The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is ________

x₁ + x₂ + ....... x₈ + 45 + 50 = 500
x₁ + x₂ + ....... x₈ = 405
x₁ + x₂ + ....... x₈ + 20 + 25 = 405 + 20 + 25
$\begin{array}{l}\mathrm{\mu }=\frac{450}{10}=45\\ \text{Given }{\mathrm{\sigma }}^2=144\\ \frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{10}-(50{)}^2=144\\ {\mathrm{x}}_1^2+\dots \dots {\mathrm{x}}_8^2+{45}^2+{50}^2=144+2500=2644\\ {\mathrm{x}}_1^2+\dots \dots {\mathrm{x}}_8^2=26440-\left({45}^2+{50}^2\right)=21915\\ {\mathrm{x}}_1^2+\dots \dots {\mathrm{x}}_8^2+{20}^2+{25}^2=21915+400+625=22940\\ {\mathrm{\sigma }}^2=\frac{22940}{10}-{45}^2=269\end{array}$