Descriptive Statistics Test 14

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

$$(\sigma _A) ^2=64$$ and $$(\sigma _B) ^2=81$$
$$\because S.D=\sqrt {\sigma ^2}$$
$$\therefore S.D_A=8$$ and $$S.D_B=9$$
Since monthly mean wages in two factories are same, and $$ S.D_A < S.D_B$$.
Thus, the factory $$B$$ has more variation in wages.
Ans-Option $$B$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here $$n=8$$.
Mean, $$u=\dfrac{\sum x_i}{n}=\dfrac{64}{8}=8$$
Value of Deviations $$(x_i-u)$$ are
$$-3,1,0,4,-2,2,-2,0$$
Variance, $$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}=\dfrac{38}{8}=4.75$$.
Ans- Option $$B$$. 

First $$10$$ multiples of $$3$$ are $$3,6,9...30$$.
This is an A.P.
$$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$$
$$\therefore sum=165$$.
Mean, $$u=\dfrac {sum}{n}=\dfrac{165}{10}$$.
Variance, $$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$$.
$$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$$.
$$\therefore \sigma ^2=346.5-272.25$$
$$\therefore \sigma ^2=74.25$$
Standard Deviation, $$S.D=\sqrt{ \sigma ^2}$$
$$\therefore S.D=\sqrt{74.25}$$
Thus, $$S.D=8.61$$
Ans-Option $$A$$.

Mean, $$u=8$$.
So, $$\dfrac{x_1+x_2...x_6}{6}=8$$.
$$\therefore x_1+x_2...x_6=48$$.
Multiplying the above equation by $$3$$ on both the sides we get,
$$3x_1+3x_2...3x_6=48\times 3$$
$$\therefore u'=\dfrac{3x_1+3x_2....3x_6}{6}=\dfrac{48 \times 3}{6}$$
$$\therefore u'=24$$
Thus, new mean, $$u'=24$$.
$$\therefore$$ New variance, $$\sigma ^2=\dfrac{(3x_1)^2+...(3x_6)^2}{6}-(24)^2$$
$$\therefore \sigma ^2=9\times \dfrac{(x_1)^2+....(x_6)^2}{6}-576$$
$$\therefore \sigma ^2=9\times \dfrac{480}{16}-576$$
$$\therefore \sigma ^2=720-576$$
$$\therefore \sigma ^2=144$$
Thus, New Variance is $$144$$.
Ans-Option $$D$$

Here $$n=8$$.
Mean, $$u=\dfrac{\sum x_i}{n}=\dfrac{64}{8}=8$$
Value of Deviations $$(x_i-u)$$ are
$$-3,1,0,4,-2,2,-2,0$$
Variance, $$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}\\=\dfrac{(-3-8)^2-(1-8)^2-(0-8)^2-(4-8)^2-(-2-8)^2-(2-8)^2-(-2-8)^2-(0-8)^2}{8}\\=\dfrac{38}{8}=4.75$$.
Standard Deviation, $$S.D=\sqrt{\sigma ^2}$$
$$\therefore S.D=\sqrt{4.75}=2.17$$
Ans- Option $$D$$. 

Coefficient of Variation, $$C.V=\dfrac{S.D}{Mean} \times 100 \%$$
So, $$C.V_A=\dfrac{S.D_A}{Mean_A} \times 100 \%$$

$$\therefore C.V_A=\dfrac{\sqrt {64}}{3500} \times 100 \%$$.
$$\therefore C.V_A=0.22857 \%$$.
Ans-Option $$A$$.

Total number of wall clocks sold = $$20 + 25 + 15 + 35 + 10 + 5 = 110$$

Since we know that if all the data items of a distribution are same. Then, the variance of items is 0.
Here, all the data items are same and equal to 10.
Hence, variance is 0.
Option C is correct.