JEE Math Practice Question

If $$\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$$ and $$\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$$ (n, a > 1) then the standard deviation of n observations x1 , x2 , ..., xn is :

1. A.$$a$$ – 1
2. B.$$n\sqrt {a - 1} $$
3. C.$$\sqrt {n\left( {a - 1} \right)} $$
4. D.$$\sqrt {a - 1} $$