Question

The mean square deviation of a set of $n$ observations $x_1,x_2,.......x_n$ about a point $c$ is defined as $\frac{1}{n}\sum _{i=1}^n(x_i-c{)}^2$. The mean square deviation about $–2$ and $2$ are $18$ and $10$ respectively, then standard deviation of this set of observations is

• A. $3$
• B. $2$
• C. $9$
• D. $1$

Answer 1

The correct option is A $3$

Mean deviation about $-2$ is
$\frac{1}{n}\sum _{i=1}^n{\{x_i-(-2\left)\right\}}^2=\mathrm{18......}\left(1\right)$
Mean deviation about $2$ is
$\frac{1}{n}\sum _{i=1}^n{\{x_i-2\}}^2=\mathrm{10......}\left(2\right)\therefore \frac{2}{n}\sum _{i=1}^n(x_i^2+2^2)=28\text{(adding(1)and(2))}\Rightarrow \frac{1}{n}\sum _{i=1}^n{x}_i^2=10\text{Also}\frac{1}{n}\sum _{i=1}^n{x}_i=1\therefore {\sigma }^2=\frac{1}{n}(\sum _{i=1}^n{x}_i^2-(\sum x_i{)}^2)=10-1=9$