2}}}$$

                    $${\text{Also, sum of squares of first n natural numbers  =  }}\dfrac{{{\text{n}}\left( {{\text{n  +  1}}} \right)\left( {{\text{2n  +  1}}} \right)}}{{\text{6}}}$$

                    $$ \Rightarrow {\text{ Variance  =  }}\dfrac{{{\text{n}}\left( {{\text{n  +  1}}} \right)\left( {{\text{2n  +  1}}} \right)}}{{{\text{6n}}}}{\text{  -  }}\dfrac{{{{{\text{(n  +  1)}}}^{\text{2}}}}}{{\text{4}}}$$

                    $$ \Rightarrow {\text{ Variance  =  }}\dfrac{{\left( {{\text{n  +  1}}} \right)\left( {{\text{2n  +  1}}} \right)}}{{\text{6}}}{\text{  -  }}\dfrac{{{{{\text{(n  +  1)}}}^{\text{2}}}}}{{\text{4}}}$$

                    $$ \Rightarrow {\text{ Variance  =  }}\dfrac{{{\text{2}}{{\text{n}}^{\text{2}}}{\text{  +  3n  +  1}}}}{{\text{6}}}{\text{  -  }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{  +  2n  +  1}}}}{{\text{4}}}$$

                    $$ \Rightarrow {\text{ Variance  =  }}\dfrac{{{\text{4}}{{\text{n}}^{\text{2}}}{\text{  +  6n  +  2  -  3}}{{\text{n}}^{\text{2}}}{\text{  -  6n  -  3}}}}{{{\text{12}}}}$$

                    $$ \Rightarrow {\text{ Variance  =  }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{  -  1}}}}{{{\text{12}}}}$$

$$\mathbf{{\text{Thus, the variance of first n natural numbers is }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{  -  1}}}}{{{\text{12}}}}.}$$

$$\textbf{Step - 1: Finding the mean}$$

                   $$\text{The } n \text{ natural numbers are } 1, 2, 3,..., n$$

                   $$\text{The Sum of } n \text{ natural number is } \dfrac{{n(n+1)}}{{2}}$$

                   $$\text{Mean} = \dfrac{ \dfrac{{n(n+1)}}{{2}}}{n} = \dfrac{{n+1}}{{2}}$$

$$\textbf{Step - 2: Finding the Variance}$$

                   $$\text{Variance} = \dfrac{{\sum(x_i)^2}}{{n}} - (\text{Mean})^2$$

                   $$ \sum(x_i)^2 = \dfrac{{1^2 +2^2 + ...+ n^2}}{{n}}$$

                   $$\text{Since } 1^2 +2^2 + ...+ n^2 = \dfrac{{n(n+1)(2n+1)}}{{6}}$$

                   $$\dfrac{{\sum(x_i)^2}}{{n}} = \dfrac{{n(n+1)(2n+1)}}{{6n}}$$

                   $$\text{Variance} =  \dfrac{{n(n+1)(2n+1)}}{{6n}} - (\dfrac{{n+1}}{{2}})^2$$

                   $$=\dfrac{{(n+1)}}{{2}}\times(\dfrac{{2n+1}}{{3}}-\dfrac{{n+1}}{{2}})$$

                   $$= \dfrac{{(n+1)}}{{2}}(\dfrac{{4n+2-3n-3}}{{6}})$$

                   $$=\dfrac{{n+1}}{{2}}\times\dfrac{{n-1}}{{6}}$$

                   $$= \dfrac{{n^2 - 1}}{{12}}$$

$${\textbf{Hence, the correct answer is Option B}.}$$