1. **Problem statement:** Given the probability density function (pdf) of a continuous random variable $Y$: $$f(y) = \begin{cases} \frac{1}{5}, & 0 \leq y < 2 \\ \frac{1}{2}y + D, & 2 \leq y < 3 \\ 0, & \text{elsewhere} \end{cases}$$ Find the value of $D$ and then compute $P(1 \leq Y < 2.5)$. 2. **Formula and rules:** - The total area under the pdf must be 1: $$\int_{-\infty}^{\infty} f(y) dy = 1$$ - Probability for an interval $[a,b)$ is: $$P(a \leq Y < b) = \int_a^b f(y) dy$$ 3. **Find $D$:** Calculate the total integral: $$\int_0^2 \frac{1}{5} dy + \int_2^3 \left( \frac{1}{2}y + D \right) dy = 1$$ Calculate each integral: $$\int_0^2 \frac{1}{5} dy = \frac{1}{5} \times (2 - 0) = \frac{2}{5}$$ $$\int_2^3 \left( \frac{1}{2}y + D \right) dy = \int_2^3 \frac{1}{2}y dy + \int_2^3 D dy$$ Calculate separately: $$\int_2^3 \frac{1}{2}y dy = \frac{1}{2} \times \frac{y^2}{2} \Big|_2^3 = \frac{1}{4}(3^2 - 2^2) = \frac{1}{4}(9 - 4) = \frac{5}{4}$$ $$\int_2^3 D dy = D (3 - 2) = D$$ Sum: $$\frac{2}{5} + \frac{5}{4} + D = 1$$ 4. **Solve for $D$:** $$D = 1 - \frac{2}{5} - \frac{5}{4} = 1 - 0.4 - 1.25 = 1 - 1.65 = -0.65$$ 5. **Calculate $P(1 \leq Y < 2.5)$:** Split the integral: $$P(1 \leq Y < 2.5) = \int_1^2 \frac{1}{5} dy + \int_2^{2.5} \left( \frac{1}{2}y + D \right) dy$$ Calculate each: $$\int_1^2 \frac{1}{5} dy = \frac{1}{5} (2 - 1) = \frac{1}{5} = 0.2$$ $$\int_2^{2.5} \left( \frac{1}{2}y - 0.65 \right) dy = \int_2^{2.5} \frac{1}{2}y dy - \int_2^{2.5} 0.65 dy$$ Calculate separately: $$\int_2^{2.5} \frac{1}{2}y dy = \frac{1}{2} \times \frac{y^2}{2} \Big|_2^{2.5} = \frac{1}{4} (2.5^2 - 2^2) = \frac{1}{4} (6.25 - 4) = \frac{2.25}{4} = 0.5625$$ $$\int_2^{2.5} 0.65 dy = 0.65 (2.5 - 2) = 0.65 \times 0.5 = 0.325$$ Sum: $$0.5625 - 0.325 = 0.2375$$ 6. **Final probability:** $$P(1 \leq Y < 2.5) = 0.2 + 0.2375 = 0.4375$$ **Answer:** - $D = -0.65$ - $P(1 \leq Y < 2.5) = 0.4375$