The arithmetic mean and standard deviation of a data of nine numbers are 13 and 5 respectively.If 3 is included as 10th item of the data,then the variance of the data of 10 numbers is

Given n=9
$\overline{x}=\frac{{\displaystyle \sum _{i=1}^9{x}_i}}{9}=13\Rightarrow \sum _{i=1}^9{x}_i=117$
$\sigma =5\Rightarrow {\sigma }^2=25\Rightarrow \frac{{\displaystyle \sum _{i=1}^9{x}_i^2}}{9}-{13}^2=25$
$$$\Rightarrow \sum _{i=1}^9{x}_i^2=9(25+169)=1746$
Now after including 10th item as 3
New mean$=\frac{117+3}{10}=\frac{120}{10}=12$
New variance =  
$=\frac{1755-1440}{10}=\frac{315}{10}=31.5$