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Let x_{i}(1\leq i\leq 10) be ten observations of a random variable X. If \sum_{i=1}^{10}(x_{i}-p)=3 and \sum_{i=1}^{10}(x_{i}-p)^2=9  where 0\neq p\in\mathbb R, then the standard deviation of these observations is :
Option: 1 \sqrt{\frac{3}{5}}
Option: 2 \frac{4}{5}
Option: 3 \frac{9}{10}
Option: 4 \frac{7}{10}

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\\\text {Variance }=\frac{\Sigma\left(\mathrm{x}_{\mathrm{i}}-\mathrm{p}\right)^{2}}{\mathrm{n}}-\left(\frac{\Sigma\left(\mathrm{x}_{\mathrm{i}}-\mathrm{p}\right)}{\mathrm{n}}\right)^{2} \\ =\frac{9}{10}-\left(\frac{3}{10}\right)^{2}=\frac{81}{100} \\ \mathrm{S.D.}=\frac{9}{10}

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