$\begin{array}{1 1}(A)\;144,12\\(B)\;5,8\\(C)\;3,8\\(D)\;5,7\end{array} $

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Given Mean $\bar {x} =8$

Variance $\sigma ^2= 4$

Number of observation =6

Step 1:

Let us assume the observations as $x_1,x_2,x_3,x_4 \;and\;x_6$

Mean $\bar{x} =\large\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$

$8= \large\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$

$48= x_1+x_2+x_3+x_4+x_5+x_6$

Step 2:

If each observation is multiplied by 3 the new mean will be

$3x_1+3x_2+3x_3+3x_4+3x_5+3x_6=48 \times 3$

$\sum \limits_{i=1}^{6} x_i =144$

$\therefore$ New Mean $\bar{x}=\large\frac{144}{6}$$=24$

Hence the new mean =24

Step 3:

Given standard deviation =4

Variance $=4^2=16$

$\therefore \large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i}{n}\bigg)^2$$=16$

$\therefore \large\frac{\sum x_i^2}{6} $$-8^2=16$

=> $\large\frac{\sum x_i^2}{6}$$=16+64$

=> $\sum x_i^2 =480$

Step 4:

New variance can be calculated as

New $ \sum x_i^2=(3x_1)^2+(3x_2)^2+(3x_3)^2+(3x_4)^2+3(x_5)^2+3(x_6)^2$

$\qquad= 9(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)$

$\qquad= 9 \times 480$

$\qquad= 4320$

New variance $=\large\frac{\sum x_1^2}{n} $$-(\bar{x})^2$

$\qquad= \large\frac{4320}{6} $$-(24)^2$

$\qquad= 720-576$

$\qquad =144$

New standard deviation $= \sqrt {144}$

$\qquad= 12$

Hence A is the correct answer.

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