Question

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

• A. $\sqrt{7}$
• B. $\sqrt{5}$
• C. $5$
• D. $2$

Answer 1

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

$\sum _{i=1}^n(x_i+1{)}^2=9n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+\sum _{i=1}^n{1}^2+\sum _{i=1}^{n}2x_i=9n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+n+2\sum _{i=1}^n{x}_i=9n...\left(1\right)$

$\sum _{i=1}^n(x_i-1{)}^2=5n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+\sum _{i=1}^n{1}^2-\sum _{i=1}^{n}2x_i=5n$
$\Rightarrow \sum _{i=1}^n{x}_i^2+n-2\sum _{i=1}^n{x}_i=5n...\left(2\right)$

Adding Equation $\left(1\right)$ and $\left(2\right)$
$2\sum _{i=1}^n{x}_i^2+2n=14n$
$\Rightarrow \sum _{i=1}^n{x}_i^2=6n$

Now, subtracting Equation $\left(2\right)$ from Equation $\left(1\right)$
$4\sum _{i=1}^n{x}_i=4n$
$\Rightarrow \sum _{i=1}^n{x}_i=n$

$\therefore$ Standard deviation $=\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^2}$
$=\sqrt{\frac{1}{n}\cdot 6n-{(\frac{1}{n}\cdot n)}^2}$
$=\sqrt{5}$