1. **Problem:** Given the pdf \(f(x) = 0.075x + 0.2\) for \(3 \leq x \leq 5\) and 0 otherwise, verify it is a valid pdf and calculate probabilities. 2. **Step 1: Verify total area under the curve is 1** The total area under a pdf must be 1, so we compute: $$\int_3^5 (0.075x + 0.2) \, dx$$ Calculate the integral: $$\int_3^5 0.075x \, dx + \int_3^5 0.2 \, dx = 0.075 \cdot \frac{x^2}{2} \Big|_3^5 + 0.2x \Big|_3^5$$ Evaluate each term: $$0.075 \cdot \frac{5^2 - 3^2}{2} + 0.2(5 - 3) = 0.075 \cdot \frac{25 - 9}{2} + 0.2 \cdot 2 = 0.075 \cdot 8 + 0.4 = 0.6 + 0.4 = 1$$ So the total area is 1, confirming it is a valid pdf. 3. **Step 2: Calculate \(P(X \leq 4)\)** Since \(X\) is continuous, \(P(X \leq 4) = P(X < 4)\). Calculate: $$P(X \leq 4) = \int_3^4 (0.075x + 0.2) \, dx = 0.075 \cdot \frac{x^2}{2} \Big|_3^4 + 0.2x \Big|_3^4$$ Evaluate: $$0.075 \cdot \frac{16 - 9}{2} + 0.2(4 - 3) = 0.075 \cdot 3.5 + 0.2 = 0.2625 + 0.2 = 0.4625$$ 4. **Step 3: Calculate \(P(3.5 \leq X \leq 4.5)\) and \(P(X > 4.5)\)** Calculate: $$P(3.5 \leq X \leq 4.5) = \int_{3.5}^{4.5} (0.075x + 0.2) \, dx = 0.075 \cdot \frac{x^2}{2} \Big|_{3.5}^{4.5} + 0.2x \Big|_{3.5}^{4.5}$$ Evaluate: $$0.075 \cdot \frac{20.25 - 12.25}{2} + 0.2(4.5 - 3.5) = 0.075 \cdot 4 + 0.2 \cdot 1 = 0.3 + 0.2 = 0.5$$ Calculate: $$P(X > 4.5) = 1 - P(X \leq 4.5) = 1 - \int_3^{4.5} (0.075x + 0.2) \, dx$$ Compute \(\int_3^{4.5}\): $$0.075 \cdot \frac{4.5^2 - 3^2}{2} + 0.2(4.5 - 3) = 0.075 \cdot \frac{20.25 - 9}{2} + 0.2 \cdot 1.5 = 0.075 \cdot 5.625 + 0.3 = 0.421875 + 0.3 = 0.721875$$ So: $$P(X > 4.5) = 1 - 0.721875 = 0.278125$$ **Summary:** - Total area under pdf = 1 - \(P(X \leq 4) = P(X < 4) = 0.4625\) - \(P(3.5 \leq X \leq 4.5) = 0.5\) - \(P(X > 4.5) = 0.278125\)