Statistics

$\sum {{x_i} = 7 \times 8 = 56} $

${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} = 16$

${{\sum {x_i^2} } \over 7} - 64 = 16$

$\sum {x_i^2 = 560} $

when 14 is omitted

$\sum {{x_i} = 56 - 14 = 42} $

New mean $ = a = {{\sum {{x_i}} } \over 6} = 7$

$\sum {x_i^2 = 560 - 196 = 364} $

new variance, $b = {{\sum {x_i^2} } \over 6} - {\left( {{{\sum {{x_i}} } \over 6}} \right)^2}$

$ = {{364} \over 6} - 49 = {{35} \over 3}$

$3b = 35$

$a + 3b - 5 = 7 + 35 - 5 = 37$

$x = \{ 11,12,13\,....,40,41\} $

$y = \{ 61,62,63\,....,90,91\} $

$\overline x = {{{{31} \over 2}(11 + 41)} \over {31}} = {1 \over 2} \times 52 = 26$

$\overline y = {{{{31} \over 2}(61 + 91)} \over {31}} = {1 \over 2} \times 152 = 76$

${\sigma ^2} = {{\sum {x_i^2 + \sum {y_i^2} } } \over {62}} - {\left( {{{\sum {x + \sum y } } \over {62}}} \right)^2}$

$ = 705$

Now,

$\left| {\overline x + \overline y - {\sigma ^2}} \right|$

$ = |26 + 76 - 705| = 603$

Given $\sum\limits_{{{i = 1} \over {20}}}^{20} {{x_i} = 15 \Rightarrow \sum\limits_{i = 1}^{20} {{x_i} = 300} } $

and $\sum\limits_{{{i = 1} \over {20}}}^{20} {x_i^2 - {{\left( {\overline x } \right)}^2} = 9 \Rightarrow \sum\limits_{i = 1}^{20} {x_i^2 = 4680} } $

Mean $ = {{{{({x_i} + \alpha )}^2} + {{({x_2} + \alpha )}^2}\, + \,.....\, + \,{{({x_{20}} + \alpha )}^2}} \over {20}} = 178$

$ \Rightarrow {{\sum\limits_{i = 1}^{20} {x_i^2 + 2\alpha \sum\limits_{i = 1}^{20} {{x_i} + 20{\alpha ^2}} } } \over {20}} = 178$

$ \Rightarrow 4680 + 600\alpha + 20{\alpha ^2} = 3560$

$ \Rightarrow {\alpha ^2} + 30\alpha + 56 = 0$

$ \Rightarrow {\alpha ^2} + 28\alpha + 2\alpha + 56 = 0$

$ \Rightarrow (\alpha + 28)(\alpha + 2) = 0$

${\alpha _{\max }} = - 2 \Rightarrow \alpha _{\max }^2 = 4.$

${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$ are in G.P.

${x_1} = 3,\,r = {1 \over 2}$

$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$

$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$

$\left\{ {\matrix{ {S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr {{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr } } \right.$

$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$

$\therefore$ $[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$

$ = 142$

Given ${{\sum\limits_{i = 1}^{10} {{x_i}} } \over {10}} = 15$ ..... (1)

$ \Rightarrow \sum\limits_{i = 1}^{10} {{x_i} = 150} $

and ${{\sum\limits_{i = 1}^{10} {x_i^2} } \over {10}} - {15^2} = 15$

$ \Rightarrow \sum\limits_{i = 1}^{10} {x_i^2 = 2400} $

Replacing 25 by 15 we get

$\sum\limits_{i = 1}^9 {{x_i} + 25 = 150} $

$ \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 125} $

$\therefore$ Correct mean $ = {{\sum\limits_{i = 1}^9 {{x_i} + 15} } \over {10}} = {{125 + 15} \over {10}} = 14$

Similarly, $\sum\limits_{i = 1}^2 {x_i^2 = 2400 - {{25}^2} = 1775} $

$\therefore$ Correct variance $ = {{\sum\limits_{i = 1}^9 {x_i^2 + {{15}^2}} } \over {10}} - {14^2}$

$ = {{1775 + 225} \over {10}} - {14^2} = 4$

$\therefore$ Correct $S.D = \sqrt 4 = 2$.

$\mu = {{\sum {{x_i}} } \over {40}} = 30 \Rightarrow \sum {{x_i} = 1200} $

${\sigma ^2} = {{\sum {x_i^2} } \over {40}} - {(30)^2} = 25 \Rightarrow \sum {x_i^2 = 37000} $

After omitting two wrong observations

$\sum {{y_i} = 1200 - 12 - 10 = 1178} $

$\sum {y_i^2 = 37000 - 144 - 100 = 36756} $

Now ${\sigma ^2} = {{\sum {y_i^2} } \over {38}} - {\left( {{{\sum {{y_i}} } \over {38}}} \right)^2}$

$ = {{36756} \over {38}} - {\left( {{{1178} \over {38}}} \right)^2} = - {31^2}$

$ = 38{\sigma ^2} = 36756 - 36518 = 238$

According to given data

${{\sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2}} } \over 7} = 20$

$ \Rightarrow \sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2} = 140} $

So for any x_i, ${({x_i} - 62)^2} \le 140$

$ \Rightarrow {x_i} > 50\,\forall i = 1,2,3,\,\,.....\,\,7$

So no student is going to score less than 50.

${{\sum {x_i^2} } \over {15}} - {8^2} = 9 \Rightarrow \sum {x_i^2 = 15 \times 73 = 1095} $

Let ${\overline x _c}$ be corrected mean ${\overline x _c}$ = 9

$\sum {x_c^2 = 1095 - 25 + 400 = 1470} $

Correct variance $ = {{1470} \over {15}} - {(9)^2} = 98 - 81 = 17$

Mean $ = {{n{{(n + 1)} \over 2}} \over n} = {{n + 1} \over 2}$

M.D. $ = {{2\left( {{{n - 1} \over 2} + {{n - 3} \over 2} + {{n - 5} \over 2} + \,\,\,...\,\,\,0} \right)} \over n} = {{5(n + 1)} \over n}$

$ \Rightarrow ((n - 1) + (n - 3) + (n - 5) + \,\,...\,\,0) = 5(n + 1)$

$ \Rightarrow \left( {{{n + 1} \over 4}} \right)\,.\,(n - 1) = 5(n + 1)$

So, $n = 21$