To compare the variability of the datasets using the standard deviation, let's first 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 the squared differences from the mean:
- \( (4 - 8)^2 = 16 \)
- \( (5 - 8)^2 = 9 \)
- \( (6 - 8)^2 = 4 \)
- \( (12 - 8)^2 = 16 \)
- \( (13 - 8)^2 = 25 \)
Sum of squared differences: \( 16 + 9 + 4 + 16 + 25 = 70 \)
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Calculate the variance: \[ \text{Variance} = \frac{70}{5 - 1} = \frac{70}{4} = 17.5 \]
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Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{17.5} \approx 4.18 \]
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 squared differences from the mean:
- \( (4 - 7)^2 = 9 \)
- \( (5 - 7)^2 = 4 \)
- \( (7 - 7)^2 = 0 \)
- \( (9 - 7)^2 = 4 \)
- \( (10 - 7)^2 = 9 \)
Sum of squared differences: \( 9 + 4 + 0 + 4 + 9 = 26 \)
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Calculate the variance: \[ \text{Variance} = \frac{26}{5 - 1} = \frac{26}{4} = 6.5 \]
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Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{6.5} \approx 2.55 \]
Summary:
- Standard deviation of Dataset A: \( \approx 4.18 \)
- Standard deviation of Dataset B: \( \approx 2.55 \)
Comparison: The standard deviation of Dataset A is larger than that of Dataset B, indicating that Dataset A has more variability.
This means the correct response based on the calculated values is: The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability.