1. **Problem Statement:** We have a population divided by age groups and poverty status. We want to find: (i) The joint probability distribution of age (X) and poverty status (Y). (ii) The marginal probability distribution of age (X). (iii) The marginal probability distribution of poverty status (Y). 2. **Given Data:** - Total population = 36,998 (in thousands) - Age groups and counts: - Under 18: Poverty = 347, No poverty = 12,680, Total = 13,027 - 18 to 64: Poverty = 517, No poverty = 19,997, Total = 20,514 - 65 & Older: Poverty = 130, No poverty = 3,327, Total = 3,457 3. **Random Variables:** - $X = \begin{cases} 9 & \text{if age} < 18 \\ 41 & \text{if } 18 \leq \text{age} < 64 \\ 82 & \text{if age} \geq 65 \end{cases}$ - $Y = \begin{cases} 0 & \text{if not living in poverty} \\ 1 & \text{if living in poverty} \end{cases}$ 4. **Formulas:** - Joint probability: $P(X=x, Y=y) = \frac{\text{count in category}}{\text{total population}}$ - Marginal probability of $X$: $P(X=x) = \sum_y P(X=x, Y=y)$ - Marginal probability of $Y$: $P(Y=y) = \sum_x P(X=x, Y=y)$ --- **(i) Joint Probability Distribution:** Calculate each joint probability by dividing counts by total population 36,998. - $P(X=9, Y=1) = \frac{347}{36998} \approx 0.00938$ - $P(X=9, Y=0) = \frac{12680}{36998} \approx 0.3426$ - $P(X=41, Y=1) = \frac{517}{36998} \approx 0.01397$ - $P(X=41, Y=0) = \frac{19997}{36998} \approx 0.5405$ - $P(X=82, Y=1) = \frac{130}{36998} \approx 0.00351$ - $P(X=82, Y=0) = \frac{3327}{36998} \approx 0.0899$ **Check sum:** $0.00938 + 0.3426 + 0.01397 + 0.5405 + 0.00351 + 0.0899 = 1.0$ (approximately) --- **(ii) Marginal Probability Distribution of Age $X$:** Sum joint probabilities over $Y$ for each $X$: - $P(X=9) = 0.00938 + 0.3426 = 0.35198$ - $P(X=41) = 0.01397 + 0.5405 = 0.55447$ - $P(X=82) = 0.00351 + 0.0899 = 0.09341$ Check sum: $0.35198 + 0.55447 + 0.09341 = 1.0$ --- **(iii) Marginal Probability Distribution of Poverty $Y$:** Sum joint probabilities over $X$ for each $Y$: - $P(Y=1) = 0.00938 + 0.01397 + 0.00351 = 0.02686$ - $P(Y=0) = 0.3426 + 0.5405 + 0.0899 = 0.97299$ Check sum: $0.02686 + 0.97299 = 0.99985 \approx 1$ --- **Final answers:** - Joint distribution $P(X,Y)$ as above. - Marginal distribution $P(X)$ as above. - Marginal distribution $P(Y)$ as above.