Question

A continuous random variable X has a probability density function \[ f(x) = \begin{cases} 3x^{2}, & \text{if } 0 \leq x \leq 1 0, & \text{otherwise} \end{cases} \] If \(P(X \leq a) = P(X > a)\), then the value of a is equal to _______

• A. \(1/33^{1/3}\)
• B. \(1/32^{1/3}\)
• C. \(1/22^{1/3}\)
• D. \(1/2^{1/3}\)

Hint:

The point $a$ where $P(X \leq a) = 0.5$ is by definition the median of the distribution.
For a power function distribution $f(x) = k x^{k-1}$ on $[0,1]$, the cumulative distribution function is $F(x) = x^k$.
Setting $F(a) = 0.5$ gives $a^k = 0.5 \implies a = (1/2)^{1/k}$ directly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to find the value of a constant \(a\) (which represents the median of the distribution) for a continuous random variable \(X\) defined by a given probability density function (PDF).

Step 2: Key Formula or Approach:
The total probability under any probability density function is equal to 1.
The condition \(P(X \leq a) = P(X > a)\) means that the value \(a\) divides the total probability density into two equal halves.
Therefore:
\[ P(X \leq a) = \int_{0}^{a} f(x) dx = 0.5 \]

Step 3: Detailed Explanation:

• Set up the integral representing the cumulative probability up to \(a\):
\[ \int_{0}^{a} 3x^{2} dx = 0.5 \]
• Evaluate the definite integral:
\[ \left[ x^{3} \right]_{0}^{a} = 0.5 \] \[ a^{3} - 0 = 0.5 \]
• Express the decimal as a fraction and solve for \(a\):
\[ a^{3} = \frac{1}{2} \] \[ a = \left(\frac{1}{2}\right)^{1/3} = \frac{1}{2^{1/3}} \]

Step 4: Final Answer:
The value of \(a\) is equal to \(\frac{1}{2^{1/3}}\).