Statistics MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{Given,}} \cr & 82 = \frac{{\left( {27 + x} \right) + \left( {31 + x} \right) + \left( {89 + x} \right) + \left( {107 + x} \right) + \left( {156 + x} \right)}}{5} \cr & \Rightarrow 82 \times 5 = 410 + 5x \cr & \Rightarrow 410 - 410 = 5x \cr & \Rightarrow x = 0 \cr & \therefore \,{\text{Required mean is,}} \cr & \overline x = \frac{{\left( {130 + x} \right) + \left( {126 + x} \right) + \left( {68 + x} \right) + \left( {50 + x} \right) + \left( {1 + x} \right)}}{5} \cr & \Rightarrow \overline x = \frac{{375 + 5x}}{5} \cr & \Rightarrow \overline x = \frac{{375 + 0}}{5} \cr & \Rightarrow \overline x = \frac{{375}}{5} \cr & \Rightarrow \overline x = 75 \cr} $$

Let the number of students of school $${\bf{I}} = x$$
$$\therefore $$  Number of students of School $${\bf{II}} = 1.5x$$
As given :
Mean of marks for school $${\bf{I}} = 82$$
and mean of marks for school $${\bf{II}} = 86$$
$$\therefore $$  Combined mean
$$\eqalign{ & = \frac{{x \times 82 + 1.5x \times 86}}{{x + 1.5x}} \cr & = \frac{{x\left( {82 + 129} \right)}}{{2.5x}} \cr & = \frac{{211}}{{2.5}} \cr & = 84.4 \cr} $$

For the numbers 2, 4, 6, 8, . . . . . ,2$$n$$
$$\eqalign{ & \overline x = \frac{{2\left[ {n\left( {n + 1} \right)} \right]}}{{2n}} = \left( {n + 1} \right) \cr & {\text{And}}\,\,Var = \frac{{\sum {{{\left( {x - \overline x } \right)}^2}} }}{{2n}} \cr & = \frac{{\sum {{x^2}} }}{n} - {\left( {\overline x } \right)^2} \cr & = \frac{{4\sum {{n^2}} }}{n} - {\left( {n + 1} \right)^2} \cr & = \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - {\left( {n + 1} \right)^2} \cr & = \frac{{2\left( {2n + 1} \right)\left( {n + 1} \right)}}{3} - {\left( {n + 1} \right)^2} \cr & = \,\left( {n + 1} \right)\,\left[ {\frac{{4n + 2 - 3n - 3}}{3}} \right] \cr & = \,\frac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{3} \cr & = \,\frac{{{n^2} - 1}}{3} \cr} $$
∴ Statement - 1 is false. Clearly, statement - 2 is true.

Median is the mean of $${25^{th}}$$ and $${26^{th}}$$ observation
$$\eqalign{ & \therefore \,\,M = \frac{{25a + 26a}}{2} = 25.5a \cr & M.D\left( M \right) = \frac{{\sum {\left| {{x_i} - M} \right|} }}{N} \cr & \Rightarrow \,\,50 = \frac{1}{{50}}\left[ {2 \times \left| a \right| \times \left( {0.5 + 1.5 + 2.5 + .....24.5} \right)} \right] \cr & \Rightarrow \,\,2500 = 2\left| a \right| \times \frac{{25}}{2}\left( {25} \right) \cr & \Rightarrow \,\,\left| a \right| = 4 \cr} $$

$$\eqalign{ & {\sigma ^2} = \frac{{\sum {{x^2}} }}{n} - {\left( {\frac{{\sum x }}{n}} \right)^2} \cr & \Rightarrow \,\,\frac{{{k^2} + 2}}{4} - {\left( {\frac{k}{4}} \right)^2} = 5 \cr & \Rightarrow \,\,4{k^2} + 8 - {k^2} = 80 \cr & \Rightarrow \,\,3{k^2} = 72 \cr & \Rightarrow \,\,k = 2\sqrt 6 \cr} $$

The total weight of seven students is
$$55 \times 7 = 385\,kg$$
The sum of the weights of six students is
$$52 + 58 + 55 + 53 + 56 + 54 = 328\,kg$$
Hence, the weight of the seventh student is
$$385 - 328 = 57\, kg.$$

$$\eqalign{ & {\text{Mean and SD }}\sigma {\text{ of the combined group are}} \cr & m = \frac{{63 \times 27.6 + 26 \times 19.2}}{{63 + 26}} = 25.1 \cr & {\text{Thus, AM is decreased by}}\,\,27.6 - 25.1 = 2.5 \cr & {\sigma ^2} = \frac{{63 \times {{\left( {7.1} \right)}^2} + 26 \times {{\left( {6.2} \right)}^2}}}{{89}} + \frac{{63{{\left( {25.1 - 27.6} \right)}^2} + 26{{\left( {25.1 - 19.2} \right)}^2}}}{{89}} \cr & \Rightarrow \sigma = 7.8\,\,\,\left( {{\text{approx}}{\text{.}}} \right) \cr} $$

$$\eqalign{ & {\text{Required mean :}} \cr & = \frac{1}{n}\sum\limits_{i = 1}^n {\left( {2{x_i} + 3} \right)} \cr & = \frac{2}{n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right) + \frac{{3n}}{n} \cr & = 2\left\{ {\frac{1}{n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)} \right\} + 3 \cr & = 2\overline x + 3 \cr} $$

Given data $$3,\,5,\,1,\,6,\,5,\,9,\,5,\,2,\,8,\,6$$      and mean, median and mode are $$x,\,y,\,z$$   respectively.
Rearranging data $$1,\,2,\,3,\,5,\,5,\,5,\,6,\,6,\,8,\,9$$
$$\eqalign{ & {\text{Mean}} = x \cr & = \frac{{1 + 2 + 3 + 5 + 5 + 5 + 6 + 6 + 8 + 9}}{{10}} \cr & = \frac{{50}}{{10}} \cr & = 5 \cr & {\text{Median}} = y \cr & = \frac{{\frac{{{n^{th}}}}{2}{\text{term}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}{\text{term}}}}{2} \cr & = \frac{{5 + 5}}{2} \cr & = 5 \cr} $$
Mode $$\left( z \right) = $$   most frequently occurring value $$= 5$$
Hence $$x = y = z.$$

Let a student gets $$x$$ marks out of $$40.$$
He gets $$\frac{{5x}}{4}$$ marks out of $$50.$$
Thus, each observation will be multiplied by $$\frac{5}{4}$$
Hence, mean is also multiplied by $$\frac{5}{4}$$
Giving mean is $$ = 38 \times \frac{5}{4} = 47.5$$