$$\eqalign{ & \sigma _x^2 = 4,\sigma _y^2 = 5,x = 2,y = 4 \cr & \frac{1}{5}\sum {x_i^2 - {{\left( 2 \right)}^2} = 4;} \frac{1}{5}\sum {y_i^2 - {{\left( 4 \right)}^2} = 5} \cr & \sum {x_i^2 = 40;\sum {y_i^2 = 105} } \cr & \Rightarrow \,\,\sum {\left( {x_i^2 + y_i^2} \right) = 145} \cr & \Rightarrow \,\,\sum {\left( {{x_i} + {y_i}} \right)} = 5\left( 2 \right) + 5\left( 4 \right) = 30 \cr} $$
Variance of combined data
$$\eqalign{ & = \frac{1}{{10}}\sum {\left( {x_i^2 + y_i^2} \right) - {{\left( {\frac{1}{{10}}\sum {\left( {{x_i} + {y_i}} \right)} } \right)}^2}} \cr & = \frac{{145}}{{10}} - 9 \cr & = \frac{{11}}{2} \cr} $$

$$\frac{{70 \times 75 + 30 \times x}}{{100}} = 72 \Rightarrow x = 65$$

$$\eqalign{ & {\text{The mean of the series}} \cr & \overline X = \frac{1}{{2n + 1}}\left\{ {a + \left( {a + d} \right) + \left( {a + 2d} \right) + ...... + \left( {a + 2nd} \right)} \right\} \cr & = \frac{1}{{2n + 1}}\left\{ {\frac{{2n + 1}}{2}\left( {2a + 2nd} \right)} \right\} \cr & = a + nd \cr & {\text{Therefore, mean deviation from mean}} \cr & = \frac{1}{{2n + 1}}\sum\limits_{r = 0}^{2n} {\left| {\left( {a + rd} \right) - \left( {a + nd} \right)} \right|} \cr & = \frac{1}{{2n + 1}}\sum\limits_{r = 0}^{2n} {\left| {r - n} \right|d} \cr & = \frac{{\left[ {2\left( {1 + 2 + ...... + n} \right) + 0} \right]d}}{{2n + 1}} \cr & = \frac{{n\left( {n + 1} \right)d}}{{2n + 1}} \cr} $$

We know that for positive real numbers $${x_1},\,{x_2},......,\,{x_n},$$    we have,
$$\eqalign{ & \frac{{\sum {x_i^2} }}{n} \geqslant {\left( {\frac{{\sum {{x_i}} }}{n}} \right)^2} \cr & \Rightarrow \frac{{400}}{n} \geqslant {\left( {\frac{{80}}{n}} \right)^2} \cr & \Rightarrow n \geqslant 16 \cr & {\text{So only possible value for }}n = 18 \cr} $$

For the second set of observations $$N = 150$$   (even), so median is the A.M. of $${75^{th}}$$ and $${76^{th}}$$ observation first set is not sorted so we have to arrange them in increasing or decreasing order.

Let $$\overline x $$ is the mean of $$n$$ observation $${x_1},\,{x_2},\,......,\,{x_n}$$
$$\eqalign{ & \Rightarrow \overline x = \frac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n} \cr & {\text{Now, }}\left( {a - b} \right){\text{ is added to each term}}{\text{.}} \cr & \therefore \,{\text{New mean}} \cr & = \frac{{{x_1} + \left( {a - b} \right) + {x_2} + \left( {a - b} \right) + ...... + {x_n} + \left( {a - b} \right)}}{n} \cr & = \frac{{{x_1} + {x_2} + ..... + {x_n}}}{n} + \frac{{n\left( {a - b} \right)}}{n} \cr & = \overline x + \left( {a - b} \right) \cr} $$

If initially all marks were $${x_i}$$ then $${\sigma _1}^2 = \frac{{\sum\limits_i {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N}$$
Now each is increased by 10
$$\eqalign{ & {\sigma _1}^2 = \frac{{\sum\limits_i {{{\left[ {\left( {{x_i} + 10} \right) - \left( {\overline x + 10} \right)} \right]}^2}} }}{N} \cr & = \frac{{\sum\limits_i {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} \cr & = {\sigma _1}^2 \cr} $$
Hence, variance will not change even after the grace marks were given.

$$n = 9$$  then median term $$ = {\left( {\frac{{9 + 1}}{2}} \right)^{th}} = {5^{th}}$$     term. Last four observations are increased by 2. The median is $${5^{th}}$$ observation which is remaining unchanged.
∴ there will be no change in median.

$$\eqalign{ & {\text{Let total number of observations}} = 18{\text{ }} \cr & {\text{Let mean of one group}} = {a_1} = 15{\text{ }} \cr & {\text{and mean of other group}} = {a_2} \cr & {\text{Given : Combined mean}} = \frac{{40}}{3} \cr & {\text{Now, combined mean is}} \cr & = \frac{{\left( {{a_1} \times 10} \right) + \left( {{a_2} \times 8} \right)}}{{18}} \cr & \Rightarrow \frac{{40}}{3} = \frac{{150 + 8{a_2}}}{{18}} \cr & \Rightarrow {a_2} = \frac{{45}}{4} \cr} $$

Clearly mean $$A = 0$$
Standard deviation $$\sigma = \sqrt {\frac{{\sum {{{\left( {x - A} \right)}^2}} }}{{2n}}} $$
$$\eqalign{ & 2 = \sqrt {\frac{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + .....{{\left( {0 - a} \right)}^2} + .....}}{{2n}}} \cr & 2 = \sqrt {\frac{{{a^2}.2n}}{{2n}}} \cr & 2 = \left| a \right| \cr} $$
Hence, $$\left| a \right| = 2$$