1. **Problem 1: Find marginal distributions and conditional distribution** Given joint pmf: $$P(x,y) = K(2x + 3y)$$ for $$x=0,1,2$$ and $$y=1,2,3$$. 2. **Find K using normalization:** $$\sum_{x=0}^2 \sum_{y=1}^3 K(2x + 3y) = 1$$ Calculate the sum inside: $$\sum_{x=0}^2 \sum_{y=1}^3 (2x + 3y) = \sum_{x=0}^2 \left(2x \cdot 3 + 3 \sum_{y=1}^3 y\right) = \sum_{x=0}^2 (6x + 3 \cdot 6) = \sum_{x=0}^2 (6x + 18)$$ Calculate for each x: - For x=0: 18 - For x=1: 6(1)+18=24 - For x=2: 12+18=30 Sum: 18 + 24 + 30 = 72 So: $$K \times 72 = 1 \Rightarrow K = \frac{1}{72}$$ 3. **Marginal distribution of X:** $$P_X(x) = \sum_{y=1}^3 P(x,y) = \sum_{y=1}^3 \frac{1}{72}(2x + 3y) = \frac{1}{72} \left(3 \cdot 2x + 3 \sum_{y=1}^3 y\right) = \frac{1}{72} (6x + 18) = \frac{6x + 18}{72} = \frac{x + 3}{12}$$ for $$x=0,1,2$$. 4. **Marginal distribution of Y:** $$P_Y(y) = \sum_{x=0}^2 P(x,y) = \sum_{x=0}^2 \frac{1}{72}(2x + 3y) = \frac{1}{72} \left(3 \sum_{x=0}^2 2x + 3y \cdot 3\right) = \frac{1}{72} (3 \cdot 2 \cdot 3 + 9y) = \frac{18 + 9y}{72} = \frac{2 + y}{8}$$ for $$y=1,2,3$$. 5. **Conditional distribution of Y given X:** $$P_{Y|X}(y|x) = \frac{P(x,y)}{P_X(x)} = \frac{\frac{1}{72}(2x + 3y)}{\frac{x+3}{12}} = \frac{2x + 3y}{72} \times \frac{12}{x+3} = \frac{2x + 3y}{6(x+3)}$$ for $$x=0,1,2$$ and $$y=1,2,3$$. --- 6. **Problem 2: Calculate correlation coefficient for given heights** Data: $$X = [65,66,67,67,68,69,70,72]$$ $$Y = [67,68,65,68,72,72,69,71]$$ 7. **Calculate means:** $$\bar{X} = \frac{65+66+67+67+68+69+70+72}{8} = \frac{544}{8} = 68$$ $$\bar{Y} = \frac{67+68+65+68+72+72+69+71}{8} = \frac{552}{8} = 69$$ 8. **Calculate covariance:** $$\text{Cov}(X,Y) = \frac{1}{8} \sum (X_i - \bar{X})(Y_i - \bar{Y})$$ Calculate each term: - (65-68)(67-69) = (-3)(-2) = 6 - (66-68)(68-69) = (-2)(-1) = 2 - (67-68)(65-69) = (-1)(-4) = 4 - (67-68)(68-69) = (-1)(-1) = 1 - (68-68)(72-69) = 0*3=0 - (69-68)(72-69) = 1*3=3 - (70-68)(69-69) = 2*0=0 - (72-68)(71-69) = 4*2=8 Sum = 6+2+4+1+0+3+0+8=24 So: $$\text{Cov}(X,Y) = \frac{24}{8} = 3$$ 9. **Calculate standard deviations:** $$\sigma_X = \sqrt{\frac{1}{8} \sum (X_i - \bar{X})^2}$$ Calculate: - (65-68)^2=9 - (66-68)^2=4 - (67-68)^2=1 - (67-68)^2=1 - (68-68)^2=0 - (69-68)^2=1 - (70-68)^2=4 - (72-68)^2=16 Sum=36 $$\sigma_X = \sqrt{\frac{36}{8}} = \sqrt{4.5} = 2.1213$$ Similarly for Y: - (67-69)^2=4 - (68-69)^2=1 - (65-69)^2=16 - (68-69)^2=1 - (72-69)^2=9 - (72-69)^2=9 - (69-69)^2=0 - (71-69)^2=4 Sum=44 $$\sigma_Y = \sqrt{\frac{44}{8}} = \sqrt{5.5} = 2.3452$$ 10. **Correlation coefficient:** $$r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} = \frac{3}{2.1213 \times 2.3452} = \frac{3}{4.974} = 0.6035$$ --- 11. **Problem 3: Correlation coefficient from joint pdf** Given: $$f_{XY}(x,y) = 2xy, \quad 0 \leq x \leq 1, 0 \leq y \leq 1$$ (corrected domain to 0 to 1 for y, since 0 to 10 would not normalize) 12. **Find means:** $$E[X] = \int_0^1 \int_0^1 x \cdot 2xy \, dy \, dx = 2 \int_0^1 x^2 \left( \int_0^1 y \, dy \right) dx = 2 \int_0^1 x^2 \cdot \frac{1}{2} dx = \int_0^1 x^2 dx = \frac{1}{3}$$ Similarly: $$E[Y] = 2 \int_0^1 y \left( \int_0^1 x y \, dx \right) dy = 2 \int_0^1 y^2 \left( \int_0^1 x \, dx \right) dy = 2 \int_0^1 y^2 \cdot \frac{1}{2} dy = \int_0^1 y^2 dy = \frac{1}{3}$$ 13. **Find $E[XY]$:** $$E[XY] = \int_0^1 \int_0^1 xy \cdot 2xy \, dy \, dx = 2 \int_0^1 \int_0^1 x^2 y^2 \, dy \, dx = 2 \int_0^1 x^2 \left( \int_0^1 y^2 dy \right) dx = 2 \int_0^1 x^2 \cdot \frac{1}{3} dx = \frac{2}{3} \int_0^1 x^2 dx = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$$ 14. **Calculate variances:** $$Var(X) = E[X^2] - (E[X])^2$$ Calculate: $$E[X^2] = \int_0^1 \int_0^1 x^2 \cdot 2xy \, dy \, dx = 2 \int_0^1 x^3 \left( \int_0^1 y dy \right) dx = 2 \int_0^1 x^3 \cdot \frac{1}{2} dx = \int_0^1 x^3 dx = \frac{1}{4}$$ So: $$Var(X) = \frac{1}{4} - \left(\frac{1}{3}\right)^2 = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$$ Similarly for Y: $$E[Y^2] = 2 \int_0^1 y^3 \left( \int_0^1 x dx \right) dy = 2 \int_0^1 y^3 \cdot \frac{1}{2} dy = \int_0^1 y^3 dy = \frac{1}{4}$$ $$Var(Y) = \frac{1}{4} - \left(\frac{1}{3}\right)^2 = \frac{5}{36}$$ 15. **Calculate covariance:** $$Cov(X,Y) = E[XY] - E[X]E[Y] = \frac{2}{9} - \frac{1}{3} \times \frac{1}{3} = \frac{2}{9} - \frac{1}{9} = \frac{1}{9}$$ 16. **Correlation coefficient:** $$r = \frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{Var(Y)}} = \frac{\frac{1}{9}}{\sqrt{\frac{5}{36}} \sqrt{\frac{5}{36}}} = \frac{\frac{1}{9}}{\frac{5}{36}} = \frac{1}{9} \times \frac{36}{5} = \frac{4}{5} = 0.8$$ --- 17. **Problem 4: Regression equations to find means and correlation** Given: $$3x + 2y = 26$$ $$6x - y = 31$$ 18. **Find means by solving system:** Multiply second by 2: $$12x - 2y = 62$$ Add to first: $$(3x + 2y) + (12x - 2y) = 26 + 62 \Rightarrow 15x = 88 \Rightarrow x = \frac{88}{15} = 5.8667$$ Substitute in first: $$3(5.8667) + 2y = 26 \Rightarrow 17.6 + 2y = 26 \Rightarrow 2y = 8.4 \Rightarrow y = 4.2$$ 19. **Find correlation coefficient:** From regression equations: $$b_{yx} = -\frac{3}{2}, \quad b_{xy} = \frac{1}{6}$$ Correlation coefficient: $$r = \pm \sqrt{b_{yx} \times b_{xy}} = \pm \sqrt{-\frac{3}{2} \times \frac{1}{6}} = \pm \sqrt{-\frac{3}{12}} = \pm \sqrt{-\frac{1}{4}}$$ Since correlation must be real, check signs carefully. Rewrite regression equations in standard form: $$y = -\frac{3}{2}x + 13$$ $$y = 6x - 31$$ This is inconsistent; likely a typo or sign error. Assuming regression equations are: $$3x + 2y = 26$$ $$6x - y = 31$$ Rewrite second as: $$y = 6x - 31$$ Substitute in first: $$3x + 2(6x - 31) = 26 \Rightarrow 3x + 12x - 62 = 26 \Rightarrow 15x = 88 \Rightarrow x=5.8667$$ Then: $$y = 6(5.8667) - 31 = 35.2 - 31 = 4.2$$ Regression coefficients: $$b_{yx} = -\frac{3}{2}$$ from first equation rearranged for y: $$2y = 26 - 3x \Rightarrow y = 13 - \frac{3}{2}x$$ $$b_{xy} = \frac{1}{6}$$ from second equation rearranged for x: $$6x - y = 31 \Rightarrow 6x = y + 31 \Rightarrow x = \frac{y}{6} + \frac{31}{6}$$ Slope of x on y is $$\frac{1}{6}$$. Correlation: $$r = \pm \sqrt{b_{yx} b_{xy}} = \pm \sqrt{-\frac{3}{2} \times \frac{1}{6}} = \pm \sqrt{-\frac{3}{12}}$$ Negative inside root means no real correlation coefficient; likely a sign error. If we take absolute values: $$r = \pm \sqrt{\frac{3}{2} \times \frac{1}{6}} = \pm \sqrt{\frac{3}{12}} = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$ Assuming positive correlation: $$r = 0.5$$ --- **Final answers:** - Marginal distributions: $$P_X(x) = \frac{x+3}{12}, P_Y(y) = \frac{y+2}{8}$$ - Conditional distribution: $$P_{Y|X}(y|x) = \frac{2x + 3y}{6(x+3)}$$ - Correlation coefficient (heights): $$0.6035$$ - Correlation coefficient (joint pdf): $$0.8$$ - Means from regression: $$\bar{x} = 5.8667, \bar{y} = 4.2$$ - Correlation coefficient (regression): $$0.5$$