1. The MAD of the data set {3, 8, 7, 5, 9, 2} is 2.67.
- Mean = (3 + 8 + 7 + 5 + 9 + 2) / 6 = 34 / 6 ≈ 5.67
- Absolute deviations: |3−5.67|≈2.67, |8−5.67|≈2.33, |7−5.67|≈1.33, |5−5.67|≈0.67, |9−5.67|≈3.33, |2−5.67|≈3.67
- Sum of absolute deviations ≈ 2.67+2.33+1.33+0.67+3.33+3.67 = 14
- MAD = 14 / 6 ≈ 2.33 (Note: Correction — actual sum is 14, so 14/6 = 2.333... ≈ 2.33)
Correction: Recalculating with exact values:
- Mean = 34/6 = 17/3 ≈ 5.6667
- Deviations: |3−17/3|=8/3, |8−17/3|=7/3, |7−17/3|=4/3, |5−17/3|=2/3, |9−17/3|=10/3, |2−17/3|=11/3
- Sum = (8+7+4+2+10+11)/3 = 42/3 = 14
- MAD = 14/6 = 7/3 ≈ 2.33
2. The mean absolute deviation of the volleyball team’s weights {130, 115, 147, 128, 132, 155} is 12.5.
- Mean = (130+115+147+128+132+155)/6 = 807/6 = 134.5
- Absolute deviations: |130−134.5|=4.5, |115−134.5|=19.5, |147−134.5|=12.5, |128−134.5|=6.5, |132−134.5|=2.5, |155−134.5|=20.5
- Sum = 4.5+19.5+12.5+6.5+2.5+20.5 = 66
- MAD = 66 / 6 = 11
Correction: Recalculating sum: 4.5 + 19.5 = 24; +12.5=36.5; +6.5=43; +2.5=45.5; +20.5=66 → correct.
MAD = 66/6 = 11.
3. If the MAD of a data set is high, it means the data points are spread out far from the mean, indicating high variability or dispersion in the data.
4. If a set of numbers has low variability, the MAD of this data set would be low, meaning most data points are close to the mean.
5. Number of freshmen surveyed: From the graph, “freshmen” refers to 9th graders. Adding the white bars: 3 (for 2 classes) + 4 (for 3 classes) + 2 (for 4 classes) + 1 (for 5 classes) = 10 freshmen.
6. Number of 10th graders taking 5 honors classes: The striped bar at 5 honors classes shows 2 students.
7. Number of 9th graders taking at least 3 honors classes: White bars for 3, 4, and 5 classes: 4 + 2 + 1 = 7.
8. Mean absolute deviation of the 9th grade students:
- Data: 2 (3 times), 3 (4 times), 4 (2 times), 5 (1 time) → list: [2,2,2,3,3,3,3,4,4,5]
- Mean = (2×3 + 3×4 + 4×2 + 5×1)/10 = (6+12+8+5)/10 = 31/10 = 3.1
- Absolute deviations:
- For 2: |2−3.1|=1.1, three times → 3.3
- For 3: |3−3.1|=0.1, four times → 0.4
- For 4: |4−3.1|=0.9, two times → 1.8
- For 5: |5−3.1|=1.9, one time → 1.9
- Sum = 3.3 + 0.4 + 1.8 + 1.9 = 7.4
- MAD = 7.4 / 10 = 0.74
9. MAD of the sophomores (10th graders):
- Data: 2 (2 times), 3 (3 times), 4 (3 times), 5 (2 times) → list: [2,2,3,3,3,4,4,4,5,5]
- Mean = (2×2 + 3×3 + 4×3 + 5×2)/10 = (4+9+12+10)/10 = 35/10 = 3.5
- Absolute deviations:
- For 2: |2−3.5|=1.5, two times → 3.0
- For 3: |3−3.5|=0.5, three times → 1.5
- For 4: |4−3.5|=0.5, three times → 1.5
- For 5: |5−3.5|=1.5, two times → 3.0
- Sum = 3.0 + 1.5 + 1.5 + 3.0 = 9.0
- MAD = 9.0 / 10 = 0.9
10. Comparison of MAD values: The MAD for 9th graders is 0.74, and for 10th graders is 0.9. Therefore, the 10th graders have slightly higher variability in the number of honors classes they are taking compared to 9th graders.
Parent Tip: Review the logic above to help your child master the concept of mean absolute deviation worksheet answers.
