1. **Problem Statement:** We have two sets of data for seven students: course averages before the final exam $X = \{69.3, 87.7, 50.5, 51.9, 80.7, 70.5, 72.4\}$ and final exam scores $Y = \{56, 89, 55, 49, 61, 66, 72\}$. We need to find the regression lines $Y$ on $X$ and $X$ on $Y$, the correlation coefficient $r$, and the angle between the regression lines. 2. **Formulas and Important Rules:** - Regression line of $Y$ on $X$: $$Y = a + bX$$ where $$b = r \frac{s_Y}{s_X}$$ and $$a = \bar{Y} - b\bar{X}$$ - Regression line of $X$ on $Y$: $$X = c + dY$$ where $$d = r \frac{s_X}{s_Y}$$ and $$c = \bar{X} - d\bar{Y}$$ - Correlation coefficient: $$r = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{(n-1)s_X s_Y}$$ - Angle $\theta$ between regression lines: $$\tan \theta = \left| \frac{b - \frac{1}{d}}{1 + b \frac{1}{d}} \right|$$ 3. **Calculate means:** $$\bar{X} = \frac{69.3 + 87.7 + 50.5 + 51.9 + 80.7 + 70.5 + 72.4}{7} = 69.71$$ $$\bar{Y} = \frac{56 + 89 + 55 + 49 + 61 + 66 + 72}{7} = 63.43$$ 4. **Calculate standard deviations:** $$s_X = \sqrt{\frac{\sum (X_i - \bar{X})^2}{n-1}} = 13.17$$ $$s_Y = \sqrt{\frac{\sum (Y_i - \bar{Y})^2}{n-1}} = 14.39$$ 5. **Calculate covariance:** $$\text{Cov}(X,Y) = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{n-1} = 178.43$$ 6. **Calculate correlation coefficient:** $$r = \frac{\text{Cov}(X,Y)}{s_X s_Y} = \frac{178.43}{13.17 \times 14.39} = 0.942$$ 7. **Calculate regression coefficients:** - For $Y$ on $X$: $$b = r \frac{s_Y}{s_X} = 0.942 \times \frac{14.39}{13.17} = 1.03$$ $$a = \bar{Y} - b \bar{X} = 63.43 - 1.03 \times 69.71 = -8.44$$ - For $X$ on $Y$: $$d = r \frac{s_X}{s_Y} = 0.942 \times \frac{13.17}{14.39} = 0.86$$ $$c = \bar{X} - d \bar{Y} = 69.71 - 0.86 \times 63.43 = 15.04$$ 8. **Regression lines:** $$Y = -8.44 + 1.03X$$ $$X = 15.04 + 0.86Y$$ 9. **Calculate angle between regression lines:** $$\tan \theta = \left| \frac{b - \frac{1}{d}}{1 + b \frac{1}{d}} \right| = \left| \frac{1.03 - \frac{1}{0.86}}{1 + 1.03 \times \frac{1}{0.86}} \right| = \left| \frac{1.03 - 1.16}{1 + 1.20} \right| = \frac{0.13}{2.20} = 0.059$$ $$\theta = \arctan(0.059) = 3.38^\circ$$ **Final answers:** - Regression line of $Y$ on $X$: $$Y = -8.44 + 1.03X$$ - Regression line of $X$ on $Y$: $$X = 15.04 + 0.86Y$$ - Correlation coefficient: $$r = 0.942$$ - Angle between regression lines: $$3.38^\circ$$