Question

The probability density function of a continuous random variable \( X \) is given by \( f(x) = k(x-1)^3 \), for \( 1 \leq x \leq 3 \). The value of \( k \) is:

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

The probability density function must integrate to 1 over the given range. Always check by solving \( \int f(x) \,dx = 1 \).

Answer 1

The Correct Option is C

Step 1: Using the Probability Density Function (PDF) property For a valid probability density function: \[ \int_{1}^{3} f(x) \,dx = 1 \] Substituting \( f(x) = k(x-1)^3 \), \[ \int_{1}^{3} k(x-1)^3 \,dx = 1 \]
Step 2: Evaluating the integral \[ k \int_{1}^{3} (x-1)^3 \,dx = 1 \] Using the standard integral formula: \[ \int (x-1)^3 \,dx = \frac{(x-1)^4}{4} \] Evaluating from \( x=1 \) to \( x=3 \): \[ \frac{(3-1)^4}{4} - \frac{(1-1)^4}{4} = \frac{16}{4} - 0 = 4 \]
Step 3: Solving for \( k \) \[ k(D) = 1 \] \[ k = \frac{1}{4} \]