Statistics

Series A = 101, 102 ............ 200

Series B = 151, 152 ............ 250

Here series B can be obtained if we change the origin of A by 50 units.

And we know the variance does not change by changing the origin.

So, $\,\,\,\,$ ${V_A} = {V_B}$

$ \Rightarrow \,\,\,\,\,{{{V_A}} \over {{V_B}}} = 1$

Given that,

Mean = 21 and median = 22

We know,

Mode + 2 Mean = 3 Median

$\therefore$ Mode = 3 $ \times $ 22 $-$ 2 $ \times $ 21

= 66 $-$ 42

= 24

As we know,

${\sigma ^2} \ge 0$

$\therefore\,\,\,$ ${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} \ge 0$

$ \Rightarrow \,\,\,{{400} \over n} - {{6400} \over {{n^2}}} \ge 0$

$ \Rightarrow \,\,\,n \ge 16$

$\therefore\,\,\,$ Possible value of n according to the option is = 18

The statements are analyzed as follows :

(a) Mode can be computed from histogram : This is correct. The mode is the value that appears most frequently in a data set. A histogram provides a graphical representation of the frequency of each data value. The data value corresponding to the highest bar in a histogram is the mode.

(b) Median is not independent of change of scale : This is correct. The median is the middle value in a data set when the values are arranged in ascending or descending order. When you change the scale (e.g., by multiplying all data points by a constant), the median changes as well.

(c) Variance is independent of change of origin and scale : This is not correct. Variance, which measures the dispersion of a set of data points, is affected by changes in scale (it is not independent of scale). If you multiply all the values in a dataset by a constant, the variance gets multiplied by the square of that constant. However, it is true that variance is independent of the change of origin (it remains the same if you add or subtract a constant from all data points in the set).

Therefore, the correct answer is Option C : only (a) and (b) are correct.

Mean $\left( A \right) = {{a - a} \over {2n}} = 0$

Given standard deviation (S.D) = 2

$\therefore\,\,\,$ $\sqrt {{{\sum {{{\left( {x - A} \right)}^2}} } \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + ..... + {{\left( {0 - a} \right)}^2}} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{{a^2} + {a^2}........2n\,times} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{2n\,.\,{a^2}} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{a^2}} = 2$

$ \Rightarrow \,\,\,\left| a \right| = 2$

Here total no of observation is 9 which is a odd number. As we know for odd number 9 the median will be the 5^th term.

Now question says, you increase largest 4 number by 2 which does not affect the 5^th term so the new median will be the same.

Given that,

N = 15, $\sum {x{}^2} = 2830,\,\sum x = 170$

As, 20 was replaced by 30 then,

$\sum x = 170 - 20 + 30 = 180$

and $\sum {{x^2}} = 2830 - 400 + 900 = 3330$

So, the corrected variance

$ = {{\sum {{x^2}} } \over N} - {\left( {{{\sum x } \over N}} \right)^2}$

$ = {{3330} \over {15}} - {\left( {{{180} \over {15}}} \right)^2}$

$ = 222 - {12^2}$

$ = 222 - 144$

$ = 78$

Given that, total students = 100

and number of boys = 70

$\therefore$ No. of girls = 100 - 70 = 30

Average of 100 students = 72

$\therefore$ Total marks of 100 students = 100 $ \times $ 72 = 7200

Average of 70 boys = 75

$\therefore$ Total marks of 70 boys = 70 $ \times $ 75 = 5250

$\therefore$ Total marks of 30 girls = 7200 $ - $ 5250 = 1950

$\therefore$ Average marks of 30 girls = ${{1950} \over {30}}$ = 65