Question

Mean and standard deviation of $100$ observations were found to be $40$ and $10$ respectively. If at the time of calculation two observations were wrongly taken as $30$ and $70$ in place of $3$ and $27$ respectively, find the correct S.D.

Answer 1

Formula to calculate $S.D$ is

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

Given $\overline{x}=40$ and $S.D=10,N=100$

$\frac{\sum x}{n}=40$

$\Rightarrow \frac{\sum x}{100}=4$

$\Rightarrow x=400$

$\therefore x=400-30-70+3+27=3930$

$\therefore$ correct mean $=\sum x/n=3930/10=39.30$

Given $S.D\left(\sigma \right)=10$

${\sigma }^2=100$

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

$\Rightarrow \frac{\sum x^2}{100}-(40{)}^2={10}^2$

$\Rightarrow \sum x^2/100={10}^2+{40}^2$

$\Rightarrow \sum x^2/100=100+1600$

$\Rightarrow \sum x^2=1700\times 100$

$\Rightarrow \sum x^2=170000$

$\therefore 30$ and $70$ should be replaced by $3$ and $27$

$\sum x^2=170000-900-4900+9+729=16+938$

$S.D\left(\sigma \right)=\sqrt{\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2}$

$=\sqrt{164938/100-{\left(\frac{3930}{100}\right)}^2}$

$=\sqrt{1649.38-1544.49}$

$=\sqrt{104.89}$

$S.D\left(\sigma \right)=10.241$

$\therefore$ The given answer is correct.