$\begin{array}{1 1}(A)\;10.10,1.99\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;5,7\end{array} $

If wrong item is omitted,

Given mean $\bar{x}=10$

Standard deviation $(\sigma) =2$

Number of observations =20

Step 1:

$\therefore \large\frac{\sum x_i}{n}$$ =\bar {x}$

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

$\sum x_i=200$

Given wrong observation is 8. and it is omitted

$\therefore \sum x_i =200-8$

$\qquad= 192$

and remaining number of observations $=19$

$\therefore $ Correct Mean $=\large\frac{\sum x_i}{n}$$=\large\frac{192}{19}$

$\qquad= 10.10$

Step 2:

Given Standard deviation $(\sigma)=2$

$\therefore $ Variance $(\sigma^2) =2^2=4$

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

$=> \large\frac{\sum x_i^2}{20} $$-10^2 =4$

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

$\qquad= 2080$

If wrong observation '8' is omitted. then,

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

$\qquad= 2080-64$

$\qquad= 2016$

Step 3:

Correct standard deviation $\sigma =\sqrt { \large\frac{2016}{19} -\bigg( \large\frac{192}{19}\bigg)^2}$

$\qquad= \sqrt { \large\frac{2016 \times 19 -(192)^2}{19 \times 19}}$

$\qquad= \large\frac{1}{19} $$\sqrt {38304-36864}$

$\qquad= \large\frac{1}{19} $$\sqrt {1440}$

$\qquad= 1.99$

Hence A is the correct answer.

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