Question

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is ______

Answer 1

$ n =10, \bar{x}=\frac{\sum x_i}{10}=15 $
$6^2=\frac{\sum x_i^2}{10}-(\bar{x})^2=15 $
$ \Rightarrow \displaystyle\sum_{i=1}^{10} x_i=150$
$ \Rightarrow \displaystyle\sum_{i=1}^9 x_i+25=150 $
$ \Rightarrow \displaystyle\sum_{i=1}^9 x_i=125 $
$ \Rightarrow \displaystyle\sum_{i=1}^9 x_i+15=140$
Actual mean $=\frac{140}{10}=14=\bar{x}_{\text {new }} $
$\displaystyle\sum_{i=1}^9 \frac{x_i^2+25^2-15^2}{10}=15$
$ \Rightarrow \displaystyle\sum_{i=1}^9 x_i^2+625=2400$
$ \displaystyle\sum_{i=1}^9 x_i^2=1775 $
$\displaystyle \sum_{i=1}^9 x_i^2+15^2=2000=\left(\sum x_i^2\right)_{\text {actual }} $
$ 6_{\text {actual }}^2=\frac{\left(\sum x_i^2\right)_{\text {actual }}-\left(\bar{x}_{\text {new }}\right)^2}{10} $
$ =\frac{2000}{10}-14^2$
$ =200-196=4 $
(S.D) actual $=6=2$