# 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.

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

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.