Question

If a continuous random variable \( X \) has the following probability density function \[ g(x) = \begin{cases} \frac{k}{4} x(2 - x), & 0 < x < 2 \\ 0, & \text{otherwise} \end{cases} \] then the value of \( k \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When solving for constants in probability density functions, ensure that the total probability integrates to 1. Use the properties of definite integrals to solve.

Answer 1

The Correct Option is C

To find the value of \( k \), we use the property that the total probability must equal 1 for a probability density function. Therefore, we integrate \( g(x) \) over the range \( 0 < x < 2 \) and set the result equal to 1: \[ \int_0^2 \frac{k}{4} x(2 - x) \, dx = 1. \] First, expand the integrand: \[ x(2 - x) = 2x - x^2. \] Now, integrate: \[ \int_0^2 \frac{k}{4} (2x - x^2) \, dx = \frac{k}{4} \left[ \int_0^2 2x \, dx - \int_0^2 x^2 \, dx \right]. \] The integrals are: \[ \int_0^2 2x \, dx = 2 \cdot \frac{x^2}{2} \Big|_0^2 = 8, \int_0^2 x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3}. \] Substitute these values into the equation: \[ \frac{k}{4} \left( 8 - \frac{8}{3} \right) = 1. \] Simplify: \[ \frac{k}{4} \cdot \frac{16}{3} = 1 $\Rightarrow$ k = \frac{3}{4} \cdot \frac{3}{16} = 3. \] Final Answer: \[ \boxed{\text{(C) } 3}. \]