1. Problem 9: Given the probability density function (pdf) \( f(x) = cx^2(2-x) \) for \( 0 \leq x \leq 2 \) and 0 otherwise. (a) To find \( c \), use the property that the total probability integrates to 1: $$ \int_0^2 cx^2(2-x) \, dx = 1 $$ Calculate the integral: $$ \int_0^2 x^2(2-x) \, dx = \int_0^2 (2x^2 - x^3) \, dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 = \frac{2(8)}{3} - \frac{16}{4} = \frac{16}{3} - 4 = \frac{16 - 12}{3} = \frac{4}{3} $$ So: $$ c \times \frac{4}{3} = 1 \implies c = \frac{3}{4} $$ (b) The mean mass \( E(X) \) is: $$ E(X) = \int_0^2 x f(x) \, dx = \int_0^2 x \cdot \frac{3}{4} x^2 (2-x) \, dx = \frac{3}{4} \int_0^2 (2x^3 - x^4) \, dx $$ Calculate the integral: $$ \int_0^2 2x^3 \, dx = \left[ \frac{2x^4}{4} \right]_0^2 = \left[ \frac{x^4}{2} \right]_0^2 = \frac{16}{2} = 8 $$ $$ \int_0^2 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^2 = \frac{32}{5} $$ So: $$ E(X) = \frac{3}{4} (8 - \frac{32}{5}) = \frac{3}{4} \times \frac{40 - 32}{5} = \frac{3}{4} \times \frac{8}{5} = \frac{24}{20} = 1.2 $$ (c) Probability mass more than 1.3 kg: $$ P(X > 1.3) = \int_{1.3}^2 f(x) \, dx = \frac{3}{4} \int_{1.3}^2 x^2 (2-x) \, dx = \frac{3}{4} \int_{1.3}^2 (2x^2 - x^3) \, dx $$ Calculate the integral: $$ \int_{1.3}^2 2x^2 \, dx = \left[ \frac{2x^3}{3} \right]_{1.3}^2 = \frac{16}{3} - \frac{2(1.3)^3}{3} = \frac{16}{3} - \frac{2(2.197)}{3} = \frac{16}{3} - \frac{4.394}{3} = \frac{11.606}{3} $$ $$ \int_{1.3}^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_{1.3}^2 = 4 - \frac{(1.3)^4}{4} = 4 - \frac{2.8561}{4} = 4 - 0.714 = 3.286 $$ So: $$ P(X > 1.3) = \frac{3}{4} \left( \frac{11.606}{3} - 3.286 \right) = \frac{3}{4} \times (3.8687 - 3.286) = \frac{3}{4} \times 0.5827 = 0.437 \text{ (approx)} $$ (d) To determine if the median \( m \) is greater than, less than, or equal to 1.3, solve: $$ \int_0^m f(x) \, dx = 0.5 $$ Calculate: $$ \int_0^m \frac{3}{4} x^2 (2-x) \, dx = \frac{3}{4} \int_0^m (2x^2 - x^3) \, dx = \frac{3}{4} \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^m = \frac{3}{4} \left( \frac{2m^3}{3} - \frac{m^4}{4} \right) = 0.5 $$ Simplify: $$ \frac{3}{4} \left( \frac{2m^3}{3} - \frac{m^4}{4} \right) = 0.5 \implies \frac{3}{4} \times \frac{8m^3 - 3m^4}{12} = 0.5 $$ $$ \frac{3(8m^3 - 3m^4)}{48} = 0.5 \implies 3(8m^3 - 3m^4) = 24 $$ $$ 24m^3 - 9m^4 = 24 \implies 8m^3 - 3m^4 = 8 $$ Try \( m=1.3 \): $$ 8(1.3)^3 - 3(1.3)^4 = 8(2.197) - 3(2.856) = 17.576 - 8.568 = 9.008 > 8 $$ Try \( m=1.2 \): $$ 8(1.2)^3 - 3(1.2)^4 = 8(1.728) - 3(2.0736) = 13.824 - 6.2208 = 7.6032 < 8 $$ Since the function is increasing in \( m \), median \( m \) is between 1.2 and 1.3, so median \( < 1.3 \). --- 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: $$ \int_{20}^{28} \frac{k}{x^2} \, dx = 1 \implies k \int_{20}^{28} x^{-2} \, dx = 1 $$ Calculate integral: $$ \int_{20}^{28} x^{-2} \, dx = \left[ -\frac{1}{x} \right]_{20}^{28} = -\frac{1}{28} + \frac{1}{20} = \frac{1}{20} - \frac{1}{28} = \frac{28 - 20}{560} = \frac{8}{560} = \frac{1}{70} $$ So: $$ k \times \frac{1}{70} = 1 \implies k = 70 $$ (b) Expected time \( E(X) \): $$ E(X) = \int_{20}^{28} x \cdot \frac{70}{x^2} \, dx = 70 \int_{20}^{28} x^{-1} \, dx = 70 \left[ \ln x \right]_{20}^{28} = 70 (\ln 28 - \ln 20) = 70 \ln \frac{28}{20} = 70 \ln 1.4 $$ Approximate: $$ \ln 1.4 \approx 0.3365 \implies E(X) \approx 70 \times 0.3365 = 23.56 $$ (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: $$ \frac{1}{20} = 0.05, \quad \frac{1}{23.56} \approx 0.04245 $$ So: $$ P(X < 23.56) = 70 (0.05 - 0.04245) = 70 \times 0.00755 = 0.5285 $$ (d) To compare mean and median: Median \( m \) satisfies: $$ \int_{20}^m \frac{70}{x^2} \, dx = 0.5 \implies 70 \left( \frac{1}{20} - \frac{1}{m} \right) = 0.5 $$ Simplify: $$ \frac{70}{20} - \frac{70}{m} = 0.5 \implies 3.5 - \frac{70}{m} = 0.5 \implies \frac{70}{m} = 3.0 \implies m = \frac{70}{3} \approx 23.33 $$ Since \( E(X) \approx 23.56 > m \approx 23.33 \), the mean is greater than the median. Final answers: - 9(a) \( c = \frac{3}{4} \) - 9(b) Mean mass \( = 1.2 \) - 9(c) Probability mass > 1.3 \( \approx 0.437 \) - 9(d) Median mass \( < 1.3 \) - 10(a) \( k = 70 \) - 10(b) Expected time \( \approx 23.56 \) - 10(c) \( P(X < E(X)) \approx 0.5285 \) - 10(d) Mean \( > \) Median