To compare the variability of the datasets using the standard deviation, we first need to calculate the standard deviation of both datasets.
Dataset A: 11, 7, 12, 8, 9, 5
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Calculate the mean: \[ \text{Mean} = \frac{(11 + 7 + 12 + 8 + 9 + 5)}{6} = \frac{52}{6} \approx 8.67 \]
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Calculate the squared differences from the mean:
- (11 - 8.67)² = 5.44
- (7 - 8.67)² = 2.78
- (12 - 8.67)² = 11.11
- (8 - 8.67)² = 0.44
- (9 - 8.67)² = 0.11
- (5 - 8.67)² = 13.49
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Calculate the variance: \[ \text{Variance} = \frac{(5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49)}{6} \approx \frac{33.37}{6} \approx 5.56 \]
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Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{5.56} \approx 2.36 \]
Dataset B: 10, 8, 10, 8, 8, 4
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Calculate the mean: \[ \text{Mean} = \frac{(10 + 8 + 10 + 8 + 8 + 4)}{6} = \frac{58}{6} \approx 9.67 \]
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Calculate the squared differences from the mean:
- (10 - 9.67)² = 0.11
- (8 - 9.67)² = 2.78
- (10 - 9.67)² = 0.11
- (8 - 9.67)² = 2.78
- (8 - 9.67)² = 2.78
- (4 - 9.67)² = 32.11
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Calculate the variance: \[ \text{Variance} = \frac{(0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.11)}{6} \approx \frac{40.67}{6} \approx 6.78 \]
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Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{6.78} \approx 2.60 \]
Summary:
- The standard deviation of Dataset A is approximately 2.36.
- The standard deviation of Dataset B is approximately 2.60.
Comparing the two standard deviations:
- Dataset A: 2.36
- Dataset B: 2.60
Conclusion:
The standard deviation of Dataset A is approximately 2.36, and the standard deviation of Dataset B is approximately 2.60. Therefore, the standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.
Thus, the correct response is: The standard deviation of Dataset A is 2.36. The standard deviation of Dataset B is 2.60. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.