$\begin{array}{1 1}(A)\; 39.30,10.24\\(B)\;1.6,1.4\\(C)\;16,10\\(D)\;6,45\end{array} $

- Formula to calculate SD is $\sigma= \sqrt { \large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n}\bigg)^2}$
- Given $\bar {X}=40 $ and $SD =10,N= 100$

Step 1:

$\large\frac{\sum x}{n} $$=40$

$\large\frac{\sum x}{100} $$=40=> \sum x=4000$

$\therefore x =4000-30-70+3+27$

$\qquad= 3930$

Correct mean $= \large\frac{\sum x}{n} =\frac{3930}{100}$

$\qquad= 39.30$

Step 2:

Given $SD(\sigma)=10$

$\sigma^2=100$

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

$\large\frac{\sum x^2}{100} $$-(40^2) =10^2$

=> $ \large\frac{\sum x^2}{100}$$=10^2+40^2$

=> $ \large\frac{\sum x^2}{100}$$=100+1600$

$\large\frac{\sum x^2}{100} $$=1700$

$\sum x^2=170000$

$\therefore 30$ and $70$ should be replaced by $3 \;and \;7$

$\sum x^2 =170000-900-4900+9+729$

$\qquad=164938$

Step 3:

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

$\qquad= \sqrt { \large\frac{164938}{100} - \bigg( \large\frac{3930}{100} \bigg)^2 }$

$\qquad= \sqrt { \large\frac{164938}{100}-1544.49}$

$\qquad =\sqrt {1649.38-1544.49}$

$\qquad= \sqrt {104.89}$

$\qquad= 10.241$

Hence A is the correct answer.

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