Given: $\sum _{\mathrm{k}=1}^{10}{\mathrm{a}}_{\mathrm{k}}=50$

$\Rightarrow {\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{10}=50 ...\left(i\right)$

$\Rightarrow {\left({\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{10}\right)}^2=2500$

$\Rightarrow \sum _{\mathrm{i}=1}^{10}{\mathrm{a}}_{\mathrm{i}}^2+2\sum _{\mathrm{k}<\mathrm{j}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{j}}=2500$

$\Rightarrow \sum _{\mathrm{i}=1}^{10}{\mathrm{a}}_{\mathrm{i}}^2+2\times 1100=2500$

$\Rightarrow \sum _{\mathrm{i}=1}^{10}{\mathrm{a}}_{\mathrm{i}}^2=2500-2\left(1100\right)$

$\Rightarrow \sum _{\mathrm{i}=1}^{10}{\mathrm{a}}_{\mathrm{i}}^2=300$

 Standard deviation, $\mathrm{\sigma }=\sqrt{\frac{\sum {\mathrm{a}}_{\mathrm{i}}^2}{\mathrm{n}}-{\left(\frac{\sum {\mathrm{a}}_{\mathrm{i}}}{\mathrm{n}}\right)}^2}$

$\Rightarrow \mathrm{\sigma }=\sqrt{\frac{\sum {\mathrm{a}}_{\mathrm{i}}^2}{10}-{\left(\frac{\sum {\mathrm{a}}_{\mathrm{i}}}{10}\right)}^2}$

$\Rightarrow \mathrm{\sigma }=\sqrt{\frac{300}{10}-{\left(\frac{50}{10}\right)}^2}$

$\Rightarrow \mathrm{\sigma }=\sqrt{30-25}=\sqrt{5}$