A normal random variable X has following probability density function $${{\text{f}}_{\text{x}}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {8\pi } }}{{\text{e}}^{ - \left\{ {\frac{{{{\left( {{\text{x}} - 1} \right)}^2}}}{8}} \right\}}},\, - \infty < {\text{x}} \leqslant \infty .$$
Then $$\int\limits_1^\infty {{{\text{f}}_{\text{x}}}\left( {\text{x}} \right){\text{dx}}} $$   is

A. 0

B. $$\frac{1}{2}$$

C. $$1 - \frac{1}{{\text{e}}}$$

D. 1

Answer: Option B