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

Answer: 2

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