Prove the following

Question:

Let $x_{i}(1 \leq i \leq 10)$ be ten observations of a random variable $X$.

If $\sum_{i=1}^{10}\left(x_{i}-p\right)=3 \quad$ and $\quad \sum_{i=1}^{10}\left(x_{i}-p\right)^{2}=9 \quad$ where

$0 \neq p \in \mathbf{R}$, then the standard deviation of these observations is :

  1. (1) $\sqrt{\frac{3}{5}}$

  2. (2) $\frac{4}{5}$

  3. (3) $\frac{9}{10}$

  4. (4) $\frac{7}{10}$


Correct Option: , 3

Solution:

S.D. $=\sqrt{\frac{\sum_{i=1}^{10}\left(x_{i}-p\right)^{2}}{10}-\left(\frac{\sum_{i=1}^{10}\left(x_{i}-p\right)}{10}\right)^{2}}$

$=\sqrt{\frac{9}{10}-\left(\frac{3}{10}\right)^{2}}=\frac{9}{10}$

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