1. **State the problem:** We have GPA data for flute players, percussionists, and saxophone players. We need to determine which statements A, B, C, and D are true based on changes in median, mean, standard deviation, and interquartile range (IQR). 2. **Given data:** - Flute GPAs: 13 players, all between 3.84 and 3.88 (assume uniform or close values). - Percussion GPAs: 3 players with GPAs 2.4, 2.8, 3.2. - Saxophone GPAs: 9 players, 4 with 3.9 and 5 with 4.0. 3. **Analyze statement A:** "If one flute player played saxophone instead, median GPA for saxophones changes by 0.05." - Current saxophone GPAs sorted: 3.9, 3.9, 3.9, 3.9, 4.0, 4.0, 4.0, 4.0, 4.0 - Median of 9 values is the 5th value: 4.0 - Add one flute player GPA (between 3.84 and 3.88), say 3.86, to saxophones: New GPAs (10 values): 3.84-3.88 (one value), 3.9 x4, 4.0 x5 - Sorted: 3.86, 3.9, 3.9, 3.9, 3.9, 4.0, 4.0, 4.0, 4.0, 4.0 - Median of 10 values is average of 5th and 6th values: (3.9 + 4.0)/2 = 3.95 - Change in median: 4.0 - 3.95 = 0.05 **Statement A is true.** 4. **Analyze statement B:** "If one flute player played saxophone instead, mean GPA for saxophones changes by 0.5." - Current saxophone mean: $$\frac{4 \times 3.9 + 5 \times 4.0}{9} = \frac{15.6 + 20}{9} = \frac{35.6}{9} \approx 3.9556$$ - Add flute GPA 3.86: New mean: $$\frac{35.6 + 3.86}{10} = \frac{39.46}{10} = 3.946$$ - Change in mean: $$3.9556 - 3.946 = 0.0096$$ Change is about 0.01, not 0.5. **Statement B is false.** 5. **Analyze statement C:** "If one saxophone player became percussionist, standard deviation of percussionists decreases." - Percussion GPAs: 2.4, 2.8, 3.2 - Saxophone GPAs: 3.9 (4 players), 4.0 (5 players) - If a saxophone player (say 3.9 or 4.0) joins percussionists, new percussion GPAs become 4 values. Calculate original percussion SD: - Mean percussion: $$\frac{2.4 + 2.8 + 3.2}{3} = \frac{8.4}{3} = 2.8$$ - Variance percussion: $$\frac{(2.4-2.8)^2 + (2.8-2.8)^2 + (3.2-2.8)^2}{3} = \frac{0.16 + 0 + 0.16}{3} = 0.1067$$ - SD percussion: $$\sqrt{0.1067} \approx 0.3266$$ Add saxophone GPA 3.9: - New GPAs: 2.4, 2.8, 3.2, 3.9 - New mean: $$\frac{2.4 + 2.8 + 3.2 + 3.9}{4} = \frac{12.3}{4} = 3.075$$ - New variance: $$\frac{(2.4-3.075)^2 + (2.8-3.075)^2 + (3.2-3.075)^2 + (3.9-3.075)^2}{4}$$ $$= \frac{0.4556 + 0.0756 + 0.0156 + 0.6806}{4} = \frac{1.2274}{4} = 0.3069$$ - New SD: $$\sqrt{0.3069} = 0.554$$ SD increased from 0.3266 to 0.554. Try adding saxophone GPA 4.0: - New mean: $$\frac{2.4 + 2.8 + 3.2 + 4.0}{4} = 3.1$$ - New variance: $$\frac{(2.4-3.1)^2 + (2.8-3.1)^2 + (3.2-3.1)^2 + (4.0-3.1)^2}{4}$$ $$= \frac{0.49 + 0.09 + 0.01 + 0.81}{4} = \frac{1.4}{4} = 0.35$$ - New SD: $$\sqrt{0.35} = 0.5916$$ SD again increased. **Statement C is false.** 6. **Analyze statement D:** "Interquartile range (IQR) is greater for flutes than percussionists." - Flute GPAs between 3.84 and 3.88, range = 0.04 - Percussion GPAs: 2.4, 2.8, 3.2 - Percussion IQR: - Q1 = 2.4 (lowest) - Q2 = 2.8 (median) - Q3 = 3.2 (highest) - IQR = Q3 - Q1 = 3.2 - 2.4 = 0.8 Since 0.8 > 0.04, **Statement D is false.** **Final answers:** Only statement A is true. **Answer:** A