1. **Problem (a):** Find the cumulative distribution function (CDF) values for $F(x) = \frac{1}{16} \left( \frac{4}{x} \right)$ at $x=0,1,2,3,4$. 2. **Important note:** The function $F(x) = \frac{1}{16} \left( \frac{4}{x} \right)$ is undefined at $x=0$ because division by zero is not allowed. 3. Calculate each value where defined: - $F(1) = \frac{1}{16} \times \frac{4}{1} = \frac{4}{16} = \frac{1}{4}$ - $F(2) = \frac{1}{16} \times \frac{4}{2} = \frac{1}{8}$ - $F(3) = \frac{1}{16} \times \frac{4}{3} = \frac{1}{12}$ - $F(4) = \frac{1}{16} \times \frac{4}{4} = \frac{1}{16}$ 4. Since $F(0)$ is undefined, we state it as such. --- 5. **Problem (b):** Given the density function $$f(x) = \begin{cases} \frac{x^2}{3} & -1 < x < 2 \\ 0 & \text{elsewhere} \end{cases}$$ Find the cumulative distribution function $F(x)$ and evaluate $P(0 < X \leq 1)$. 6. The CDF $F(x)$ is the integral of the density function from the lower limit up to $x$: $$F(x) = \int_{-1}^x \frac{t^2}{3} dt = \frac{1}{3} \int_{-1}^x t^2 dt = \frac{1}{3} \left[ \frac{t^3}{3} \right]_{-1}^x = \frac{1}{9} (x^3 - (-1)^3) = \frac{1}{9} (x^3 + 1)$$ for $-1 < x < 2$. 7. For $x \leq -1$, $F(x) = 0$ and for $x \geq 2$, $F(x) = 1$ (since total probability is 1). 8. To find $P(0 < X \leq 1)$, compute: $$P(0 < X \leq 1) = F(1) - F(0) = \frac{1}{9} (1^3 + 1) - \frac{1}{9} (0^3 + 1) = \frac{1}{9} (2) - \frac{1}{9} (1) = \frac{1}{9}$$ --- **Final answers:** - (a) $F(0)$ undefined, $F(1) = \frac{1}{4}$, $F(2) = \frac{1}{8}$, $F(3) = \frac{1}{12}$, $F(4) = \frac{1}{16}$ - (b) $F(x) = 0$ for $x \leq -1$, $F(x) = \frac{1}{9} (x^3 + 1)$ for $-1 < x < 2$, $F(x) = 1$ for $x \geq 2$, and $P(0 < X \leq 1) = \frac{1}{9}$