Question

The mean square deviations of a set of observations $x_1,x_2,\cdots ,x_n$ about a point c is defined to be $\frac{1}{n}{\sum }_{i=1}^n(x_i-c{)}^2$
The mean square deviations about -1 and +1 of a set of observations are 7 and 3, respectively. Find the standard deviation of this set of observations.

• A. $\sqrt{3}$
• B. 3
• C. $\sqrt{5}$
• D. 5

Answer 1

The correct option is A

$\sqrt{3}$

Mean square deviations,$=\frac{1}{n}{\sum }_{i=1}^n(x_i-c{)}^2$ about c.
Also, given that mean square deviation about – 1 and + 1 are 7 and 3, respectively.
$\Rightarrow \frac{1}{n}{\sum }_{i=1}^n(x_i+1{)}^2=7\text{and}\frac{1}{n}{\sum }_{i=1}^n(x_i-1{)}^2=3\Rightarrow {\sum }_{i=1}^n{x}_i^2+2{\sum }_{i=1}^n{x}_i+n=7n\text{and}{\sum }_{i=1}^n{x}_i^2-2{\sum }_{i=1}^n{x}_i+n=3n\Rightarrow {\sum }_{i=1}^n{x}_i^2+2{\sum }_{i=1}^n{x}_i=6n\text{and}{\sum }_{i=1}^n{x}_i^2-2{\sum }_{i=1}^n{x}_i=2n$
$\Rightarrow {\sum }_{i=1}^n{x}_i=n\Rightarrow \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}=1$
$\therefore$ Standard deviation
$=\sqrt{\frac{1}{n}{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}=\sqrt{\frac{1}{n}{\sum }_{i=1}^n(x_i-1{)}^2}=\sqrt{3}$