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 $\sigma=10$

$\therefore \quad \frac{\Sigma x_{i}}{n}=40$

$\Rightarrow \quad \frac{\Sigma x_{i}}{100}=40$

$\Rightarrow \quad \Sigma x_{i}=4000$

Now, Corrected $\Sigma x_{i}=4000-30-70+3+27=3930$

Corrected mean $=\frac{2930}{100}=39.3$

Now, $\sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}=\frac{\Sigma x_{i}^{2}}{n}-(40)^{2}$

$\Rightarrow \quad 100=\frac{\Sigma x_{i}^{2}}{100}-1600$

$\Rightarrow \quad \Sigma x_{i}^{2}=170000$

Now, $\quad$ Corrected $\Sigma x_{i}^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}=164938$

Corrected $\sigma=\sqrt{\frac{164938}{100}-(39.3)^{2}}=\sqrt{1649.38-1544.49}=\sqrt{104.9}$

$=10.24$