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.

Given,$n=100, \bar{x}=40$ and $\sigma=10$.
$\Sigma x_{i} = n \times \bar{x} = 100 \times 40 = 4000$.
Corrected $\Sigma x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930$.
Corrected mean $\bar{x}_{\text{new}} = \frac{3930}{100} = 39.3$.
We know $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$.
$100 = \frac{\Sigma x_{i}^{2}}{100} - (40)^{2} \Rightarrow \Sigma x_{i}^{2} = 100(100 + 1600) = 170000$.
Corrected $\Sigma x_{i}^{2} = 170000 - (30)^{2} - (70)^{2} + (3)^{2} + (27)^{2} = 170000 - 900 - 4900 + 9 + 729 = 164938$.
Corrected $\sigma_{\text{new}} = \sqrt{\frac{\Sigma x_{i}^{2}}{n} - (\bar{x}_{\text{new}})^{2}} = \sqrt{\frac{164938}{100} - (39.3)^{2}}$.
$= \sqrt{1649.38 - 1544.49} = \sqrt{104.89} \approx 10.24$.