1. Problem 9: Given the probability density function (pdf) \( f(x) = c x^2 (2 - x) \) for \( 0 \leq x \leq 2 \) and 0 otherwise. (a) To find \( c \), use the property that the total probability integrates to 1: $$\int_0^2 c x^2 (2 - x) \, dx = 1$$ Expand the integrand: $$x^2 (2 - x) = 2x^2 - x^3$$ Integrate: $$c \int_0^2 (2x^2 - x^3) \, dx = c \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 = c \left( \frac{2 \cdot 8}{3} - \frac{16}{4} \right) = c \left( \frac{16}{3} - 4 \right) = c \left( \frac{16 - 12}{3} \right) = \frac{4c}{3}$$ Set equal to 1: $$\frac{4c}{3} = 1 \implies c = \frac{3}{4}$$ (b) Find the mean (expected value) \( E(X) \): $$E(X) = \int_0^2 x f(x) \, dx = \int_0^2 x \cdot \frac{3}{4} x^2 (2 - x) \, dx = \frac{3}{4} \int_0^2 x^3 (2 - x) \, dx$$ Expand: $$x^3 (2 - x) = 2x^3 - x^4$$ Integrate: $$\frac{3}{4} \left[ \frac{2x^4}{4} - \frac{x^5}{5} \right]_0^2 = \frac{3}{4} \left( \frac{2 \cdot 16}{4} - \frac{32}{5} \right) = \frac{3}{4} \left( 8 - \frac{32}{5} \right) = \frac{3}{4} \cdot \frac{40 - 32}{5} = \frac{3}{4} \cdot \frac{8}{5} = \frac{6}{5} = 1.2$$ (c) Find \( P(X > 1.3) \): $$P(X > 1.3) = 1 - P(X \leq 1.3) = 1 - \int_0^{1.3} \frac{3}{4} x^2 (2 - x) \, dx$$ Calculate the integral: $$\int_0^{1.3} x^2 (2 - x) \, dx = \int_0^{1.3} (2x^2 - x^3) \, dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^{1.3} = \frac{2 (1.3)^3}{3} - \frac{(1.3)^4}{4}$$ Numerical values: $$1.3^3 = 2.197, \quad 1.3^4 = 2.8561$$ So: $$\frac{2 \times 2.197}{3} - \frac{2.8561}{4} = \frac{4.394}{3} - 0.714 = 1.4647 - 0.714 = 0.7507$$ Multiply by \( \frac{3}{4} \): $$\frac{3}{4} \times 0.7507 = 0.563$$ Therefore: $$P(X > 1.3) = 1 - 0.563 = 0.437$$ (d) To determine if the median \( m \) is greater than, less than, or equal to 1.3, solve: $$\int_0^m \frac{3}{4} x^2 (2 - x) \, dx = 0.5$$ Using the integral formula from (a): $$\frac{3}{4} \left( \frac{2m^3}{3} - \frac{m^4}{4} \right) = 0.5$$ Simplify: $$\frac{3}{4} \left( \frac{2m^3}{3} - \frac{m^4}{4} \right) = \frac{m^3}{2} - \frac{3m^4}{16} = 0.5$$ Try \( m = 1.3 \): $$\frac{(1.3)^3}{2} - \frac{3 (1.3)^4}{16} = \frac{2.197}{2} - \frac{3 \times 2.8561}{16} = 1.0985 - 0.535 = 0.5635 > 0.5$$ Since the CDF at 1.3 is greater than 0.5, the median \( m < 1.3 \). --- 2. Problem 10: Given pdf \( f(x) = \frac{k}{x^2} \) for \( 20 \leq x \leq 28 \) and 0 otherwise. (a) Find \( k \) using total probability: $$\int_{20}^{28} \frac{k}{x^2} \, dx = 1$$ Integrate: $$k \left[ -\frac{1}{x} \right]_{20}^{28} = k \left( -\frac{1}{28} + \frac{1}{20} \right) = k \left( \frac{1}{20} - \frac{1}{28} \right)$$ Calculate difference: $$\frac{1}{20} - \frac{1}{28} = \frac{28 - 20}{560} = \frac{8}{560} = \frac{1}{70}$$ Set equal to 1: $$k \times \frac{1}{70} = 1 \implies k = 70$$ (b) Find expected value \( E(X) \): $$E(X) = \int_{20}^{28} x \cdot \frac{70}{x^2} \, dx = 70 \int_{20}^{28} \frac{1}{x} \, dx = 70 [\ln x]_{20}^{28} = 70 (\ln 28 - \ln 20) = 70 \ln \frac{28}{20} = 70 \ln 1.4$$ Numerical value: $$\ln 1.4 \approx 0.3365 \implies E(X) \approx 70 \times 0.3365 = 23.56$$ (c) Calculate \( P(X < E(X)) = P(X < 23.56) \): $$P(X < 23.56) = \int_{20}^{23.56} \frac{70}{x^2} \, dx = 70 \left[ -\frac{1}{x} \right]_{20}^{23.56} = 70 \left( -\frac{1}{23.56} + \frac{1}{20} \right) = 70 \left( \frac{1}{20} - \frac{1}{23.56} \right)$$ Calculate difference: $$\frac{1}{20} = 0.05, \quad \frac{1}{23.56} \approx 0.04245$$ Difference: $$0.05 - 0.04245 = 0.00755$$ Multiply: $$70 \times 0.00755 = 0.5285$$ (d) Compare mean and median: Median \( m \) satisfies: $$\int_{20}^m \frac{70}{x^2} \, dx = 0.5$$ Calculate: $$70 \left( \frac{1}{20} - \frac{1}{m} \right) = 0.5 \implies \frac{1}{20} - \frac{1}{m} = \frac{0.5}{70} = \frac{1}{140}$$ Solve for \( m \): $$\frac{1}{m} = \frac{1}{20} - \frac{1}{140} = \frac{7 - 1}{140} = \frac{6}{140} = \frac{3}{70}$$ Thus: $$m = \frac{70}{3} \approx 23.33$$ Since \( E(X) \approx 23.56 > m \approx 23.33 \), the mean is greater than the median.