1. **State the problem:** We have two data sets $X_i = [9, 12, 5, 6, 9, 14, 3, 6, 12, 14]$ and $Y_i = [10, 15, 9, 10, 14, 17, 5, 8, 16, 16]$. We need to calculate: - a) Means $\bar{x}$ and $\bar{y}$ - b) Variances of $X$ and $Y$ - c) Standard deviations of $X$ and $Y$ - d) Covariance between $X$ and $Y$ 2. **Calculate means:** $$\bar{x} = \frac{1}{10} \sum_{i=1}^{10} X_i = \frac{9+12+5+6+9+14+3+6+12+14}{10} = \frac{90}{10} = 9$$ $$\bar{y} = \frac{1}{10} \sum_{i=1}^{10} Y_i = \frac{10+15+9+10+14+17+5+8+16+16}{10} = \frac{120}{10} = 12$$ 3. **Calculate variances:** Variance formula: $$s^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{x})^2$$ For $X$: Calculate each $(X_i - \bar{x})^2$: $$(9-9)^2=0, (12-9)^2=9, (5-9)^2=16, (6-9)^2=9, (9-9)^2=0, (14-9)^2=25, (3-9)^2=36, (6-9)^2=9, (12-9)^2=9, (14-9)^2=25$$ Sum: $0+9+16+9+0+25+36+9+9+25=138$ $$s_x^2 = \frac{138}{9} = 15.3333$$ For $Y$: Calculate each $(Y_i - \bar{y})^2$: $$(10-12)^2=4, (15-12)^2=9, (9-12)^2=9, (10-12)^2=4, (14-12)^2=4, (17-12)^2=25, (5-12)^2=49, (8-12)^2=16, (16-12)^2=16, (16-12)^2=16$$ Sum: $4+9+9+4+4+25+49+16+16+16=152$ $$s_y^2 = \frac{152}{9} = 16.8889$$ 4. **Calculate standard deviations:** $$s_x = \sqrt{15.3333} \approx 3.916$$ $$s_y = \sqrt{16.8889} \approx 4.109$$ 5. **Calculate covariance:** Covariance formula: $$\text{cov}(X,Y) = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{x})(Y_i - \bar{y})$$ Calculate each product: $$(9-9)(10-12)=0, (12-9)(15-12)=3 \times 3=9, (5-9)(9-12)=-4 \times -3=12, (6-9)(10-12)=-3 \times -2=6, (9-9)(14-12)=0, (14-9)(17-12)=5 \times 5=25, (3-9)(5-12)=-6 \times -7=42, (6-9)(8-12)=-3 \times -4=12, (12-9)(16-12)=3 \times 4=12, (14-9)(16-12)=5 \times 4=20$$ Sum: $0+9+12+6+0+25+42+12+12+20=138$ $$\text{cov}(X,Y) = \frac{138}{9} = 15.3333$$ **Final answers:** - Means: $\bar{x} = 9$, $\bar{y} = 12$ - Variances: $s_x^2 = 15.3333$, $s_y^2 = 16.8889$ - Standard deviations: $s_x \approx 3.916$, $s_y \approx 4.109$ - Covariance: $15.3333$