Question

The mean square deviation of 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. $1$
• D. none of these

Answer 1

The correct option is A $3$

Here, $\frac{1}{n}\sum _{i=1}^n{(x_i+2)}^2=18$
$\Rightarrow \sum _{i=1}^n(x_i^2+4x_i+4)=18n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+4\sum _{i=1}^n{x}_i+4\sum _{i=1}^{n}1=18n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+4\sum _{i=1}^n{x}_i=18n-4n=14n....\left(1\right)$
And $\frac{1}{n}\sum _{i=1}^n{(x_i-2)}^2=10$
$\Rightarrow \sum _{i=1}^n(x_i^2-4x_i+4)=10n$
$\Rightarrow \sum _{i=1}^n{x}_i^2-4\sum _{i=1}^n{x}_i+4\sum _{i=1}^{n}1=10n$
$\Rightarrow \sum _{i=1}^n{x}_i^2-4\sum _{i=1}^n{x}_i=10n-4n=6n....\left(2\right)$
Solving (1) and (2) we get, $\sum _{i=1}^n{x}_i^2=10n,\sum _{i=1}^n{x}_i=n$
$\therefore$ standard deviation of this observation is $=\sigma =\sqrt{\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2}=\sqrt{9}=3$