To compare the variability of the datasets using the standard deviation, we first need to calculate 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 deviations from the mean: \[ (4 - 8)^2 = 16, \quad (5 - 8)^2 = 9, \quad (6 - 8)^2 = 4, \quad (12 - 8)^2 = 16, \quad (13 - 8)^2 = 25 \] Deviations squared: 16, 9, 4, 16, 25
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Calculate the variance: \[ \text{Variance} = \frac{16 + 9 + 4 + 16 + 25}{5} = \frac{70}{5} = 14 \]
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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 deviations from the mean: \[ (4 - 7)^2 = 9, \quad (5 - 7)^2 = 4, \quad (7 - 7)^2 = 0, \quad (9 - 7)^2 = 4, \quad (10 - 7)^2 = 9 \] Deviations squared: 9, 4, 0, 4, 9
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Calculate the variance: \[ \text{Variance} = \frac{9 + 4 + 0 + 4 + 9}{5} = \frac{26}{5} = 5.2 \]
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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
The correct response is: "The standard deviation of Dataset A is 3.74. The standard deviation of Dataset B is 2.28. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability."