Question

Let $xi(1\le i\le 10)$ be ten observations of a random variable $X$. If $\sum _{i=1}^{10}(x_i-p)=3and\sum _{i=1}^{10}{(x_i-p)}^2=9$ where $0\ne p\in R$, then the standard deviation of these observations is

• A. $\frac{7}{10}$
• B. $\frac{9}{10}$
• C. $\sqrt{\frac{3}{5}}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B

$\frac{9}{10}$

Explanation for the correct option

It is given that $n=10$

We know that

$\begin{array}{rcl}\text{S.D}.& =& \sqrt{\mathrm{variance}}\\ & =& \sqrt{\frac{\sum _{}{(x_i-\overline{x})}^2}{n-1}}\end{array}$

On solving, we get:

$\begin{array}{rcl}\therefore S.D.& =& \sqrt{\frac{{\displaystyle \sum _{}{(x_i-p)}^2}}{n}-{\left(\frac{\sum _{}(x_i-p)}{n}\right)}^2}\\ & =& \sqrt{\frac{9}{10}-{\left(\frac{3}{10}\right)}^2}\left[\because (x_i-p)=3and{(x_i-p)}^2=9\right]\\ & =& \sqrt{\frac{81}{100}}\\ & \Rightarrow & S.D=\frac{9}{10}\end{array}$

Hence, the correct answer is option B.