If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9} $$    and $$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2} = 45} ,$$    then the standard deviation of the 9 items $${x_1},{x_2},.....,{x_9}$$    is:

Solution :
Given $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9} $$
$$ \Rightarrow \,\,\sum\limits_{i = 1}^9 {{x_i} = 54\,\,\,\,\,.....\left( {\text{i}} \right)} $$
Also, $$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2} = 45} $$
$$ \Rightarrow \,\,\sum\limits_{i = 1}^9 {{x_i}^2 - 10\sum\limits_{i = 1}^9 {{x_i} + 9\left( {25} \right) = 45\,\,\,\,.....\left( {{\text{ii}}} \right)} } $$
From (i) and (ii) we get,
$$\sum\limits_{i = 1}^9 {x_i^2 = 360} $$
Since, variance $$ = \frac{{\sum {{x_i}^2} }}{9} - {\left( {\frac{{\sum {{x_i}} }}{9}} \right)^2}$$
$$\eqalign{ & = \frac{{360}}{9} - {\left( {\frac{{54}}{9}} \right)^2} \cr & = 40 - 36 \cr & = 4 \cr} $$
∴ Standard deviation $$ = \sqrt {{\text{Variance}}} = 2$$