$$\eqalign{ & {\text{First }}50{\text{ even natural numbers are}}\,2,\,4,\,6,......,\,100 \cr & {\text{Variance}} = \frac{{\sum {x_i^2} }}{N} - {\left( {\overline x } \right)^2} \cr & \Rightarrow {\sigma ^2} = \frac{{{2^2} + {4^2} + ...... + {{100}^2}}}{{50}} - {\left( {\frac{{2 + 4 + ...... + 100}}{{50}}} \right)^2} \cr & \Rightarrow {\sigma ^2} = \frac{{4\left( {{1^2} + {2^2} + {3^2} + ...... + {{50}^2}} \right)}}{{50}} - {\left( {51} \right)^2} \cr & \Rightarrow {\sigma ^2} = 4\left( {\frac{{50 \times 51 \times 101}}{{50 \times 6}}} \right) - {\left( {51} \right)^2} \cr & \Rightarrow {\sigma ^2} = 3434 - 2601 \cr & \Rightarrow {\sigma ^2} = 833 \cr} $$

$$\eqalign{ & {\text{We have, }}r = \mathop {\max }\limits_{i \ne j} \left| {{x_i} - {x_j}} \right|{\text{ and}} \cr & {S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \cr & {\text{Now, consider}} \cr & {\left( {{x_i} - \overline x } \right)^2} = {\left( {{x_i} - \frac{{{x_1} + {x_2} + ...... + {x_n}}}{n}} \right)^2} \cr & = \frac{1}{{{n^2}}}\left[ {\left( {{x_i} - {x_1}} \right) + \left( {{x_i} - {x_2}} \right) + ...... + \left( {{x_i} - {x_i} - 1} \right)} \right] + \left[ {\left( {{x_i} - {x_i} + 1} \right) + ...... + \left( {{x_i} - {x_n}} \right)} \right] \leqslant \frac{1}{{{n^2}}}{\left[ {\left( {n - 1} \right)r} \right]^2} \cr & \Rightarrow {\left( {{x_i} - \overline x } \right)^2} \leqslant {r^2} \cr & \Rightarrow \sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \leqslant n{r^2} \cr & \Rightarrow \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \leqslant \frac{{n{r^2}}}{{\left( {n - 1} \right)}} \cr & \Rightarrow S \leqslant r\sqrt {\frac{n}{{n - 1}}} \cr} $$

Mean of the scores $$ = \frac{{202}}{{15}}$$
Mean of the correct scores $$ = \frac{{200}}{{15}}$$
i.e., Mean changes.
Median is same for both cases i.e., $$14.$$
Mode is proportional to mean.

The new observations are obtained by adding $$20$$  to each. Hence, $$\sigma $$ does not change.

KEY CONCEPT : If each observation is multiplied by $$k,$$ mean gets multiplied by $$k$$ and variance gets multiplied by $${k^2}.$$ Hence the new mean should be $$2\overline x $$ and new variance should be $${k^2}{\sigma ^2}.$$

$$\eqalign{ & {\text{Mean of series}}\left( {{x_1},\,{x_2},\,{x_3},......,\,{x_n}} \right) \cr & \overline x = \frac{{{x_1} + {x_2} + {x_3}, +......,\,{x_n}}}{n} \cr & \Rightarrow {x_1} + {x_2} + {x_3} + ...... + \,{x_n} = n\overline x \cr & {\text{Now we will replace }}{x_2}{\text{ by }}\lambda \cr & {\text{So number of elements in series will not change}}{\text{.}} \cr & {\text{New series will include }}\lambda {\text{ and exclude }}{x_2} \cr & {\text{Hence new series sum :}} \cr & \left( {{x_1} + {x_2} + ...... \,{x_n}} \right) - {x_2} + \lambda = n\overline x + \lambda - {x_2} \cr & {\text{Now new mean}} \cr & = \frac{{n\overline x + \lambda - {x_2}}}{n} = \frac{{n\overline x - {x_2} + \lambda }}{n} \cr} $$

$$\eqalign{ & {\text{Use }}{\sigma ^2} = \frac{{{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)}}{{{n_1} + {n_2}}} \cr & {\text{where }}{d_1} = {m_1} - a,\,{d_2} = {m_2} - a, \cr & a\,{\text{ being the mean}}\,{\text{of the whole group}} \cr & {\text{Let }}{m_2} = {\text{ mean of the second group}} \cr & \therefore \,15.6 = \frac{{100 \times 15 + 150 \times {m_2}}}{{250}} \Rightarrow {m_2} = 16 \cr & {\text{Thus,}} \cr & 13.44 = \frac{{\left( {100 \times 95 + 150 \times {\sigma ^2}} \right) + 100 \times {{\left( {0.6} \right)}^2} + 150 \times {{\left( {0.4} \right)}^2}}}{{250}} \cr & \Rightarrow \sigma = 4 \cr} $$

$$\eqalign{ & \frac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }} = 46.5 \cr & \Rightarrow \frac{{15 \times 10 + 30 \times 40 + M \times 30 + 75 \times 10 + 90 \times 10}}{{10 + 40 + 30 + 10 + 10}} = 46.5 \cr & \therefore \,M = 55 \cr & {\text{So the class intervals can be}} \cr & 10 - 20,\,20 - 40,\,40 - 70,\,70 - 80,\,80 - 100 \cr} $$

$$\because $$ Variance $$ = {\sigma ^2} = \frac{{\sum {x_i^2} }}{N} - {\left( {\overline x } \right)^2}$$
$$\eqalign{ & \Rightarrow \,\,18 = \frac{{\sum {x_i^2} }}{5} - {\left( {150} \right)^2} \cr & \Rightarrow \,\,\sum {x_i^2 = 90 + 112590} \cr & = 112590 \cr} $$
Then, variance of the height of six students
$$\eqalign{ & V' = \frac{{112590 + {{\left( {156} \right)}^2}}}{6} - {\left( {\frac{{750 + 156}}{6}} \right)^2} \cr & = 22821 - 22801 \cr & = 20 \cr} $$

$$\eqalign{ & {\text{Mean}} = \frac{{1 \times 5 + 2 \times 4 + 3 \times 6 + 4 \times f}}{{5 + 4 + 6 + f}} \cr & {\text{i}}{\text{.e}}{\text{., }}3 = \frac{{5 + 8 + 18 + 4f}}{{15 + f}} \cr & \Rightarrow 45 + 3f = 31 + 4f \cr & \Rightarrow 45 - 31 = f \cr & \Rightarrow f = 14 \cr} $$