A data consists of n observations: x_1,x_2,,x_n.

Question

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

Accepted Answer

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}}  = 9n $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2}  + 2\sum\limits_{i = 1}^n {{x_i}}  + n = 9n\,\,\,\,\,...\,(1)$$

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}}  = 5n $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2}  - 2\sum\limits_{i = 1}^n {{x_i}}  + n = 5n\,\,\,\,\,...\,(2)$$

Performing (1) + (2), we get

$$2\sum\limits_{i = 1}^n {x_i^2}  + 2n = 14n$$

$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$

Performing (1) - (2), we get

$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}}  = 4n$$

$$ \Rightarrow $$ $ \Rightarrow \sum\limits_{i = 1}^n {{x_i}}  = n$S.D($\sigma $)$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$\sigma $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$\sigma $$ = \sqrt 5 $

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