Question

If \( f(x) = \frac{x^2 + 2{18} \), for \( -2<x<4 \), and 0 otherwise, is the p.d.f. of a random variable \( X \), then the value of \( P(|X|<2) \) is}

• A. \( \frac{5}{9} \)
• B. \( \frac{4}{9} \)
• C. \( \frac{2}{9} \)
• D. \( \frac{1}{9} \)

Hint:

When calculating probabilities for continuous random variables, integrate the p.d.f. over the desired interval.

Answer 1

The Correct Option is B

Step 1: Calculate the probability.
The probability \( P(|X|<2) \) is the area under the p.d.f. from \( -2 \) to \( 2 \). \[ P(|X|<2) = \int_{-2}^{2} \frac{x^2 + 2}{18} \, dx \] Step 2: Integrate the function.
First, calculate the integral of \( \frac{x^2 + 2}{18} \) over the interval \( [-2, 2] \): \[ \int_{-2}^{2} \frac{x^2 + 2}{18} \, dx = \frac{1}{18} \left[ \int_{-2}^{2} x^2 \, dx + \int_{-2}^{2} 2 \, dx \right] \] After integrating, we get \( \frac{4}{9} \).
Step 3: Conclusion.
The correct answer is (B) \( \frac{4}{9} \).