Question

The mean square deviation of a set of $n$ observation $x_1,x_2,...x_n$ about a point $c$ is defined as $\frac{1}{n}\sum _{i=1}^n{\left(x_i-c\right)}^2$.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)

We have $\frac{1}{n}\sum _{i=1}^n{\left(x_i+2\right)}^2=18$ and
$\frac{1}{n}\sum _{i=1}^n{\left(x_i-2\right)}^2=10$
$\Rightarrow \sum _{i=1}^n{\left(x_i+2\right)}^2=18n$ and
$\sum _{i=1}^n{\left(x_i-2\right)}^2=10n$
$\Rightarrow \sum _{i=1}^n{\left(x_i+2\right)}^2+\sum _{i=1}^n{\left(x_i-2\right)}^2=28n$
and $\sum _{i=1}^n{\left(x_i+2\right)}^2-\sum _{i=1}^n{\left(x_i-2\right)}^2=8n$
$\Rightarrow 2\sum _{i=1}^n{\left(x_i+4\right)}^2=28n2\sum _{i=1}^{n}4x_i=8n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+4n=14n\sum _{i=1}^n{x}_i=n$
$\Rightarrow \sum _{i=1}^n{x}_i^2=10n\sum _{i=1}^n{x}_i=n$
$\therefore \sigma =\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^2}$
$=\sqrt{\frac{10n}{n}-{\left(\frac{n}{n}\right)}^2}=3$