Question

For the following probability density function of a random variable X, find (a) P(X<1) and (b) P(|X|<1).
\[ f(x) = \begin{cases} \frac{x + 2}{18}, & -2<x<4
0, & \text{otherwise} \end{cases} \]

Hint:

For PDF, ensure integral equals 1; compute probabilities by integrating over specified intervals.

Answer 1

Verify \( f(x) \) is a valid PDF: \[ \int_{-2}^4 \frac{x + 2}{18} \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-2}^4 = \frac{1}{18} \left( \left( \frac{16}{2} + 8 \right) - \left( \frac{4}{2} - 4 \right) \right) = \frac{1}{18} (16 - (-2)) = 1. \] (a) \( P(X<1) \): \[ P(X<1) = \int_{-2}^1 \frac{x + 2}{18} \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-2}^1 = \frac{1}{18} \left( \left( \frac{1}{2} + 2 \right) - \left( \frac{4}{2} - 4 \right) \right) = \frac{1}{18} \left( \frac{5}{2} - (-2) \right) = \frac{9}{36} = \frac{1}{4}. \] (b) \( P(|X|<1) = P(-1<X<1) \): \[ \int_{-1}^1 \frac{x + 2}{18} \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-1}^1 = \frac{1}{18} \left( \left( \frac{1}{2} + 2 \right) - \left( \frac{1}{2} - 2 \right) \right) = \frac{1}{18} \left( \frac{5}{2} - \left( -\frac{3}{2} \right) \right) = \frac{4}{18} = \frac{2}{9}. \] Answer: (a) \( \frac{1}{4} \), (b) \( \frac{2}{9} \).