(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$