Question

If \(X\) is a random variable with probability density function \[ f(x)=\frac{1}{\sqrt{12\pi}}\,e^{-\frac{(x-1)^2}{12}}, \qquad 0<x<\infty, \] then the value of \[ \int_{1}^{\infty} f(x)\,dx \] is:

• A. 0
• B. \(\frac{1}{2}\)
• C. \(1 - \frac{1}{2}\)
• D. 1

Hint:

For a normal distribution, the mean divides the area into two equal halves.
Thus, \(P(X \ge \mu) = 0.5\).

Answer 1

The Correct Option is B