Question

The mean ans dtandard 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 in each of the following cases :

(i) If worng item is omitted

(ii) If it is replaced by 12.

Answer 1

(i) I fwrong item is omitted

Given $\overline{x}=10and\sigma =2,n=20$

$\Rightarrow \frac{\sum x_i}{20}=10$

$\Rightarrow \sum x_i=200$

If observation 8 is omitted, then

$\sum x_i=200-8=192$

Now, remaining number of observation = 19

$\Rightarrow Correctmean=\frac{\sum x_i}{n}=\frac{192}{19}=10.10$

Again, $\sigma =2\Rightarrow {\sigma }^2=4$

$\Rightarrow \frac{\sum x_i^2}{20}-(10{)}^2=4$

$\Rightarrow \sum x_i^2=(4+100)\times 20=104\times 20=2080$

If observation 8 is omitted , then

$\sum x_i^2=2080-64=2016$

Now, correct standard deviation $\sigma =\sqrt{\frac{2016}{19}-{\left(\frac{192}{19}\right)}^2}$

= $\sqrt{\frac{2016\times 19-(192{)}^2}{19\times 19}}$

=$\frac{1}{19}\sqrt{38304-36864}$

= $\frac{1}{19}\sqrt{1440}=\frac{37.95}{19}=1.99$

(ii) if it is replaced by 12.

Given $\overline{x}=10and\sigma =2,n=20$

$\Rightarrow \frac{\sum x}{20}=10$

$\Rightarrow \frac{\sum x}{=}200$

If observatio 8 is replaced by 12, then

$\sum x=200-8+12=192+12=204$

Correct mean = $\frac{204}{20}=\frac{102}{10}10.2$

Again , $\sigma =2\Rightarrow {\sigma }^2=4$

$\Rightarrow \frac{\sum x^2}{n}-(\overline{x}{)}^2=4$

$\Rightarrow \frac{\sum x^2}{20}-(10{)}^2=4$

$\Rightarrow \sum x^2=2080$

If observation 8 is replaced by 12, then

$\sum x^2=2080-(8{)}^2+(12{)}^2$

= 2080-64+144

=2224-64=2160

Now, correct standard deviation

$\sigma =\frac{\sum x^2}{n}-(\overline{x}{)}^2$

$=\sqrt{\frac{2016}{20}-{\left(\frac{204}{20}\right)}^2}$

$=\sqrt{\frac{2016\times 20-(204{)}^2}{20\times 20}}$

$=\frac{1}{200}\sqrt{43200-41616}$

$=\frac{1}{20}\sqrt{1584}=\frac{39.79}{20}=1.98$