1. **Problem 1: Calculate the mean, range, and standard deviation for ABC University data** Given data: $120, 110, 100, 200, 10, 90, 100, 100$ 2. **Mean formula:** $$\text{Mean} = \frac{\sum x_i}{n}$$ where $x_i$ are the data points and $n$ is the number of data points. 3. Calculate the mean: $$\sum x_i = 120 + 110 + 100 + 200 + 10 + 90 + 100 + 100 = 830$$ Number of data points $n = 8$ $$\text{Mean} = \frac{830}{8} = 103.75$$ 4. **Range formula:** $$\text{Range} = \text{Maximum value} - \text{Minimum value}$$ Maximum value = 200, Minimum value = 10 $$\text{Range} = 200 - 10 = 190$$ 5. **Standard deviation formula (sample):** $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ where $\bar{x}$ is the mean. 6. Calculate each squared deviation: $$(120 - 103.75)^2 = 264.06$$ $$(110 - 103.75)^2 = 39.06$$ $$(100 - 103.75)^2 = 14.06$$ $$(200 - 103.75)^2 = 9289.06$$ $$(10 - 103.75)^2 = 8780.06$$ $$(90 - 103.75)^2 = 189.06$$ $$(100 - 103.75)^2 = 14.06$$ $$(100 - 103.75)^2 = 14.06$$ Sum of squared deviations: $$264.06 + 39.06 + 14.06 + 9289.06 + 8780.06 + 189.06 + 14.06 + 14.06 = 18503.5$$ 7. Calculate standard deviation: $$s = \sqrt{\frac{18503.5}{7}} = \sqrt{2643.36} \approx 51.41$$ --- 8. **Problem 2: Calculate the mean, range, and standard deviation for XYZ University data** Given data: $200, 180, 30, 20, 10, 160, 150, 80$ 9. Calculate the mean: $$\sum x_i = 200 + 180 + 30 + 20 + 10 + 160 + 150 + 80 = 830$$ Number of data points $n = 8$ $$\text{Mean} = \frac{830}{8} = 103.75$$ 10. Calculate the range: Maximum value = 200, Minimum value = 10 $$\text{Range} = 200 - 10 = 190$$ 11. Calculate squared deviations: $$(200 - 103.75)^2 = 9289.06$$ $$(180 - 103.75)^2 = 5790.06$$ $$(30 - 103.75)^2 = 5406.06$$ $$(20 - 103.75)^2 = 7006.06$$ $$(10 - 103.75)^2 = 8780.06$$ $$(160 - 103.75)^2 = 3164.06$$ $$(150 - 103.75)^2 = 2139.06$$ $$(80 - 103.75)^2 = 564.06$$ Sum of squared deviations: $$9289.06 + 5790.06 + 5406.06 + 7006.06 + 8780.06 + 3164.06 + 2139.06 + 564.06 = 42138.5$$ 12. Calculate standard deviation: $$s = \sqrt{\frac{42138.5}{7}} = \sqrt{6020.5} \approx 77.6$$ --- 13. **Problem 3: Compare the two distributions** - Both groups have the same mean ($103.75$) and range ($190$). - However, the standard deviation for ABC University is approximately $51.41$, while for XYZ University it is approximately $77.6$. - A smaller standard deviation means data points are closer to the mean, indicating more homogeneity. **Conclusion:** ABC University students' spending is more homogeneous than XYZ University students' spending. **Final answers:** - ABC University: Mean = $103.75$, Range = $190$, Standard Deviation $\approx 51.41$ - XYZ University: Mean = $103.75$, Range = $190$, Standard Deviation $\approx 77.6$ - More homogeneous group: ABC University