Question

If the probability density function of a continuous random variable \( X \) is \( f(x) \), \( -\infty<x<\infty \), and if the median of \( X \) is \( M \), then:

• A. \( \displaystyle \int_{-M}^{M} f(x)\,dx = \frac{1}{2} \)
• B. \( \displaystyle \int_{-\infty}^{M} f(x)\,dx = \int_{M}^{\infty} f(x)\,dx = \frac{1}{2} \)
• C. \( \displaystyle \int_{-\infty}^{-M} f(x)\,dx = \int_{-M}^{M} f(x)\,dx = \int_{M}^{\infty} f(x)\,dx \)
• D. \( \displaystyle \int_{-\infty}^{\infty} f(x)\,dx = \frac{1}{2} \)

Hint:

The median of a continuous distribution splits the total probability into two equal parts. Always integrate the PDF up to the median to check if the area is \( \frac{1{2 \).

Answer 1

The Correct Option is B

Step 1: Understand the definition of the median.
The median \( M \) of a continuous random variable \( X \) is defined as the value such that: \[ P(X \leq M) = \frac{1}{2} \]
Step 2: Convert this probability to an integral.
Since \( f(x) \) is the probability density function (PDF), we write: \[ P(X \leq M) = \int_{-\infty}^{M} f(x)\,dx = \frac{1}{2} \]
Step 3: Use total probability.
The total area under the PDF must be 1: \[ \int_{-\infty}^{\infty} f(x)\,dx = 1 \]
Step 4: Deduce the remaining probability.
Then, \[ P(X>M) = 1 - \int_{-\infty}^{M} f(x)\,dx = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow \int_{M}^{\infty} f(x)\,dx = \frac{1}{2} \] Conclusion:
Both halves of the PDF (on either side of the median) carry equal probability: \[ \int_{-\infty}^{M} f(x)\,dx = \int_{M}^{\infty} f(x)\,dx = \frac{1}{2} \]