The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Then the correct standard deviation when the wrong item is omitted is

Mean $=10\Rightarrow \frac{{\sum }_{i=1}^{20}{x}_i}{20}=10\Rightarrow \sum _{i=1}^{20}{x}_i=200$

S.D.$=2\Rightarrow {\sigma }^2=4$

$\Rightarrow \frac{{\sum }_{i=1}^{20}{x}_i^2}{20}-(10{)}^2=4\Rightarrow \sum _{i=1}^{20}{x}_i^2=2080$

Thus, incorrect $\left(\sum _{i=1}^{20}{x}_i\right)=200$ and incorrect $\left(\sum _{i=1}^{20}{x}_i^2\right)=2080$ On omitting 8, we are left with 19 observations.

$\begin{array}{l}\therefore \mathrm{correct}\left(\sum _{i=1}^{19} x_i\right)=\mathrm{incorrect}\left(\sum _{i=1}^{20} x_i\right)-8\\ =(200-8)=192\end{array}$

Thus correct $\left(\sum _{i=1}^{19} x_i\right)=192$

$\therefore$Correct mean=$=\frac{192}{19}$

Also, correct $\left(\sum _{i=1}^{19} x_i^2\right)=$incorrect $\left(\sum _{i=1}^{19} x_i^2\right)-64$

$\begin{array}{l}=(2080-64)=2016\\ \therefore \text{ Correct S.D. }=\sqrt{\frac{1}{19}\left(\text{ correct }\sum _{i=1}^{19} x_i^2\right)-(\text{ correct mean }{)}^2}\\ =\sqrt{\left(\frac{1}{19}\times 2016\right)-{\left(\frac{192}{19}\right)}^2}=\sqrt{\frac{1440}{361}}=\frac{(12\times 3.162)}{19}=1.997\end{array}$