# The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively . Later on it was found that three observations were incorrect. Which are recorded by 21,21, and 18 . Find the mean and standard deviation, if the incorrect observations are omitted.

$\begin{array}{1 1}(A)\;10.10,1.99\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;20,3.03\end{array}$

Given , number of observations =100
Mean $\bar {x}=20$
Standard deviation $\sigma=3$
Step 1:
$\therefore \large\frac{\sum x_i}{n}$$=\bar {x} \qquad= \large\frac{\sum x_i}{100}$$=20$
$\qquad= \sum x_i =2000$
The incorrect observations $21,21$ and $18$ are omitted.
$\therefore$ Correct mean of 97 observations
$\qquad= \large\frac{1940}{97} $$=20 Step 2: Standard deviation \sigma=3 => \sqrt {\large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i}{n}\bigg)^2}$$=\sigma$
$\sqrt {\large\frac{\sum x_i ^2}{100 }- \normalsize (20)^2}=3$
$\large\frac{\sum x_i^2}{100} $$-400 =9 \large\frac{\sum x_i^2}{100}$$=409$
=> $\sum x_i^2 =40900$
By omitting the correct observations 21,21,18
$\sum x+i^2 =40900 -(21)^2 -(21)^2-(18)^2$
$\qquad= 40900 -441-441-324$
$\qquad= 39694$
Correct Standard deviation of 97 observations
$\sigma= \sqrt { \large\frac{39694}{97} - \bigg( \large\frac{1940}{97} \bigg)^2}$
$\qquad= \sqrt {409.2 -(20)^2}$
$\qquad= \sqrt { 409.2 -400 }$
$\qquad= \sqrt {9.2}$
$\qquad= 3.03$
Hence D is the correct answer.