Please solve Q. No.3

Dear Student,
The mean of the following set of numbers is 
$$\begin{gathered} \mu =\frac{25+50+45+30+70+42+36+48+34+60}{10} \\ \mu =\frac{440}{10}=44 \end{gathered}$$
The formula for standard deviation is given by
$$\begin{gathered} \sigma =\sqrt{\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum {(x_i-\mu )}^2}}}}{n}} \\ {x}_i=Each value in the sample \\ \mu = mean of the sample \\ n = total number of samples \end{gathered}$$
Thus, the summation turns out to be:-
${(25-44)}^2+{(50-44)}^2+{(45-44)}^2+{(30-44)}^2+{(70-44)}^2+{(42-44)}^2+{(36-44)}^2+{(48-44)}^2+{(34-44)}^2+{(60-44)}^2$
$$\begin{gathered} 361+36+1+196+676+4+64+16+100+256 \\ i.e 1710 \end{gathered}$$
$Thus, \sigma =\sqrt{\frac{1710}{10}}=\sqrt{171}=13.07$
Regards.