We are given the probability density function (PDF) of a continuous random variable \( X \): \[ f(x) = \frac{x + 2}{18}, \quad -2 < x < 4 \] and we need to find \( P(|X| < 1) \), which means: \[ P(-1 < X < 1) = \int_{-1}^{1} f(x) \,dx \] Step 1: Set Up the Integral \[ P(-1 < X < 1) = \int_{-1}^{1} \frac{x + 2}{18} \,dx \] Step 2: Compute the Integral
Splitting the integral: \[ P(-1 < X < 1) = \frac{1}{18} \int_{-1}^{1} (x + 2) \,dx \] Computing each term separately: \[ \int (x + 2) \,dx = \frac{x^2}{2} + 2x \] Evaluating from \( -1 \) to \( 1 \): \[ \left[ \frac{x^2}{2} + 2x \right]_{-1}^{1} \] Step 3: Evaluate at Limits At \( x = 1 \): \[ \frac{1^2}{2} + 2(1) = \frac{1}{2} + 2 = \frac{5}{2} \] At \( x = -1 \): \[ \frac{(-1)^2}{2} + 2(-1) = \frac{1}{2} - 2 = -\frac{3}{2} \] Difference: \[ \frac{5}{2} - \left(-\frac{3}{2} \right) = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4 \] Multiplying by \( \frac{1}{18} \): \[ P(-1 < X < 1) = \frac{4}{18} = \frac{2}{9} \] Final Answer: \( \frac{2}{9} \)