1. **Problem Statement:** We have a random variable $X$ with probability density function (pdf) $$f(x) = \begin{cases} \frac{1}{6}x, & 2 < x \leq 4 \\ 0, & \text{otherwise} \end{cases}$$ We need to (a) verify that $f(x)$ satisfies the properties of a pdf, and (b) find $P(2 < X \leq 3)$. 2. **Properties of a pdf:** - $f(x) \geq 0$ for all $x$. - The total area under the pdf curve must be 1, i.e., $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$ 3. **Check non-negativity:** For $2 < x \leq 4$, $f(x) = \frac{1}{6}x$ which is positive since $x > 2$. For other $x$, $f(x) = 0$. So, $f(x) \geq 0$ everywhere. 4. **Check total area:** Calculate $$\int_{2}^{4} \frac{1}{6}x \, dx = \frac{1}{6} \int_{2}^{4} x \, dx = \frac{1}{6} \left[ \frac{x^2}{2} \right]_{2}^{4} = \frac{1}{6} \left( \frac{16}{2} - \frac{4}{2} \right) = \frac{1}{6} (8 - 2) = \frac{6}{6} = 1$$ This confirms the total area under $f(x)$ is 1. 5. **Conclusion for (a):** $f(x)$ satisfies the properties of a pdf. 6. **Find $P(2 < X \leq 3)$:** Calculate $$P(2 < X \leq 3) = \int_{2}^{3} \frac{1}{6}x \, dx = \frac{1}{6} \left[ \frac{x^2}{2} \right]_{2}^{3} = \frac{1}{6} \left( \frac{9}{2} - \frac{4}{2} \right) = \frac{1}{6} \times \frac{5}{2} = \frac{5}{12}$$ **Final answers:** - (a) $f(x)$ is a valid pdf. - (b) $P(2 < X \leq 3) = \frac{5}{12}$.