Statistics MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{Let,}}\,\,\sum\nolimits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9} \cr & \Rightarrow \sum\nolimits_{i = 1}^9 {{x_i} - } \sum\nolimits_{i = 1}^9 {5 = 9} \cr & \Rightarrow \sum\nolimits_{i = 1}^9 {{x_i} - } \left( {9 \times 5} \right) = 9 \cr & \Rightarrow \sum {{x_i}} - 45 = 9 \cr & \Rightarrow \sum {{x_i}} = 54 \cr & {\text{Similarly,}} \cr & \sum {x_i^2 - 10 \times 54 + 25 \times 9 = 45} \cr & \Rightarrow \sum {x_i^2 = 360} \cr & \Rightarrow \sigma = \sqrt {\frac{{360}}{9} - {{\left( {\frac{{54}}{9}} \right)}^2}} \cr & \Rightarrow \sigma = \sqrt {\frac{{324}}{{81}}} \cr & \Rightarrow \sigma = 2 \cr} $$

$$\eqalign{ & {\text{Sum of all observations}} \cr & = 20 \times 15 = 300 \cr & {\text{Sum of correct observations}} \cr & = 300 - \left( {3 + 6} \right) + \left( {8 + 4} \right) = 303 \cr & {\text{Correct mean}} = \frac{{303}}{{20}} = 15.15 \cr} $$

$$\eqalign{ & {\text{Given }}M = \frac{{a + b + c + d + e}}{5} \cr & \Rightarrow a + b + c + d + e = 5M \cr & \Rightarrow a + b + c + d + e - 5M = 0 \cr & \Rightarrow \left( {a - M} \right) + \left( {b - M} \right) + \left( {c - M} \right) + \left( {d - M} \right) + \left( {e - M} \right) = 0 \cr & {\text{Hence, Required value}} = 0 \cr} $$

$$\eqalign{ & {\text{Given,}}\,169 = \frac{{\sum x }}{{50}} \cr & {\text{Correct value is }}143 \cr & {\text{Hence, sum is short by }}9 \cr & \therefore \,M = \frac{{169 \times 50 + 9}}{{50}} = 169 + \frac{9}{{50}} = 169.18 \cr} $$

$$\eqalign{ & {\text{Let}}\,{x_1}{\text{,}}\,{x_2}{\text{,}}......{\text{,}}{x_n}\,{\text{be }}n{\text{ observations}}{\text{.}} \cr & {\text{Then, }}\overline x = \frac{1}{n}\sum {{x_i}} \,;\,{\text{Let }}{y_i} = \frac{{{x_i}}}{\alpha } + 10 \cr & {\text{then, }}\frac{1}{n}\sum\limits_{i = 1}^n {{y_i}} = \frac{1}{\alpha }\left( {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right) + \frac{1}{n}\left( {10n} \right) \cr & \Rightarrow \overline y = \frac{1}{\alpha }\overline x + 10 = \frac{{\overline x + 10\alpha }}{\alpha } \cr} $$

Sum of $$100$$  items $$ = 49 \times 100 = 4900$$
Sum of items added $$= 60 + 70 + 80 = 210$$
Sum of items replaced $$= 40 + 20 + 50 = 110$$
New sum $$= 4900 + 210 - 110 = 5000$$
$$\therefore $$  Correct mean $$ = \frac{{5000}}{{100}} = 50$$

$$\eqalign{ & n = 180, \cr & p = P\left( {{\text{getting a six}}} \right) = \frac{1}{6}{\text{ and }}q = \frac{5}{6} \cr & {\text{So, S}}{\text{.D}}{\text{.}} = \sqrt {npq} \cr & = \sqrt {180 \times \frac{1}{6} \times \frac{5}{6}} \cr & = \sqrt {5 \times 5} \cr & = 5 \cr} $$

Given $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9} $$
$$ \Rightarrow \,\,\sum\limits_{i = 1}^9 {{x_i} = 54\,\,\,\,\,.....\left( {\text{i}} \right)} $$
Also, $$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2} = 45} $$
$$ \Rightarrow \,\,\sum\limits_{i = 1}^9 {{x_i}^2 - 10\sum\limits_{i = 1}^9 {{x_i} + 9\left( {25} \right) = 45\,\,\,\,.....\left( {{\text{ii}}} \right)} } $$
From (i) and (ii) we get,
$$\sum\limits_{i = 1}^9 {x_i^2 = 360} $$
Since, variance $$ = \frac{{\sum {{x_i}^2} }}{9} - {\left( {\frac{{\sum {{x_i}} }}{9}} \right)^2}$$
$$\eqalign{ & = \frac{{360}}{9} - {\left( {\frac{{54}}{9}} \right)^2} \cr & = 40 - 36 \cr & = 4 \cr} $$
∴ Standard deviation $$ = \sqrt {{\text{Variance}}} = 2$$

$$\eqalign{ & {\text{Let Number of girls student be }}x \cr & {\text{Sum of marks}}\, = 25 \times 40 + x \times 48 \cr & {\text{Total students}} = 25 + x \cr & \therefore \,43 = \frac{{25 \times 40 + x \times 48}}{{x + 25}} \cr & \Rightarrow 43x + 43 \times 25 = 25 \times 40 + x \times 48 \cr & \Rightarrow 5x = 3 \times 25 \cr & \Rightarrow x = 15 \cr} $$

Given that $${x_1}\, < \,{x_2} < {x_3}\, < \,.....\,\, < {x_{201}}$$
∴ Median of the given observation $$ = {\frac{{201 + 1}}{2}^{th}}$$   items
= $${101^{th}}$$  item
$$ = {x_{101}}$$
Now, deviations will be minimum if taken from the median
∴ Mean deviation will be min if $$k = {x_{101}}.$$