JEE Main 2026 — 4 April, Evening Shift — Question 32

$\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}+2\right)^{2}=180$ $\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^{2}+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}+\sum_{\mathrm{i}=1}^{10} 4=180$

\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^{2}+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=180-40 \end{gathered}$$ Also $\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-1\right)^{2}=90$ $$\begin{aligned} & \sum_{i=1}^{10} x_{i}^{2}-2 \sum_{i=1}^{10} x_{i}+\sum_{i=1}^{10} 1=90 \\& \sum_{i=1}^{10} x_{i}^{2}-2 \sum_{i=1}^{10} x_{i}=90-10 \\& \sum_{i=1}^{10} x_{i}^{2}+4 \sum_{i=1}^{10} x_{i}=140 \\& \sum_{i=1}^{10} x_{i}^{2}-2 \sum_{i=1}^{10} x_{i}=80 \\& \sum_{i=1}^{10} x_{i}^{2}=100 \text { and } \sum_{i=1}^{10} x_{i}=10 \\& \sum^{2}=\frac{10}{10} x_{i}^{2} \\& N \\& =\frac{100}{10}-\left(\frac{\sum_{i=1}^{10} x_{i}}{N}\right)^{2} \\& \sigma^{2}=10-1=9 \\& \Rightarrow \sigma=3 \end{aligned}$$