$\begin{aligned}
&\text { Mean }=\frac{\sum x_i}{n}=\frac{\sum_{i=1}^{20} x_i}{20}=10 \\
&\Rightarrow \sum_{i=1}^{20} x_i=20 \times 10=200
\end{aligned}$
(i) Since, the wrong item whose observation 8 has been omitted, the correct mean $=\frac{\sum_{i=1}^{20} x-8}{19}$ because on omitting the wrong item, the total number of observations left are 19.
Now, $\frac{\sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2}{n}=\frac{\sum_{i=1}^{20} x_i^2}{n}-\bar{x}^2$
Since, incorrect mean and variance are given as 10 and 4 ,
so $\frac{\sum_{i=1}^{20} x_i^2}{20}-10^2=4$
i.e., $\sum_{i=1}^{20} x_i^2=(4+100) 20=104 \times 20=2080$
Correct variance $=\frac{\sum_{i=1}^{19}\left(x_i-10.11\right)^2}{19}$
i.e., $\frac{\sum_{i=1}^{20} x_i^2-8^2}{19}-(10.11)^2$
$\frac{2080-64}{19}-(10.11)^2=\frac{2016}{19}-(10.11)^2$
$=106.105-102.212=3.893$
Correct S.D $=\sqrt{3.893}=1.97$
(ii) The wrong item which was 8 had been replaced by 12 , as such the correct mean
$=\frac{\sum_{i=1}^{20} x_i-8+12}{20}=\frac{\sum_{i=1}^{20} x_i}{20}+\frac{4}{20}$
$=\frac{200}{20}+\frac{1}{5}=10+0.2=10.2$
Since, incorrect mean and variance are given as 10 and 4 ,
So
$\begin{aligned}
&\frac{\sum_{i=1}^{20} x_i^2}{20}-10^2=4 \Rightarrow \frac{\sum_{i=1}^{20} x_i^2}{20}=4+100=104 \\
&\Rightarrow \sum_{i=1}^{20} x_i^2=20 \times 104 \quad \text { i.e } \sum_{i=1}^{20} x_i^2=2080 \\
&\Rightarrow \text { Correct value of } \sum_i x_i^2=2080-8^2+12^2 \\
&=2080-64+144=2160
\end{aligned}$
Correct variance $=\frac{\sum_{i=1}^{20} x_i^2}{20}-10.2^2$
$=\frac{2160}{20}-(10.2)^2=108-104.04=3.96$
$\therefore \quad$ Correct $\mathrm{SD}=\sqrt{\text { variance }}$
or $\quad \sqrt{3.96}=1.99(\mathrm{appx})$