Question

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 were recorded as $21,21$ and $18$. Find the mean and standard deviation if the incorrect observation are omitted

Answer 1

Given, number of observation $n=100$
Incorrect mean $\left(\overline{x}\right)=20$

We know that

$\overline{x}=\frac{\sum x_i}{n}$
$\Rightarrow 20=\frac{1}{100}\sum _{i=1}^{100}{x}_i$
$\Rightarrow \sum _{i=l}^{100}{x}_i=20\times 100=2000$
So, the incorrect sum of observations $=2000$.

Given incorrect observations are $21,21,18$ and these has to be omitted.
So, correct sum of observations $=2000-21-21-18=2000-60=1940$
Correct mean $=\frac{Correctsum}{100-3}=\frac{1940}{97}=20$

Given, incorrect standard deviation $\sigma =3$

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

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

$\Rightarrow 3^2=\frac{\sum x_i^2}{100}-(20{)}^2$
$\Rightarrow Incorrect\sum x_i^2=100(9+400)=40900$

Correct $\sum _{i=l}^n{x}_i^2=Incorrect\sum _{i=l}^n{x}_i^2-(21{)}^2-(21{)}^2-(18{)}^2$
$=40900-441-441-324$
$=39694$

$\therefore$ Correct standard deviation = $\sqrt{\frac{correct\sum x_i^2}{n}-(Correctmean{)}^2}$
$=\sqrt{\frac{39694}{97}-(20{)}^2}$
$=\sqrt{409.216-400}$
$=\sqrt{9.216}$
$=3.036$