To compare the variability of the two datasets using the standard deviation, we first need to calculate the mean and then the standard deviation for each dataset.
Dataset A: 4, 5, 6, 12, 13
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Calculate the Mean: \[ \text{Mean} = \frac{(4 + 5 + 6 + 12 + 13)}{5} = \frac{40}{5} = 8 \]
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Calculate the Variance: \[ \text{Variance} = \frac{((4 - 8)^2 + (5 - 8)^2 + (6 - 8)^2 + (12 - 8)^2 + (13 - 8)^2)}{5} \] Breaking it down:
- \((4 - 8)^2 = (-4)^2 = 16\)
- \((5 - 8)^2 = (-3)^2 = 9\)
- \((6 - 8)^2 = (-2)^2 = 4\)
- \((12 - 8)^2 = (4)^2 = 16\)
- \((13 - 8)^2 = (5)^2 = 25\)
Now summing these: \[ 16 + 9 + 4 + 16 + 25 = 70 \]
The variance: \[ \text{Variance} = \frac{70}{5} = 14 \]
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Calculate the Standard Deviation: \[ \text{Standard Deviation} = \sqrt{14} \approx 3.74 \]
Dataset B: 4, 5, 7, 9, 10
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Calculate the Mean: \[ \text{Mean} = \frac{(4 + 5 + 7 + 9 + 10)}{5} = \frac{35}{5} = 7 \]
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Calculate the Variance: \[ \text{Variance} = \frac{((4 - 7)^2 + (5 - 7)^2 + (7 - 7)^2 + (9 - 7)^2 + (10 - 7)^2)}{5} \] Breaking it down:
- \((4 - 7)^2 = (-3)^2 = 9\)
- \((5 - 7)^2 = (-2)^2 = 4\)
- \((7 - 7)^2 = (0)^2 = 0\)
- \((9 - 7)^2 = (2)^2 = 4\)
- \((10 - 7)^2 = (3)^2 = 9\)
Now summing these: \[ 9 + 4 + 0 + 4 + 9 = 26 \]
The variance: \[ \text{Variance} = \frac{26}{5} = 5.2 \]
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Calculate the Standard Deviation: \[ \text{Standard Deviation} = \sqrt{5.2} \approx 2.28 \]
Summary of Results
- Standard Deviation of Dataset A: 3.74
- Standard Deviation of Dataset B: 2.28
Conclusion
Dataset A has a higher standard deviation (3.74) compared to Dataset B (2.28), indicating that Dataset A has greater variability in its values.