The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is

Mean  $=10\Rightarrow \frac{{\displaystyle \sum x_i}}{20}=10\Rightarrow \sum x_i=200.$
Corrected $\sum x_i=200-9+11=202\Rightarrow$  corrected mean  $=\frac{202}{20}=10.1$
Variance  $=4\Rightarrow \frac{{\displaystyle \sum x_i^2}}{20}-{10}^2=4\Rightarrow \frac{{\displaystyle \sum x_i^2}}{20}=104\Rightarrow \sum x_i^2=2080.$
Corrected  $\sum x_i^2=2080-9^2+{11}^2=2120$
Corrected variance  $=\frac{2120}{20}-{\left(10.1\right)}^2=106-102.01=3.99$