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)\\

\sum_{i=1}^{n} (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........(2)\\
(1) + (2) \Rightarrow 2 \sum x_i^2=12n \Rightarrow \sum x_i^2=6n\\
(1) - (2) \Rightarrow 4 \sum x_i = 4n \Rightarrow \sum x_i=n\\

\therefore \frac{\sum x_i}{n}=1=\bar x\\
\sigma =\sqrt{\frac{\sum (x_i-\bar x)^2}{n}}=\sqrt{\frac{\sum (x_i-1)^2}{n}}=\sqrt{\frac{5n}{n}}=\sqrt{5}\)