34. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

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

$N={\displaystyle \sum _{}^{}f}=27$

$\left(\frac{N}{2}\right)=\frac{27}{2}=13.5$

So, we have median lies in the class 12 – 16

I₁ = 12, f = 8, h = 4, c.f. = 13

So, here we apply formula

$M=I_1+\frac{\frac{N}{2}-c.f.}{f}\times h=12+\frac{13.5-13}{8}\times 4$

$=12+\frac{5}{2}$

$M=\frac{24.5}{2}=12.25$

20 M = 20 × 12.25

= 245

  $\frac{a+b+68+44+40+60}{6}=55$

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 $4\lambda or4\lambda +1$