Question

For observations $x_1,x_2,x_3,..........,x_n,$ if ${\sum }_{i=1}^n(x_i+1{)}^2=9n$ and ${\sum }_{i=1}^n(x_i-1{)}^2=5n.$, then standard deviation of the data is

• A. $\sqrt{3}$
• B. $\sqrt{5}$
• C. $\sqrt{2}$
• D. $\sqrt{10}$

Answer 1

The correct option is B $\sqrt{5}$

Given that
${\sum }_{i=1}^n(x_i+1{)}^2=9n={\sum }_{i=1}^n(x_i^2+2x_i+1)=9n\Rightarrow \sum x_i^2+2\sum x_i+n=9n\Rightarrow \sum x_i^2+2\sum x_i=8n........(1\left){\sum }_{i=1}^n\right(x_i-1{)}^2={\sum }_{i=1}^n(x_i^2-2x_i+1)=5n\Rightarrow \sum x_i^2-2\sum x_i+n=5n\Rightarrow \sum x_i^2-2\sum x_i=4n........\left(2\right)\left(1\right)+\left(2\right)\Rightarrow 2\sum x_i^2=12n\Rightarrow \sum x_i^2=6n\left(1\right)-\left(2\right)\Rightarrow 4\sum x_i=4n\Rightarrow \sum x_i=n\sigma =\sqrt{\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2}=\sqrt{6-1}=\sqrt{5}$