Statistics MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{Mean}} = \frac{{101 + d\left( {1 + 2 + 3 + ...... + 100} \right)}}{{101}} \cr & = 1 + \frac{{d \times 100 \times 101}}{{101 \times 2}} \cr & = 1 + 50d \cr & \because \,{\text{Mean deviation from the mean}} = 255 \cr & \Rightarrow \frac{1}{{101}}\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + 2d} \right) - \left( {1 + 50d} \right)} \right| + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right] = 225 \cr & \Rightarrow 2d\left[ {1 + 2 + 3 + ...... + 50} \right] = 101 \times 225 \cr & \Rightarrow 2d \times \frac{{50 \times 51}}{2} = 101 \times 225 \cr & \Rightarrow d = \frac{{101 \times 255}}{{50 \times 51}} \cr & \Rightarrow d = 10.1 \cr} $$

Total number of students
$$= 40 + 35 + 45 + 42 = 162$$
Total marks obtained
$$\eqalign{ & = \left( {40 \times 50} \right) + \left( {35 \times 60} \right) + \left( {45 \times 55} \right) + \left( {42 \times 45} \right) \cr & = 8465 \cr} $$
Overall average of marks per students
$$ = \frac{{8465}}{{162}} = 52.25$$

$$\eqalign{ & {\text{Total sum of }}13{\text{ observations}} \cr & = 14 \times 13 = 182 \cr & {\text{Sum of }}14{\text{ observations}} \cr & = 7 \times 12 + 7 \times 16 \cr & = 84 \times 112 \cr & = 196 \cr & {\text{So, the }}{7^{th}}{\text{ observations}}\,{\text{is}} \cr & = 196 - 182 \cr & = 14 \cr} $$

$$\eqalign{ & M = \frac{{{x_1} + {x_2} + {x_3}......{x_n}}}{n} \cr & nM = {x_1} + {x_2} + {x_3} + ...... + {x_{n - 1}} + {x_n} \cr & {\text{i}}{\text{.e}}{\text{., }}nM - {x_n} = {x_1} + {x_2} + {x_3} + ...... + {x_{n - 1}} \cr & \frac{{nM - {x_n} + x'}}{n} = \frac{{{x_1} + {x_2} + {x_3} + ...... + {x_{n - 1}} + x'}}{n} \cr & \therefore {\text{ New average }} = \frac{{nM - {x_n} + x'}}{n} \cr} $$

$$\eqalign{ & {\text{Mean }}E\left( x \right) = \sum {rP\left( r \right)} = \sum {r.\frac{1}{{{2^r}}}} \cr & \therefore \,E\left( x \right) = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + ...... \cr & \Rightarrow \frac{1}{2}E\left( x \right) = \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... \cr & \therefore \,\frac{1}{2}E\left( x \right) = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... \cr & \Rightarrow \frac{1}{2}E\left( x \right) = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} \cr & \therefore \,E\left( x \right) = 2 \cr} $$

$$\eqalign{ & {\text{Here for each }}{x_i} = i, \cr & {\text{weight}}\,{w_i} = {i^2} + i \cr & {\text{Hence, the required mean}} \cr & = \frac{{\sum {{w_i}{x_i}} }}{{\sum {{w_i}} }} \cr & = \frac{{\sum\limits_{i = 1}^n {i\left( {{i^2} + i} \right)} }}{{\sum\limits_{i = 1}^n {\left( {{i^2} + i} \right)} }} \cr & = \frac{{\sum\limits_{i = 1}^n {{i^3}} + \sum\limits_{i = 1}^n {{i^2}} }}{{\sum\limits_{i = 1}^n {{i^2}} + \sum\limits_{i = 1}^n i }} \cr & = \frac{{\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}}} \cr & = \frac{{\frac{{n\left( {n + 1} \right)}}{2}\left\{ {\frac{{n\left( {n + 1} \right)}}{2} + \frac{{2n + 1}}{3}} \right\}}}{{\frac{{n\left( {n + 1} \right)}}{2}\left\{ {\frac{{2n + 1}}{3} + 1} \right\}}} \cr & = \frac{{3{n^2} + 7n + 2}}{{2\left( {2n + 4} \right)}} \cr & = \frac{{\left( {3n + 1} \right)\left( {n + 2} \right)}}{{4\left( {n + 2} \right)}} \cr & = \frac{{3n + 1}}{4} \cr} $$

$$\eqalign{ & {\text{Mean}} = 40,\,{\text{S}}{\text{.D}}{\text{.}} = 8 \cr & \therefore \,{\text{Coeft of variation}} \cr & = \frac{8}{{40}} \times 100 \cr & = \frac{1}{5} \times 100 \cr & = 20\% \cr} $$

$$\eqalign{ & {\text{Given }}\sum x = 170,\,\sum {{x^2} = 2830} \cr & {\text{Increase in }}\sum x = 10, \cr & {\text{then, }}\sum x ' = 170 + 10 = 180 \cr & {\text{Increase in }}\sum {{x^2} = 900 - 400 = 500} , \cr & {\text{then, }}\sum x {'^2} = 2830 + 500 = 3330 \cr & \therefore \,{\text{Variance is :}} \cr & = \frac{1}{n}\sum x {'^2} - {\left( {\frac{{\sum x '}}{n}} \right)^2} \cr & = \frac{{3330}}{{15}} - {\left( {\frac{{180}}{{15}}} \right)^2} \cr & = 222 - 144 \cr & = 78 \cr} $$

$$\eqalign{ & {\text{Let }}{x_1},\,{x_2}......{x_n}{\text{ be }}n{\text{ items}}{\text{.}} \cr & {\text{Then, }}\overline x = \frac{1}{n}\sum {{x_i}} \cr & {\text{Let }}{y_1} = {x_1} + 1,\,{y_2} = {x_2} + 2,\,{y_3} = {x_3} + 3,......,{y_n} = {x_n} + n \cr & {\text{Then the mean of the new series is :}} \cr & \frac{1}{n}\sum {{y_i}} = \frac{1}{n}\sum\limits_{i = 1}^n {\left( {{x_i} + \frac{1}{i}} \right)} \cr & \Rightarrow \frac{1}{n}\sum {{y_i}} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i} + \frac{1}{n}\left( {1 + 2 + 3 + ...... + n} \right)} \cr & \Rightarrow \frac{1}{n}\sum {{y_i}} = \overline x + \frac{1}{n}.\frac{{n\left( {n + 1} \right)}}{2} \cr & \Rightarrow \frac{1}{n}\sum {{y_i}} = \overline x + \frac{{n + 1}}{2} \cr} $$

$$\sum {x = 170} ,\sum {{x^2} = 2830} $$      increase in $$\sum {x = 10,} $$   then $$\sum {x' = 170 + 10 = 180} $$
Increase in $$\sum {{x^2}} = 900 - 400 = 500$$      then $$\sum {x{'^2} = 28} 30 + 500 = 3330$$
Variance $$ = \frac{1}{n}\sum {x{'^2} - {{\left( {\frac{1}{n}\sum {x'} } \right)}^2}} $$
$$\eqalign{ & = \frac{1}{{15}} \times 3330 - {\left( {\frac{1}{{15}} \times 180} \right)^2} \cr & = 222 - 144 = 78 \cr} $$