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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.

$\begin{array}{1 1}(A)\;10.2,1.98\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;5,7\end{array}$

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.