If the data is $x_1,x_2, \dots x_{10}$ such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000; then the standard deviation of this data is

Solution:

Given, $\frac{x_1+x_2+x_3+x_4}{4}=11$

$\Rightarrow x_1+x_2+x_3+x_4=44\to \left(i\right)$

Also, $\frac{x_5+x_6+x_7+x_8+x_9+x_{10}}{6}=96\to \left(ii\right)$

from (i) and (ii),$\sum _{i=1}^{10} x_i=44+96=140$

Now, Variance $\left({\sigma }^2\right)=\frac{\sum _{i=1}^{10} x_i^2}{10}-{\left(\frac{\sum _{i=1}^{10} x_i}{10}\right)}^2$

$\begin{array}{l}=\frac{2000}{10}-{\left(\frac{140}{10}\right)}^2\left[\text{ Given }\sum _{i=1}^{10} x_i^2=2000\right]\\ =200-196=4\end{array}$

$\therefore$ Standard deviation $\left(\sigma \right)=2$