Question

Mean and standard deviation of 100 observations were found to be 40 and 10,respectively.if at the time of calclation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard devation

Answer 1

Given :$n=100.\overline{x}=40,\sigma =10$

$\because \frac{\sum x_i}{n}=\overline{x}$

in correct $\sum x_i=40\times 100=4000$

$\because$observation were wrongly taken as 30 and 70 in place of 3 and 27 respectively.

$\Rightarrow$correct$\sum x_i=4000-30-70+3+27=3930$

$\therefore$correct mean$\frac{3930}{100}=39.3$

Now,

${\sigma }_{incorrect}^2=\frac{\sum x_{incorrect}^2}{n}-{\left(\frac{\sum x_{iincorrect}}{n}\right)}^2$

$100=\frac{\sum x_{incorrect}^2}{100}-(40{)}^2$

$\sum x_{iincorrect}^2=(100+1600)100$

incorrect$\sum x_{iincorrect}^2=170000$

$\Rightarrow \sum x_{iincorrect}^2=170000-(30{)}^2-(70{)}^2+\left(3^2\right)+(27{)}^2$

$\Rightarrow \sum x_{iincorrect}^2=164938$

correct $\sigma =\sqrt{\frac{164938}{100}-(39.3{)}^2}=10.24$

Hence, the correct standard devation is $10.24$