Question

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

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

Hint:

For continuous random variables, the probability density function must be normalized such that the total probability equals 1. Use this condition to solve for unknown constants in the function.

Answer 1

The Correct Option is B

To find the value of $K$, we use the fact that the probability density function (PDF) of a continuous random variable must satisfy the normalization condition: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] This condition ensures that the total probability is 1. Since $f(x) = 0$ for $x$ outside the interval $[0, 1]$, we can simplify the integral as follows: \[ \int_0^1 K(1 - x^3) \, dx = 1 \] Now, compute the integral: \[ \int_0^1 K(1 - x^3) \, dx = K \left[ \int_0^1 1 \, dx - \int_0^1 x^3 \, dx \right] \] The first integral is: \[ \int_0^1 1 \, dx = x \Big|_0^1 = 1 \] The second integral is: \[ \int_0^1 x^3 \, dx = \frac{x^4}{4} \Big|_0^1 = \frac{1}{4} \] Thus, the equation becomes: \[ K \left[ 1 - \frac{1}{4} \right] = 1 \] \[ K \times \frac{3}{4} = 1 \] Solving for $K$: \[ K = \frac{4}{3} \] Therefore, the value of $K$ is $\dfrac{4}{3}$.