$\begin{array}{1 1}(A)\;10.2,1.98\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;5,7\end{array} $

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If it is replaced by 12,

Given mean $\bar{x}=10$

Standard deviation $(\sigma) =2$

Number of observations =20

Step 1:

$\therefore \large\frac{\sum x_i}{20} $$=10$

$\sum x_i=200$

If observation 8 is replayed by 12

$\sum x = 200-8+12$

$\qquad= 192+12=204$

$\therefore \sum x =204$

Correct mean $=\large\frac{204}{20}$

$\qquad= 10.2$

Step 2:

Given standard deviation $\sigma=2$

Variance $\sigma ^2=4$

$\large\frac{\sum x_i^2}{n} -\bigg(\large\frac{\sum x_i}{n}\bigg)^2$$=4$

$\therefore \large\frac{\sum x_i^2}{20} $$=4+100$

=> $\sum x_i^2 = 104 \times 20$

$\qquad= 2080$

If the observation 8 is replaced by 12,then

$\sum x_i^2 =2080 -8^2+12^2$

$\qquad = 2080 -64+144$

$\qquad= 2224-64$

$\qquad= 2160$

Step 3:

Correct standard deviation

$\sigma= \sqrt {\large\frac{\sum x_i^2}{n} -\bigg(\large\frac{\sum x_i}{n}\bigg)^2}$

$\qquad= \sqrt { \large\frac{2160}{20} - \bigg( \large\frac{204}{20} \bigg)^2}$

$\qquad= \sqrt {\large\frac{2160 \times 20 -(204)^2}{20 \times 20}}$

$\qquad= \large\frac{1}{20} $$\sqrt{43200-41616}$

$\qquad= \large\frac{1}{20} $$ \sqrt { 1584}$

$\qquad= 1.98$

Hence A is the correct answer.

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