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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 wrong item is omitted.

$\begin{array}{1 1}(A)\;10.10,1.99\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;5,7\end{array} $

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+1 vote
If wrong item is omitted,
Given mean $\bar{x}=10$
Standard deviation $(\sigma) =2$
Number of observations =20
Step 1:
$\therefore \large\frac{\sum x_i}{n}$$ =\bar {x}$
$\therefore \large\frac{\sum x_i}{20} $$=10$
$\sum x_i=200$
Given wrong observation is 8. and it is omitted
$\therefore \sum x_i =200-8$
$\qquad= 192$
and remaining number of observations $=19$
$\therefore $ Correct Mean $=\large\frac{\sum x_i}{n}$$=\large\frac{192}{19}$
$\qquad= 10.10$
Step 2:
Given Standard deviation $(\sigma)=2$
$\therefore $ Variance $(\sigma^2) =2^2=4$
$\therefore \large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i}{n}\bigg)^2$$=4$
$=> \large\frac{\sum x_i^2}{20} $$-10^2 =4$
$=>\sum x_i^2 =104 \times 20$
$\qquad= 2080$
If wrong observation '8' is omitted. then,
$\sum x_i^2 =2080 -8^2$
$\qquad= 2080-64$
$\qquad= 2016$
Step 3:
Correct standard deviation $\sigma =\sqrt { \large\frac{2016}{19} -\bigg( \large\frac{192}{19}\bigg)^2}$
$\qquad= \sqrt { \large\frac{2016 \times 19 -(192)^2}{19 \times 19}}$
$\qquad= \large\frac{1}{19} $$\sqrt {38304-36864}$
$\qquad= \large\frac{1}{19} $$\sqrt {1440}$
$\qquad= 1.99$
Hence A is the correct answer.
answered Jul 2, 2014 by meena.p

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