To determine which dataset is more spread out, we need to calculate the mean absolute deviation (MAD) for each dataset. The mean absolute deviation is the average of the absolute differences between each data point and the mean.
Step 1: Calculate the mean absolute deviation for Dataset #1
Dataset #1: {5, 8, 9, 10, 12, 12, 14}
- Mean = 10 (given).
- Calculate the absolute differences from the mean:
\[ |5 - 10| = 5 \ |8 - 10| = 2 \ |9 - 10| = 1 \ |10 - 10| = 0 \ |12 - 10| = 2 \ |12 - 10| = 2 \ |14 - 10| = 4 \ \]
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Absolute differences: {5, 2, 1, 0, 2, 2, 4}
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Calculate the mean absolute deviation:
\[ \text{MAD} = \frac{|5| + |2| + |1| + |0| + |2| + |2| + |4|}{7} = \frac{16}{7} \approx 2.3 \]
Step 2: Calculate the mean absolute deviation for Dataset #2
Dataset #2: {5, 6, 8, 9, 10, 10, 22}
- Mean = 10 (given).
- Calculate the absolute differences from the mean:
\[ |5 - 10| = 5 \ |6 - 10| = 4 \ |8 - 10| = 2 \ |9 - 10| = 1 \ |10 - 10| = 0 \ |10 - 10| = 0 \ |22 - 10| = 12 \ \]
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Absolute differences: {5, 4, 2, 1, 0, 0, 12}
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Calculate the mean absolute deviation:
\[ \text{MAD} = \frac{|5| + |4| + |2| + |1| + |0| + |0| + |12|}{7} = \frac{24}{7} \approx 3.4 \]
Step 3: Compare the mean absolute deviations
- MAD for Dataset #1: approximately 2.3
- MAD for Dataset #2: approximately 3.4
Conclusion:
Dataset #2 is more spread out because it has a mean absolute deviation of approximately 3.4 while Dataset #1 has a mean absolute deviation of approximately 2.3.