Question

A continuous random variable \(X\) has a probability density function \( f(x) = 3x^2, \ 0 \leq x \leq 1 \). If \( P(X \leq a) = P(X>a) \), then the value of 'a' is

• A. \( \left( \frac{1}{2} \right)^{1/3} \)
• B. \( \left( \frac{1}{3} \right)^{1/3} \)
• C. \( \left( \frac{1}{4} \right)^{1/3} \)
• D. \( \left( \frac{1}{5} \right)^{1/3} \)

Hint:

When total probability is split equally, equate the integral to \( \frac{1}{2} \) and solve for the variable.

Answer 1

The Correct Option is A

We are given that \( P(X \leq a) = P(X>a) \). This implies the total probability is equally split, so:
\[ P(X \leq a) = \int_0^a 3x^2 \, dx = \frac{1}{2} \]
\[ \Rightarrow \left[ x^3 \right]_0^a = \frac{1}{2} \Rightarrow a^3 = \frac{1}{2} \Rightarrow a = \left( \frac{1}{2} \right)^{1/3} \]