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 $\sum x_{i}^{2} = 2080 - 9^{2} + 11^{2} = 2120$

$∴$ Correct variance $= \frac{1}{20} (Corrected \sum x_{i}) - {(\frac{Corrected \sum x_{i}}{20})}^{2}$

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

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