Mean and variance of $20$ observations are $10$ and $4$ respectively. It was found that in place of $11, 9$ was taken by mistake. The correct variance, is

Let $x_1,x_2,\dots ,x_{20}$ be $20$ observations. Then,

           uncorrected $\sum x_i=20\times 10=200$ and $\frac{1}{20}\sum x_i^2-(10{)}^2=4$

$\Rightarrow$uncorrected $\sum x_i^2=2080$

    Corrected $\sum x_i=200-9+11=202$

     Corrected $\sum x_i^2=2080-9^2+{11}^2=2120$

 $\therefore$Correct variance $=\frac{1}{20}\left(\text{ Corrected }\sum x_i\right)-{\left(\frac{\text{ Corrected }\sum x_i}{20}\right)}^2$

                                  $=\frac{2120}{20}-{\left(\frac{202}{20}\right)}^2=106-102.01=3.99$