Question

If f(x) is a continous real valued random variable defined over the interval $(-\infty ,+\infty )$ and its occurance is defined by the density function given as $f\left(x\right)=\frac{1}{\sqrt{2\pi }b}exp\{-\frac{1}{2}{\left(\frac{x-a}{b}\right)}^2\}$ where a and b are the statistical attributes of the random variable {x}. The value of the integral ${\int }_{-\infty }^a\frac{1}{\sqrt{2\pi }b}exp\{-\frac{1}{2}{\left(\frac{x-a}{b}\right)}^2\}dx\text{is}$

• A. 1
• B. 0.5
• C. $\pi$
• D. $\pi /2$

Answer 1

The correct option is B 0.5

It is a normal distribution with mean a and strandard deviation b. Normal distribution is symmetric about mean.


So

${\int }_{-\infty }^a\frac{1}{\sqrt{2\pi }b}{e}^{-\frac{1}{2}{\left(\frac{x-a}{b}\right)}^{2}dx}=\frac{1}{2}$