Question

If \( X \) is a continuous random variable with the probability density function
\[ f(x) = \begin{cases} cx^3, & 0 \le x \le 2 \\ 0, & \text{otherwise} \end{cases} \] then \( P\left( \frac{1}{2} < X < \frac{3}{2} \right) \) is ..........

• A. \( \frac{5}{16} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{5}{8} \)
• D. \( \frac{1}{16} \)

Hint:

Always normalize the probability density function by ensuring the total area under the curve equals 1. Then use definite integration over the given range to find the required probability.

Answer 1

The Correct Option is A

First, normalize the PDF to find the constant \( c \):
\[ \int_{0}^{2} cx^3 dx = 1 ⇒ c \int_{0}^{2} x^3 dx = 1 ⇒ c \left[ \frac{x^4}{4} \right]_{0}^{2} = 1 ⇒ c \cdot \frac{16}{4} = 1 ⇒ 4c = 1 ⇒ c = \frac{1}{4} \] Now compute the required probability: \[ P\left( \frac{1}{2} < X < \frac{3}{2} \right) = \int_{\frac{1}{2}}^{\frac{3}{2}} f(x) dx = \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{4} x^3 dx \] \[ = \frac{1}{4} \int_{\frac{1}{2}}^{\frac{3}{2}} x^3 dx = \frac{1}{4} \left[ \frac{x^4}{4} \right]_{\frac{1}{2}}^{\frac{3}{2}} = \frac{1}{4} \left( \frac{(3/2)^4}{4} - \frac{(1/2)^4}{4} \right) \] \[ = \frac{1}{4} \left( \frac{81/16}{4} - \frac{1/16}{4} \right) = \frac{1}{4} \cdot \left( \frac{80}{64} \right) = \frac{1}{4} \cdot \frac{5}{4} = \frac{5}{16} \]