For the following data table

xi	fi
0 – 4 4 – 8 8 – 12 12 – 16 16 – 20	2 4 7 8 6

Find the value of 20 M (where M is median of the data)

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

$\frac{60^2+60^2+44^2+58^2+68^2+{\alpha }^2+{\beta }^2+56^2}{8}={\left(58\right)}^2=66.2$            

$\frac{7200+1936+3364+4624+3136+{\alpha }^2+{\beta }^2}{8}=3364=66.2$             

$2532.5+\frac{{\alpha }^2+{\beta }^2}{8}-3364=66.2$            

α² + β² = 897.7 × 8

= 7181.6

  $\frac{a+b+68+44+40+60}{6}=55$

212 + a + b = 330

⇒ a + b = 118

$\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\overline{x}\right)}^2=194$          

$\frac{a^2+b^2+{\left(68\right)}^2+{\left(44\right)}^2+{\left(40\right)}^2+{\left(60\right)}^2}{6}={\left(55\right)}^2=194$

= 3219

11760 + a² + b² = 19314

⇒ a² + b² = 19314 – 11760

= 7554

(a + b)² –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

Kindly go throuigh the solution

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

->(a - b) (a - b + 4) = 0

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

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance = $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$      

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4\lambda or4\lambda +1from.$  

As each square form is $4\lambda or4\lambda +1$  

Variance of 2001, 2003, 2006, 2007, 2009, 2010 is same as variance of 1, 3, 6, 7, 9, 10

mean =  $\frac{1+3+6+7+9+10}{6}=6$

var (x) =  $\frac{1^2+3^2+6^2+7^2+9^2+10^2}{6}?36$

var (x) = 10