If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as: $${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$ where ‘a’ and ‘b’ are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ – \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$ A. 1 B. 0.5 C. $$\pi $$ D. $$\frac{\pi }{2}$$

The correct answer is $\boxed{\frac{\pi}{2}}$.

The integral $$\int_{ – \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$ is the probability that a continuous, real valued random variable defined over the interval (-$\infty$, +$\infty$) and its occurrence is defined by the density function given as: $${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$ will take a value less than or equal to $a$.

This probability can be calculated using the following formula:

$$P\left( {X \le {\text{a}}} \right) = \int_{ – \infty }^{\text{a}} {\text{f}}\left( {\text{x}} \right){\text{dx}} = \int_{ – \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$

The integral can be evaluated using the following steps:

1. Substitute $u = \frac{x – a}{b}$.
2. The integral becomes:

$$\int_{ – \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} = \int_{ – \infty }^{\frac{{a – {\text{a}}}}{{\text{b}}}} {\frac{1}{{\sqrt {2\pi }}}}{{\text{e}}^{\frac{{ – 1}}{2}{{u}^2}}}{\text{du}} = \frac{1}{{\sqrt {2\pi }}}\int_{ – \infty }^{\frac{{a – {\text{a}}}}{{\text{b}}}} {{\text{e}}^{\frac{{ – 1}}{2}{{u}^2}}}{\text{du}}$$

1. The integral in step 2 is a standard integral that can be evaluated using the following formula:

$$\int_{ – \infty }^{\infty } {{\text{e}}^{\frac{{ – 1}}{2}{{u}^2}}}{\text{du}} = \sqrt {2\pi }$$

1. Substituting the value of the integral in step 3 into step 2 gives:

$$\int_{ – \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{{\text{x}} – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} = \frac{1}{{\sqrt {2\pi }}}\int_{ – \infty }^{\frac{{a – {\text{a}}}}{{\text{b}}}} {{\text{e}}^{\frac{{ – 1}}{2}{{u}^2}}}{\text{du}} = \sqrt {2\pi } \frac{1}{{\sqrt {2\pi }}}\frac{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{a – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{{\text{b}}} = \frac{{\text{e}}^{\frac{{ – 1}}{2}{{\left( {\frac{{a – {\text{a}}}}{{\text{b}}}} \right)}^2}}}{{\text{b}}}$$

1. The final step is to substitute $a$ for $x$. This gives:

$$P