QJEEM24Ma6ea1: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6ea1 Solution: Let be 10 observations such that and . Then the standard…

\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}

$\begin{aligned}& =\sqrt{\frac{\sum \mathrm{a}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\& =\sqrt{30-25}=\sqrt{5}\end{aligned}$