Question

For the following probability density function of a random variable \( X \), find: \[ f(x) = \frac{x + 2}{18}, \quad \text{for} \, -2<x<4 \]

Hint:

To find probabilities from a probability density function, integrate the PDF over the desired range.

Answer 1

Step 1: To find \( P(X<1) \), integrate the probability density function from \( -2 \) to \( 1 \): \[ P(X<1) = \int_{-2}^1 \frac{x + 2}{18} \, dx \] Step 2: Solve the integral: \[ P(X<1) = \frac{1}{18} \int_{-2}^1 (x + 2) \, dx \] \[ = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-2}^1 \] \[ = \frac{1}{18} \left( \left( \frac{1^2}{2} + 2(1) \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right) \right) \] \[ = \frac{1}{18} \left( \left( \frac{1}{2} + 2 \right) - \left( \frac{4}{2} - 4 \right) \right) \] \[ = \frac{1}{18} \left( \frac{5}{2} - \left( 2 - 4 \right) \right) \] \[ = \frac{1}{18} \left( \frac{5}{2} + 2 \right) = \frac{1}{18} \times \frac{9}{2} = \frac{9}{36} = \frac{1}{4} \] Thus, \( P(X<1) = \frac{1}{4} \). Step 3: To find \( P(|X|<1) \), we integrate from \( -1 \) to \( 1 \): \[ P(|X|<1) = \int_{-1}^1 \frac{x + 2}{18} \, dx \] Step 4: Solve the integral: \[ P(|X|<1) = \frac{1}{18} \int_{-1}^1 (x + 2) \, dx \] \[ = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-1}^1 \] \[ = \frac{1}{18} \left( \left( \frac{1^2}{2} + 2(1) \right) - \left( \frac{(-1)^2}{2} + 2(-1) \right) \right) \] \[ = \frac{1}{18} \left( \left( \frac{1}{2} + 2 \right) - \left( \frac{1}{2} - 2 \right) \right) \] \[ = \frac{1}{18} \left( \frac{5}{2} - \left( \frac{1}{2} - 2 \right) \right) = \frac{1}{18} \times \frac{9}{2} = \frac{9}{36} = \frac{1}{4} \] Thus, \( P(|X|<1) = \frac{1}{4} \).