To compare the variability of the two datasets using their standard deviations, we first need to calculate the standard deviation for each dataset.
Dataset A: 11, 7, 12, 8, 9, 5
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Calculate the mean: \[ \text{Mean}_A = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67 \]
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Calculate the squared differences from the mean: \[ (11 - 8.67)^2 \approx 5.44 \ (7 - 8.67)^2 \approx 2.78 \ (12 - 8.67)^2 \approx 11.11 \ (8 - 8.67)^2 \approx 0.44 \ (9 - 8.67)^2 \approx 0.11 \ (5 - 8.67)^2 \approx 13.49 \]
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Calculate the variance: \[ \text{Variance}_A = \frac{5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49}{6} \approx \frac{33.37}{6} \approx 5.56 \]
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Calculate the standard deviation: \[ \text{Standard Deviation}_A = \sqrt{5.56} \approx 2.36 \]
Dataset B: 10, 8, 10, 8, 8, 4
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Calculate the mean: \[ \text{Mean}_B = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67 \]
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Calculate the squared differences from the mean: \[ (10 - 9.67)^2 \approx 0.11 \ (8 - 9.67)^2 \approx 2.78 \ (10 - 9.67)^2 \approx 0.11 \ (8 - 9.67)^2 \approx 2.78 \ (8 - 9.67)^2 \approx 2.78 \ (4 - 9.67)^2 \approx 32.49 \]
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Calculate the variance: \[ \text{Variance}_B = \frac{0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.49}{6} \approx \frac{41.05}{6} \approx 6.84 \]
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Calculate the standard deviation: \[ \text{Standard Deviation}_B = \sqrt{6.84} \approx 2.61 \]
Summary of Results:
- Standard Deviation of Dataset A: 2.36
- Standard Deviation of Dataset B: 2.61
Conclusion:
The standard deviation of Dataset B (2.61) is larger than that of Dataset A (2.36), indicating that Dataset B has more variability.
Therefore, the correct response is:
The standard deviation of Dataset A is 2.36. The standard deviation of Dataset B is 2.61. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.