Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

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 $4\lambda or4\lambda +1$