Question

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

• A. $3$
• B. $2$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$3$

We have $\frac{1}{n}{\sum }_{i=1}^m{(x_i+2)}^2=18;\frac{1}{n}{\sum }_{i=1}^m{(x_i-2)}^2=10$
$\Rightarrow {\sum }_{i=1}^n{(x_i+2)}^2=18n;{\sum }_{i=1}^n{(x_i-2)}^2=10n$
$\Rightarrow {\sum }_{i=1}^n{(x_i+2)}^2+{\sum }_{i=1}^n{(x_i-2)}^2=28n;{\sum }_{i=1}^n{(x_i-2)}^2-{\sum }_{i=1}^n{(x_i-2)}^2=8n$
$\Rightarrow 2{\sum }_{i=1}^n{(x_i+4)}^2=28n;2{\sum }_{i=1}^{n}4x_i=8n$
$\Rightarrow {\sum }_{i=1}^n{x_i}^2=10n;{\sum }_{i=1}^n{x_i}^{}=n$
Therefore, $\sigma =\sqrt{\frac{1}{n}{\sum }_{i=1}^n{x_i}^2-{\left(\frac{1}{n}{\sum }_{i=1}^n{x_i}^2\right)}^2}=\sqrt{\frac{10n}{n}-{\left(\frac{n}{n}\right)}^2}=3$