1. **Problem Statement:** Given the regression equations $$3x + 2y - 26 = 0$$ and $$0.67x + y - 51 = 0$$ and variance of $$x = 25$$, find: - (ci) the mean values of $$x$$ and $$y$$ - (cii) the standard deviations of $$x$$ and $$y$$ - (ciii) the coefficient of correlation between $$x$$ and $$y$$ 2. **Step 1: Find the means of $$x$$ and $$y$$** The regression lines can be written as: $$y = -\frac{3}{2}x + 13$$ and $$y = -0.67x + 51$$ The means $$\bar{x}$$ and $$\bar{y}$$ lie at the intersection of these two lines. Set the right sides equal: $$-\frac{3}{2}x + 13 = -0.67x + 51$$ Multiply decimals to fractions for clarity: $$0.67 \approx \frac{2}{3}$$ (approximate) Solve for $$x$$: $$-1.5x + 13 = -0.67x + 51$$ $$-1.5x + 0.67x = 51 - 13$$ $$-0.83x = 38$$ $$x = \frac{38}{-0.83} \approx -45.78$$ This negative value is suspicious; let's solve exactly using fractions: Rewrite equations: $$3x + 2y = 26$$ $$0.67x + y = 51$$ Multiply second equation by 2: $$1.34x + 2y = 102$$ Subtract first equation from this: $$(1.34x + 2y) - (3x + 2y) = 102 - 26$$ $$1.34x - 3x = 76$$ $$-1.66x = 76$$ $$x = \frac{76}{-1.66} \approx -45.78$$ Again negative, which is unusual for age data, but we proceed mathematically. Find $$y$$ using second equation: $$y = 51 - 0.67x = 51 - 0.67(-45.78) = 51 + 30.68 = 81.68$$ So, $$\bar{x} \approx -45.78, \quad \bar{y} \approx 81.68$$ 3. **Step 2: Find standard deviations of $$x$$ and $$y$$** Given variance of $$x$$ is $$\sigma_x^2 = 25$$, so $$\sigma_x = \sqrt{25} = 5$$ From regression equations, slopes are: $$b_{yx} = -\frac{3}{2} = -1.5$$ (regression of $$y$$ on $$x$$) $$b_{xy} = -0.67$$ (regression of $$x$$ on $$y$$) Recall the relation: $$b_{yx} b_{xy} = r^2$$ where $$r$$ is the correlation coefficient. Calculate $$r^2$$: $$r^2 = (-1.5)(-0.67) = 1.005$$ Since $$r^2$$ cannot exceed 1, this suggests rounding errors; assume $$r^2 = 1$$ approximately. Calculate $$\sigma_y$$ using: $$b_{yx} = r \frac{\sigma_y}{\sigma_x}$$ Assuming $$r = -1$$ (negative because slopes are negative), $$-1.5 = -1 \times \frac{\sigma_y}{5} \Rightarrow \sigma_y = 7.5$$ 4. **Step 3: Coefficient of correlation $$r$$** From above, $$r \approx -1$$ indicating a perfect negative linear correlation. **Final answers:** - Mean values: $$\bar{x} \approx -45.78$$, $$\bar{y} \approx 81.68$$ - Standard deviations: $$\sigma_x = 5$$, $$\sigma_y = 7.5$$ - Coefficient of correlation: $$r \approx -1$$ Note: Negative mean for $$x$$ is unusual for age data, possibly due to incomplete or inconsistent data.