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