Question

If \( X \) is a continuous random variable whose probability density function is given by \[ f(x) = \begin{cases} k(4x - 2x^2), & 0<x<2 \\ 0, & \text{otherwise} \end{cases} \quad \text{then the value of } k \text{ is:} \]

• A. \( \frac{3}{8} \)
• B. \( \frac{5}{8} \)
• C. \( \frac{7}{8} \)
• D. \( 1 \)

Hint:

To determine a constant in a PDF, always integrate over the support and set the integral equal to 1.

Answer 1

The Correct Option is A

To find the value of \( k \), we need to ensure that the probability density function (PDF) integrates to 1 over its entire range. The PDF is given by: \[ f(x) = \begin{cases} k(4x - 2x^2), & 0 < x < 2 \\ 0, & \text{otherwise} \end{cases} \] We will integrate \( f(x) \) from 0 to 2 and set the result equal to 1: \[ \int_{0}^{2} k(4x - 2x^2) \, dx = 1 \] First, find the antiderivative of \( 4x - 2x^2 \): \[ \int (4x - 2x^2) \, dx = \left[ 2x^2 - \frac{2}{3}x^3 \right] \] Now, evaluate this from 0 to 2: \[ \left[ 2(2)^2 - \frac{2}{3}(2)^3 \right] - \left[ 2(0)^2 - \frac{2}{3}(0)^3 \right] = \left[ 8 - \frac{16}{3} \right] = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \] Set the result equal to 1 after multiplying by \( k \): \[ k \cdot \frac{8}{3} = 1 \] Solve for \( k \): \[ k = \frac{3}{8} \] Therefore, the value of \( k \) is \(\frac{3}{8}\).