For \(10\) observations \({x}_{1},{x}_{2},\ldots ,{x}_{10}\), if \(\sum _{i=1}^{10}{\left({x}_{i}+2\right)}^{2}=180\) an…

\(\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}+2\right)^2=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}+\sum_{\mathrm{i}=1}^{10} 4=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=180-40=140 \ldots\ (1)\)
\( \text { Also } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-1\right)^2=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i+\sum_{i=1}^{10} 1=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i=90-10=80 \ldots\ (2)\)
From (1) and (2) we get:
\( \sum_{i=1}^{10} x_i^2=100 \text { and } \sum_{i=1}^{10} x_i=10 \)
\( \sigma^2=\frac{\sum_{i=1}^{10} x_i^2}{N}-\left(\frac{\sum_{i=1}^{10} x_i}{N}\right)^2\)
\( =\frac{100}{10}-\left(\frac{10}{10}\right)^2\)
\( \sigma^2=10-1=9 \)
\( \Rightarrow \sigma=3\)