Question

The mean and standard deviation of $20$ observation 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 wrong item is omitted
(ii) If it is replaced by $12$

Answer 1

(i) Given, number of observations $n=20$
Incorrect mean $=10$
Incorrect standard deviation $=2$
$\overline{x}=\frac{1}{n}\sum _{i=1}^{20}{x}_i$
$10=\frac{1}{20}\sum _{i=1}^{20}{x}_i$
$\Rightarrow \sum _{i=1}^{20}{x}_i=200$
So, the incorrect sum of observations $=200$
Correct sum of observation $=200-8=192$

$\Rightarrow$ Correct mean $=\frac{Correctsum}{19}=\frac{192}{19}=10.1$

Standard deviation $\sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-\frac{1}{n^2}{\left(\sum _{i=1}^n{x}_i\right)}^2}=\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\overline{x}\right)}^2}$

$\Rightarrow 2=\sqrt{\frac{1}{20}Incorrrect\sum _{i=1}^n{x}_i^2-(10{)}^2}$
$\Rightarrow 4=\frac{1}{20}Incorrect\sum _{i=1}^n{x}_i^2-100$
$\Rightarrow Incorrect\sum _{i=1}^n{x}_i^2=2080$
$\therefore$ Correct $\sum _{i=1}^n{x}_i^2=Incorrect\sum _{i=1}^n{x}_i^2-{\left(8\right)}^2$
$=2080-64$
$=2016$
$\therefore$ Correct Standard deviation $=\sqrt{\frac{Correct\sum x_i^2}{n}-(CorrectMean{)}^2}$
=$\sqrt{\frac{2016}{19}-(10.1{)}^2}$
$\sqrt{106.1-102.01}$
= $\sqrt{4.09}$
= 2.02

(ii) When $8$ is replaced by $12$
Incorrect sum of observation $=200$
$\therefore$ Correct sum of observations $=200-8+12=204$
$\therefore$ Correct mean = $\frac{Correctsum}{20}=\frac{204}{20}=10.2$

Standard deviation $\sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-\frac{1}{n^2}{\left(\sum _{i=1}^n{x}_i\right)}^2}=\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\overline{x}\right)}^2}$
$\Rightarrow$$2=\sqrt{\frac{1}{20}Incorrect\sum _{i=1}^n{x}_i^2-(10{)}^2}$
$\Rightarrow$ $4=\frac{1}{20}Incorrect\sum _{i=1}^n{x}_i^2-100$
$\Rightarrow Incorrect\sum _{i=1}^n{x}_i^2=2080$
$\therefore Correct\sum _{i=1}^n{x}_i^2=Incorrect\sum _{i=1}^n{x}_i^2-(8{)}^2+(12{)}^2$
$=2080-64+144$
$=2160$
$Correctstandarddeviation=\sqrt{\frac{Correct\sum x_i^2}{n}-(Correctmean{)}^2}$
= $\sqrt{\frac{2160}{20}-(10.2{)}^2}$
$=\sqrt{108-104.04}$
$=\sqrt{3.96}$
$=1.98$