Questions: Example 2 Consider the function f with the rule: f(x) = 1.5(1-x^2) if 0 ≤ x ≤ 1 0 if x > 1 or x < 0 a Sketch the graph of f. b Show that f is a probability density function. c Find Pr(X > 0.5), where the random variable X has probability density function f. Solution a For 0 ≤ x ≤ 1, the graph of y=f(x) is part of a parabola with intercepts at (0,1.5) and (1,0).

Solution Steps

• For \( 0 \leq x \leq 1 \), the function \( f(x) = 1.5(1 - x^2) \) is a downward-opening parabola.
• The graph has intercepts at \( (0, 1.5) \) and \( (1, 0) \).
• For \( x > 1 \) or \( x < 0 \), \( f(x) = 0 \).

• A function \( f \) is a probability density function if:
  1. \( f(x) \geq 0 \) for all \( x \).
  2. The integral of \( f(x) \) over all possible values of \( x \) is 1.
• For \( 0 \leq x \leq 1 \), \( f(x) = 1.5(1 - x^2) \geq 0 \).
• For \( x > 1 \) or \( x < 0 \), \( f(x) = 0 \).
• Calculate the integral: \[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{1} 1.5(1 - x^2) \, dx \] \[ = 1.5 \int_{0}^{1} (1 - x^2) \, dx \] \[ = 1.5 \left[ x - \frac{x^3}{3} \right]_{0}^{1} \] \[ = 1.5 \left( 1 - \frac{1}{3} \right) \] \[ = 1.5 \left( \frac{2}{3} \right) \] \[ = 1 \]
• Therefore, \( f(x) \) is a probability density function.

• Calculate the probability: \[ \Pr(X > 0.5) = \int_{0.5}^{1} f(x) \, dx \] \[ = \int_{0.5}^{1} 1.5(1 - x^2) \, dx \] \[ = 1.5 \int_{0.5}^{1} (1 - x^2) \, dx \] \[ = 1.5 \left[ x - \frac{x^3}{3} \right]_{0.5}^{1} \] \[ = 1.5 \left( \left(1 - \frac{1}{3}\right) - \left(0.5 - \frac{0.5^3}{3}\right) \right) \] \[ = 1.5 \left( \frac{2}{3} - \left(0.5 - \frac{0.125}{3}\right) \right) \] \[ = 1.5 \left( \frac{2}{3} - \left(0.5 - 0.0417\right) \right) \] \[ = 1.5 \left( \frac{2}{3} - 0.4583 \right) \] \[ = 1.5 \left( 0.2083 \right) \] \[ = 0.3125 \]

Final Answer