Question:
Let $x_{i}(1 \leq i \leq 10)$ be ten observations of a random
variable $X .$ If $\sum_{i=1}^{10}\left(x_{i}-p\right)=3$ and $\sum_{i=1}^{10}\left(x_{i}-p\right)^{2}=9$
where $0 \neq \mathrm{p} \in \mathrm{R}$, then the standard deviation of these observations is:
Correct Option: , 3
Solution:
Variance $=\frac{\Sigma\left(\mathrm{x}_{\mathrm{i}}-\mathrm{p}\right)^{2}}{\mathrm{n}}-\left(\frac{\Sigma\left(\mathrm{x}_{\mathrm{i}}-\mathrm{p}\right)}{\mathrm{n}}\right)^{2}$
$=\frac{9}{10}-\left(\frac{3}{10}\right)^{2}=\frac{81}{100}$
S.D. $=\frac{9}{10}$
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