Statistics

σ²=variance
µ=mean
σ² = Σ (x? -µ)²/n
µ=17
⇒ Σ (ax+b)/17 = 17
⇒ 9a+b=17
σ²=216
⇒ Σ (ax+b-17)²/17 = 216
⇒ Σa² (x-9)²/17 = 216
⇒ a² (81-18*9+3*35) = 216
⇒ a² (24) = 216 ⇒ a²=9 ⇒ a=3 (a>0)
⇒ From (1), b=-10
So, a+b=-7

Mean = 10 = (3+7+9+12+13+20+x+y)/8
16 = x + y
Variance σ² = 25 = (Σx?²/8) - (mean)²
25 = (3²+7²+9²+12²+13²+20²+x²+y²)/8 = 100
x² + y² = 148
(x+y)² = x² + y² + 2xy
256 = 148 + 2xy
x . y = 54

5+3+7+a+b = 25 ⇒ a+b=10
S.D. = √(((5²+3²+7²+a²+b²)/5) - 5²) = 2
√((a²+b²+83)/5) - 25 = 4 ⇒ a²+b² = 62
⇒ (a+b)² - 2ab = 62 ⇒ ab = 19
So equation whose roots are a and b is x² - 10x + 19 = 0

Mean = (6+10+7+13+a+12+b+12)/8 = (60+a+b)/8 = 9 ⇒ a+b=12.
Variance = (Σx? ²/n) - (mean)² = 37/4.
Σx? ²/8 - 81 = 37/4.
Σx? ² = 8 (81+9.25) = 8 (90.25) = 722.
Σx? ² = 36+100+49+169+a²+144+b²+144 = 642+a²+b².
642+a²+b²=722 ⇒ a²+b²=80.
(a+b)²=144 ⇒ a²+b²+2ab=144 ⇒ 80+2ab=144 ⇒ 2ab=64.
(a-b)² = a²+b²-2ab = 80-64 = 16.

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$         

(i) & (ii)  ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

=> (α - β) (α - β + 4) = 0

Since   $\alpha \ne \beta so\left|\alpha -\beta \right|=4$

$ForMedian=L+\frac{\frac{N}{2}-c,}{f}xh$

$14=12+\frac{\frac{a+b}{2}+13-\left(a+b\right)}{12}*6$              

->a + b = 18                                   -(ii)

Solving (i) & (ii) a = 8 & b = 10

->(a – b)² = 4

For every a, there  must be a² – 2. So, there will be infinitely many pairs (a, b)

Variance = $\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n}=\frac{\sum \left({x_i}^2+{\overline{x}}^2-2\overline{x}xi\right)}{n}$  

$=\frac{\sum {x_i}^2+{\overline{x}}^2\sum 1-2\overline{x}\sum x_i}{n}$

$=\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(\frac{n\left(n+1\right)}{2n}\right)}^2.n-\frac{2\left(n+1\right)}{2n}.\frac{n\left(n+1\right)}{2}}{n}$ $=\frac{n^2-1}{12}$

Now, $\frac{n^2-1}{12}=14son=13$