Question

The p.d.f. of a continuous random variable \(X\) is \(f(x)\).
\[ f(x)= \begin{cases} \dfrac{x^2}{18}, & -3 \le x \le 3 \\ 0, & \text{otherwise} \end{cases} \] Then find \(P(|X|<2)\).

• A. \(1/27\)
• B. \(2/13\)
• C. \(8/27\)
• D. \(4/27\)

Hint:

For modulus probability: \[ |X|<a \Rightarrow -a<X<a \] Then use: \[ P(a<X<b)=\int_a^b f(x)\,dx \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Probability for continuous RV is the area under the PDF curve.
Step 2: Key Formula or Approach:
\(P[|X| \lt 2] = \int_{-2}^{2} f(x) dx\).
Step 3: Detailed Explanation:
\[ \int_{-2}^{2} \frac{x^2}{18} dx = 2 \int_{0}^{2} \frac{x^2}{18} dx = \frac{1}{9} \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{27} \] Step 4: Final Answer:
Probability is \(8/27\).