Question

If \( X \) is a continuous random variable with the probability density function \[ f(x) = \begin{cases} K(1 - x^3), & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases} \] Then, the value of \( K \) is ..........

• A. \( \dfrac{3}{4} \)
• B. \( \dfrac{4}{3} \)
• C. \( \dfrac{1}{3} \)
• D. 3

Hint:

The total area under a probability density function over its entire domain must be equal to 1.

Answer 1

The Correct Option is B

Step 1: Use the property of a probability density function (PDF): \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \( f(x) = 0 \) outside \( (0, 1) \), we only integrate over \( (0, 1) \):
\[ \int_0^1 K(1 - x^3) \, dx = 1 \] Step 2: Evaluate the integral: \[ K \int_0^1 (1 - x^3) \, dx = K \left[ x - \frac{x^4}{4} \right]_0^1 = K\left(1 - \frac{1}{4}\right) = K \cdot \frac{3}{4} \] Step 3: Set the integral equal to 1: \[ K \cdot \frac{3}{4} = 1 ⇒ K = \frac{4}{3} \] \[ \boxed{K = \frac{4}{3}} \]