Question

A random variable \(X\) has the following pdf:
\[ f(x) = \frac{1}{\sqrt{12\pi}} e^{-\frac{(x-1)^2}{6}}; \quad -\infty < x < \infty \]
\(= 0\), otherwise.
Then \(\int_{1}^{\infty} f(x)dx = ?\)

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

Hint:

For any normal distribution, the probability of obtaining a value greater than the mean is always exactly 0.5.
Once you identify that the lower limit of integration is the mean of the distribution, you can write the answer \(\frac{1}{2}\) immediately without performing any integration.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem asks us to evaluate the integral of a normal probability density function over a specific range.
We can solve this easily by identifying the parameters of the normal distribution and using its symmetry properties.

Step 2: Key Formula or Approach:
The PDF of a normal distribution with mean \(\mu\) and variance \(\sigma^2\) is:
\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]
Since any probability density function is symmetric about its mean \(\mu\), the total area under the curve is 1, and the area on either side of the mean is exactly 0.5:
\[ \int_{\mu}^{\infty} f(x) dx = \int_{-\infty}^{\mu} f(x) dx = \frac{1}{2} \]

Step 3: Detailed Explanation:
Let us compare our given PDF with the standard normal density equation:
Our PDF is:
\[ f(x) = \frac{1}{\sqrt{12\pi}} e^{-\frac{(x-1)^2}{6}} \]
We rewrite the constant term under the square root and the exponent to match the standard form:
\[ \sqrt{12\pi} = \sqrt{6 \times 2\pi} = \sqrt{3} \sqrt{2\pi} \]
This indicates that:
\[ \sigma = \sqrt{3} \]
Next, look at the exponent:
\[ -\frac{(x-1)^2}{6} = -\frac{(x-1)^2}{2(3)} = -\frac{1}{2} \left(\frac{x-1}{\sqrt{3}}\right)^2 \]
Comparing this with the exponent in the standard form:
\[ -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2 \]
We identify the parameters of the distribution:
- Mean, \(\mu = 1\)
- Standard Deviation, \(\sigma = \sqrt{3}\)
Thus, our variable follows a normal distribution: \(X \sim N(1, 3)\).
The integral we want to evaluate is:
\[ I = \int_{1}^{\infty} f(x) dx \]
Since our lower limit of integration is 1, which is exactly the mean \(\mu = 1\), this integral represents the probability of the event \(X \ge \mu\):
\[ I = P(X \ge 1) \]
Because the normal distribution curve is perfectly symmetric about its mean, the area under the curve from the mean to positive infinity is exactly half of the total area under the PDF curve:
\[ I = 0.5 = \frac{1}{2} \]
This matches the second option.

Step 4: Final Answer:
Therefore, the correct option is (B).