1. **Problem Statement:** Given the probability density function (PDF) $$f_X(x) = \frac{x}{250} \text{ for } 20 \leq x \leq 30, \quad f_X(x) = 0 \text{ otherwise}$$ We need to: (a) Sketch and verify it is a valid PDF. (b) Find the cumulative distribution function (CDF) $F_X(x)$. (c) Calculate $P(25 \leq X \leq 30)$. (d) Calculate $P(X > 28)$. (e) Find the mean $\mu_X$. (f) Find the variance $\sigma^2_X$. 2. **Verification of PDF:** A function is a valid PDF if: - $f_X(x) \geq 0$ for all $x$. - The total area under the curve is 1, i.e., $\int_{-\infty}^{\infty} f_X(x) dx = 1$. Since $f_X(x) = \frac{x}{250} \geq 0$ for $20 \leq x \leq 30$ and zero otherwise, the first condition holds. Calculate the integral: $$\int_{20}^{30} \frac{x}{250} dx = \frac{1}{250} \int_{20}^{30} x dx = \frac{1}{250} \left[ \frac{x^2}{2} \right]_{20}^{30} = \frac{1}{250} \left( \frac{30^2}{2} - \frac{20^2}{2} \right) = \frac{1}{250} \times \frac{900 - 400}{2} = \frac{1}{250} \times \frac{500}{2} = \frac{250}{250} = 1$$ Thus, $f_X(x)$ is a valid PDF. 3. **Cumulative Distribution Function (CDF) $F_X(x)$:** By definition: $$F_X(x) = P(X \leq x) = \int_{-\infty}^x f_X(t) dt$$ - For $x < 20$, $F_X(x) = 0$ since PDF is zero before 20. - For $20 \leq x \leq 30$: $$F_X(x) = \int_{20}^x \frac{t}{250} dt = \frac{1}{250} \left[ \frac{t^2}{2} \right]_{20}^x = \frac{1}{500} (x^2 - 400)$$ - For $x > 30$, $F_X(x) = 1$ since total probability is 1. So, $$F_X(x) = \begin{cases} 0 & x < 20 \\ \frac{x^2 - 400}{500} & 20 \leq x \leq 30 \\ 1 & x > 30 \end{cases}$$ 4. **Calculate $P(25 \leq X \leq 30)$:** Use the CDF: $$P(25 \leq X \leq 30) = F_X(30) - F_X(25) = 1 - \frac{25^2 - 400}{500} = 1 - \frac{625 - 400}{500} = 1 - \frac{225}{500} = 1 - 0.45 = 0.55$$ 5. **Calculate $P(X > 28)$:** $$P(X > 28) = 1 - F_X(28) = 1 - \frac{28^2 - 400}{500} = 1 - \frac{784 - 400}{500} = 1 - \frac{384}{500} = 1 - 0.768 = 0.232$$ 6. **Find the mean $\mu_X$:** Mean is: $$\mu_X = E[X] = \int_{20}^{30} x f_X(x) dx = \int_{20}^{30} x \cdot \frac{x}{250} dx = \int_{20}^{30} \frac{x^2}{250} dx = \frac{1}{250} \left[ \frac{x^3}{3} \right]_{20}^{30} = \frac{1}{250} \left( \frac{30^3}{3} - \frac{20^3}{3} \right) = \frac{1}{250} \times \frac{27000 - 8000}{3} = \frac{19000}{750} = \frac{190}{7.5} = 25.3333$$ 7. **Find the variance $\sigma^2_X$:** Variance is: $$\sigma^2_X = E[X^2] - (E[X])^2$$ Calculate $E[X^2]$: $$E[X^2] = \int_{20}^{30} x^2 f_X(x) dx = \int_{20}^{30} x^2 \cdot \frac{x}{250} dx = \int_{20}^{30} \frac{x^3}{250} dx = \frac{1}{250} \left[ \frac{x^4}{4} \right]_{20}^{30} = \frac{1}{250} \left( \frac{30^4}{4} - \frac{20^4}{4} \right) = \frac{1}{250} \times \frac{810000 - 160000}{4} = \frac{650000}{1000} = 650$$ Now, $$\sigma^2_X = 650 - (25.3333)^2 = 650 - 642.7778 = 7.2222$$ **Final answers:** - $P(25 \leq X \leq 30) = 0.55$ - $P(X > 28) = 0.232$ - $\mu_X = 25.3333$ - $\sigma^2_X = 7.2222$