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Home  >>  CBSE XI  >>  Math  >>  Statistics
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The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation If it is replaced by 12.

$\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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1 Answer

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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.
answered Jul 2, 2014 by meena.p
 

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