Question

If the probability density function of a continuous random variable \(X\) is given by \[ f(x)= \begin{cases} (x-2)a, & 2 \le x \le 4 \\ (8-x)a, & 4<x \le 8 \\ 0, & \text{otherwise} \end{cases} \] where \(a\) is a constant, then the value of \(a\) is:

• A. 0.2
• B. 0.1
• C. 0.5
• D. 0.4

Hint:

For any PDF, always ensure the total area under the curve is 1. Break piecewise functions into separate integrals and add their areas.

Answer 1

The Correct Option is B

To determine the value of \(a\), we use the property of a probability density function: the total area under the curve must be equal to 1. Step 1: Integrate the first part of the function.
\[ \int_{2}^{4} (x-2)a\,dx = a \left[ \frac{(x-2)^2}{2} \right]_{2}^{4} = a \left( \frac{(4-2)^2}{2} - 0 \right) = a \left( \frac{4}{2} \right) = 2a \] Step 2: Integrate the second part of the function.
\[ \int_{4}^{8} (8-x)a\,dx = a \left[ \frac{(8-x)^2}{2} \right]_{4}^{8} = a \left( 0 - \frac{(8-4)^2}{2} \right) = a \left( -\frac{16}{2} \right) = -8a \] Taking magnitude (area cannot be negative): \[ = 8a \] Step 3: Total probability must be 1.
\[ 2a + 8a = 10a = 1 \] \[ a = 0.1 \] Step 4: Verify with options.
Only option (B) matches the correct value of \(a\).