(i) When the wrong item is omitted,

We are given,

The number of observations, $n = 20$.

The mean before rechecking $ = 10$.

The standard deviation before rechecking $ = 2$.

Mean is given by,

$\overline {\text{X}}  = \frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}} $

$10 = \frac{1}{{20}}\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}} $

$\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}}  = 200$

So, the sum of observations before rechecking $ = 200$

Thus, the correct sum of observations $ = 200 – 8$

$ = 192$

Hence, the correct mean $ = \frac{{correct{\text{ }}sum}}{{19}}$

$ = \frac{{192}}{{19}}$

$ = 10.1$

Also, the standard deviation is given by,

$\sigma  = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{X}}_1}}  – \frac{1}{{{{\text{n}}^2}}}{{\left( {\sum\limits_{i = 1}^{\text{n}} {\text{X}} } \right)}^2}} $

\[2 = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {X_1^2}  – {{(\bar X)}^2}} \]

Squaring on both sides and simplifying we get,

\[4 = \frac{1}{{20}}{\text{ Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  – 100{\text{ }}\]

\[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  = 2080\]

Now, \[Correct\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  = Incorrect\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  – {(8)^2}\]

\[ = 2080 – 64\]

\[ = 2016\]

Therefore, the correct standard deviation is,

$ = \sqrt {\frac{{{\text{ Correct }}\sum {{\text{X}}_1^2} }}{{\text{n}}} – {{({\text{ Correct Mean }})}^2}} $

$ = \sqrt {\frac{{2016}}{{19}} – {{(10.1)}^2}} $

$ = \sqrt {1061.1 – 102.1} $

$ = 2.02$

(ii) when the wrong item is replaced by $12$,

We are given,

The sum of observations before rechecking $ = 200$

The correct sum of observations $ = 200 – 8 + 12$

$ = 204$

So, the correct mean $ = \frac{{correct{\text{ }}sum}}{{20}}$

$ = \frac{{204}}{{20}}$

$ = 10.2$

Also, the standard deviation is given by,

$\sigma  = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{X}}_1}}  – \frac{1}{{{{\text{n}}^2}}}{{\left( {\sum\limits_{i = 1}^{\text{n}} {\text{X}} } \right)}^2}} $

\[2 = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {X_1^2}  – {{(\bar X)}^2}} \]

Squaring on both sides and simplifying we get,

\[4 = \frac{1}{{20}}{\text{ Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  – 100{\text{ }}\]

\[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  = 2080\]

Now, \[Correct\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  = Incorrect\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  – {(8)^2} + {(12)^2}\]

\[ = 2080 – 64 + 144\]

\[ = 2160\]

Therefore, the correct standard deviation is,

$ = \sqrt {\frac{{{\text{ Correct }}\sum {{\text{X}}_1^2} }}{{\text{n}}} – {{({\text{ Correct Mean }})}^2}} $

$ = \sqrt {\frac{{2160}}{{20}} – {{(10.2)}^2}} $

$ = \sqrt {108 – 104.04} $

$ = 1.98$