Let $\mathrm{a}_1, \mathrm{a}_2, \ldots \mathrm{a}_{10}$ be 10 observations such that $\sum\limits_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum\limits_{\forall \mathrm{k} < \mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{10}$ is equal to :

<p>$$\begin{aligned} & \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 \\ & \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 \quad \text{.... (i)}\\ & \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 \quad \text{.... (ii)}\\ & \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 . \\ & \left(\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}\right)^2=2500 \\ & \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2+2 \sum_{\mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=2500 \\ & \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=2500-2(1100) \\ & \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=300 \text {, Standard deviation ‘ } \sigma \text { ’ } \end{aligned}$$</p> <p>$$\begin{aligned} & =\sqrt{\frac{\sum \mathrm{a}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\ & =\sqrt{30-25}=\sqrt{5} \end{aligned}$$</p>