Let  $x_1,x_2,x_3,\mathrm{.....},x_{100}$  are 100 observations such that  $\sum x_i=0,i=1,2,3,\mathrm{...},100,$  $\sum _{1\le i<j\le 100}\left|x_i{x}_j\right|=80000$  & mean deviation from their mean is 5, then the greatest integer less than or equal to square root of their standard deviation is equal to ____

$\sum x_i=0$

 Mean = 0
Mean deviation from mean =  $\sum _{i=1}^1\frac{1}{n}|x_i-\overline{x}|=\sum _{i=1}^{100}\frac{1}{100}|x_i-0|$
$$\begin{gathered} 5=\frac{1}{100}(\left|x_1\right|+\left|x_2\right|+\mathrm{....}+\left|x_{100}\right|) \\ 500=(\left|x_1\right|+\left|x_2\right|+\mathrm{....}+\left|x_{100}\right|) \\ 250000=\sum _{i=1}^{100}{x}_i^2+2\left(80000\right) \\ 90000=\sum _{i=1}^{100}{x}_i^2 \end{gathered}$$

Standard deviation =  $\sqrt{\frac{{\displaystyle \sum _{i=1}^{100}{(x_i-\overline{x})}^2}}{100}}=\sqrt{\frac{90000}{100}}=30$

$\left[\sqrt{\text{Standard deviation}}\right]=\left[\sqrt{30}\right]=5$