Statistics MCQ Questions & Answers in Statistics and Probability | Maths

Median $$= 46$$  which lies in $$40 – 50$$   class
Median $$ = l + h\frac{{\left\{ {\frac{n}{2} - {C_f}} \right\}}}{f}$$
where $$f =$$  frequency of median-class
$${C_f} = $$  cumulative frequency of the class preceding the median class.
$$\eqalign{ & \therefore \,46 = 40 + 10\frac{{\left[ {\frac{{229}}{2} - \left( {x + 42} \right)} \right]}}{{65}} \cr & {\text{where }}x = {\text{ frequency of class }}30 - 40 \cr & \Rightarrow x = 33.5 = 34 \cr} $$

Let the other two observations be $$'a$$ and $$'b'$$
$$\eqalign{ & \therefore {\text{ Mean}} = \frac{{2 + 4 + 6 + a + b}}{5} \cr & \Rightarrow 4 = \frac{{12 + a + b}}{5} \cr & \Rightarrow a + b = 8\,; \cr & {\text{Variance}} = \frac{1}{n}\sum {{x^2} - {x^{ - 2}}} = 5.2 \cr & \Rightarrow \frac{1}{5}\left( {4 + 16 + 36 + {a^2} + {b^2}} \right) - 16 = 5.2 \cr & \Rightarrow {a^2} + {b^2} = 50 \cr} $$
From the options, it is clear that the two observations are $$1$$ and $$7.$$

$$\eqalign{ & {\text{Using the weighted HM formula}} \cr & \frac{1}{H} = \frac{1}{N}\sum\limits_{i = 1}^n {\frac{{{f_i}}}{{{x_i}}}} {\text{ or }}H = \frac{1}{{\frac{1}{N}\sum\limits_i {\frac{{{f_i}}}{{{x_i}}}} }} \cr & \therefore \,{\text{Average speed}} \cr & = \frac{{400}}{{100\left( {\frac{1}{{100}} + \frac{1}{{200}} + \frac{1}{{300}} + \frac{1}{{400}}} \right)}} \cr & = 192\,m/h \cr} $$

Sum of 16 observations $$ = 16 \times 16$$
$$ = 256$$
Sum of resultant 18 observations $$ = 256 - 16 + \left( {3 + 4 + 5} \right)$$
$$ = 252$$
Mean of observations $$ = \frac{{252}}{{18}}$$
$$ = 14$$

Mean of $$5$$ numbers $$= 30$$
$$\therefore $$  Total sum of $$5$$ numbers $$ = 30 \times 5 = 150$$
After excluded one number
Mean of $$4$$ numbers will be $$= 28$$
$$\therefore $$  Total sum of $$4$$ numbers $$ = 4 \times 28 = 112$$
Thus, excluded number
$$=$$ (sum of $$5$$ numbers $$-$$ sum of $$4$$ numbers)
$$= 150 - 112 = 38$$

Total student= 100 ;
for 70 stds. total marks $$ = 75 \times 70 = 5250$$
⇒ Total marks of girls $$= 7200 - 5250 = 1950$$
Average of girls $$ = \frac{{1950}}{{30}} = 65$$

$$\eqalign{ & {\text{Mean of }}a,\,b,\,8,\,5,\,10{\text{ is }}6 \cr & \Rightarrow \frac{{a + b + 8 + 5 + 10}}{5} = 6 \cr & \Rightarrow a + b = 7......\left( {\text{i}} \right) \cr & {\text{Variance of }}a,\,b,\,8,\,5,\,10{\text{ is }}6.80\, \cr & \Rightarrow \frac{{{{\left( {a - 6} \right)}^2} + {{\left( {b - 6} \right)}^2} + {{\left( {8 - 6} \right)}^2} + {{\left( {5 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}}{5} = 6.80 \cr & \Rightarrow {a^2} - 12a + 36 + {\left( {1 - a} \right)^2} + 21 = 34\,\,\,\left[ {{\text{using equation }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow 2{a^2} - 14a + 24 = 0 \cr & \Rightarrow {a^2} - 7a + 12 = 0 \cr & \Rightarrow a = 3{\text{ or }}4 \cr & \Rightarrow b = 4{\text{ or }}3 \cr} $$
$$\therefore $$  The possible values of $$a$$ and $$b$$ are
$$a = 3{\text{ and }}b = 4{\text{ or }}a = 4{\text{ and }}b = 3$$

$${\sigma ^2} = \frac{{5.18 + 3.24}}{{5 + 3}} + \left( {8 - 8} \right)\frac{{15}}{{64}} = 20.25$$

$$\eqalign{ & {\text{Let }}{x_1},\,{x_2},......,\,{x_{20}}\,{\text{be the given observations}}{\text{.}} \cr & {\text{Given, }}\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \overline x } \right)}^2} = 5} \cr & {\text{To find variance of 2}}{x_1},\,2{x_2},.2{x_3}......,\,2{x_{20}}, \cr & {\text{Let }}\overline x {\text{ denotes the mean of new observation,}} \cr & {\text{Clearly, }}\overline x {\text{ }} = \frac{{\sum\limits_{i = 1}^{20} {2{x_i}} }}{{20}} = \frac{{2\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 2\overline x \cr & {\text{Now, variance of new observation}} \cr & = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - \overline x } \right)}^2}} \cr & = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - 2\overline x } \right)}^2}} \cr & = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {4{{\left( {{x_i} - \overline x } \right)}^2}} \cr & = 4\left( {\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \overline x } \right)}^2}} } \right) \cr & = 4 \times 5 \cr & = 20 \cr} $$

$$\eqalign{ & {\text{Corrected }}\sum x = 40 \times 200 - 50 + 40 = 7990 \cr & \therefore \,{\text{Corrected}}\,\,\overline x = \frac{{7990}}{{200}} = 39.95 \cr & {\text{Incorrect}}\,\sum {{x^2}} = n\left[ {{\sigma ^2} + {{\overline x }^2}} \right] \cr & = 200\left[ {{{15}^2} + {{40}^2}} \right] \cr & = 365000 \cr & {\text{Corrected}}\,\sum {{x^2}} = 365000 - 2500 + 1600 = 364100 \cr & \therefore \,{\text{Corrected}}\,\,\sigma = \sqrt {\frac{{364100}}{{200}} - {{\left( {39.95} \right)}^2}} \cr & = \sqrt {\left( {1820.5 - 1596} \right)} \cr & = \sqrt {224.5} \cr & = 14.98 \cr} $$