1. **Problem Statement:** We have four datasets representing different scenarios. For each, we need to calculate mean, median, and mode, analyze the impact of outliers, and interpret which measure best represents the "typical" value. --- ### Dataset 2: Blood Pressure Reductions Data: 4, 5, 6, 7, 7, 7, 8, 9, 10, 12, 14, 15, 15, 16, 17, 18, 18, 19, 50, 55 2. **Formulas:** - Mean: $$\text{mean} = \frac{\sum x_i}{n}$$ - Median: Middle value when data is sorted - Mode: Most frequent value 3. **Calculations:** - Sorted data already given. - Mean: $$\frac{4+5+6+7+7+7+8+9+10+12+14+15+15+16+17+18+18+19+50+55}{20} = \frac{327}{20} = 16.35$$ - Median: Average of 10th and 11th values (12 and 14): $$\frac{12+14}{2} = 13$$ - Mode: 7 (appears 3 times) 4. **Interpretation:** - Mode is 7, median 13, mean 16.35. - Outliers 50 and 55 increase the mean significantly. 5. **Impact of Outliers:** - Outliers skew the mean upward. - Median and mode less affected. 6. **Removing 55:** - New sum: $$327 - 55 = 272$$ - New mean: $$\frac{272}{19} \approx 14.32$$ - New median remains between 10th and 11th values (12 and 14): 13 --- ### Dataset 3: Basketball Scores Data: 8, 10, 12, 14, 16, 50, 10, 12, 14, 16 1. Mean: $$\frac{8+10+12+14+16+50+10+12+14+16}{10} = \frac{162}{10} = 16.2$$ 2. Median: Sorted data: 8,10,10,12,12,14,14,16,16,50 Median is average of 5th and 6th: $$\frac{12+14}{2} = 13$$ 3. Mode: 10, 12, 14, 16 all appear twice; modes are 10, 12, 14, 16 (multimodal) 4. Player's claim of average 20 points is inaccurate; mean is 16.2, median 13. 5. Removing 50: - New sum: $$162 - 50 = 112$$ - New mean: $$\frac{112}{9} \approx 12.44$$ - New median: middle value of sorted 9 values (8,10,10,12,12,14,14,16,16) is 12 --- ### Dataset 4: House Prices (millions) Data: 1.2, 1.3, 1.5, 1.7, 1.8, 2.0, 2.1, 2.3, 3.5, 5.0, 10.0 1. Mean: $$\frac{1.2+1.3+1.5+1.7+1.8+2.0+2.1+2.3+3.5+5.0+10.0}{11} = \frac{32.4}{11} \approx 2.945$$ 2. Median: 6th value (middle of 11 values) is 2.0 3. Mode: No repeats, so no mode 4. Typical price: Median (2.0) better represents typical price because mean is skewed by 10.0 5. To make prices seem more expensive, highlight mean (2.945) 6. Removing 10.0: - New sum: $$32.4 - 10.0 = 22.4$$ - New mean: $$\frac{22.4}{10} = 2.24$$ - New median: average of 5th and 6th values (1.8 and 2.0): $$\frac{1.8+2.0}{2} = 1.9$$ --- ### Dataset 5: GPAs Data: 2.1, 2.3, 2.5, 2.8, 3.0, 3.1, 3.2, 3.3, 3.3, 3.4, 3.4, 3.5, 3.6, 3.7, 4.0, 4.0, 4.0, 4.0, 4.0 1. Mean: $$\frac{\sum GPA}{19} = \frac{63.6}{19} \approx 3.347$$ 2. Median: 10th value (middle) is 3.4 3. Mode: 4.0 (appears 5 times) 4. For top 50%, median is best measure because it splits data evenly. 5. Adding 1.5: - New sum: $$63.6 + 1.5 = 65.1$$ - New mean: $$\frac{65.1}{20} = 3.255$$ (mean decreases) - New median: average of 10th and 11th values (3.3 and 3.4): $$\frac{3.3+3.4}{2} = 3.35$$ (median slightly decreases) --- **Summary:** - Mean is sensitive to outliers. - Median is robust and better for skewed data. - Mode shows most frequent value but may not exist or be unique. - For "typical" values, median is often preferred when outliers exist.