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

Step 1: Understanding the Concept:
The given pdf is a normal distribution with mean \(\mu = 1\) and variance \(\sigma^2 = 6\) (since \(\sigma^2 = 6\) because \(2\sigma^2 = 12\)).
The total area under the pdf is 1.

Step 2: Key Approach:
The integral \(\int_{1}^{\infty} f(x) dx\) represents the probability that \(X \ge 1\).
Since the distribution is symmetric about its mean 1, the area to the right of the mean is 0.5.

Step 3: Detailed Explanation:
The normal distribution is symmetric about its mean \(\mu = 1\).
Therefore, the probability that \(X \ge 1\) is exactly 0.5.
Thus, \(\int_{1}^{\infty} f(x) dx = 0.5 = \frac{1}{2}\).
Option (B) is \(\frac{1}{2}\).
Option (C) is \(1 - \frac{1}{2} = \frac{1}{2}\) as well, but it's written as \(1 - \frac{1}{2}\).
Option (B) is more direct.