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Q)

Mean and standard deviation of $100$ observation were found to be $40$ and $10$ respectively . If at the times of calculation two observations were wrongly takes as $30$ and $70$ in place of $3$ and $27$ respectively. Find the correct standard deviation.

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

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A)
Toolbox:
• 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.