To compare the variability of the datasets A and B using standard deviation, we will calculate the standard deviation for both datasets.
Step 1: Calculate the Mean
For Dataset A: \[ \text{Mean (A)} = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67 \]
For Dataset B: \[ \text{Mean (B)} = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67 \]
Step 2: Calculate the Variance
For Dataset A:
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Calculate each squared deviation 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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Sum the squared deviations. \[ \text{Sum} \approx 5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49 \approx 33.37 \]
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Calculate the variance. \[ \text{Variance (A)} = \frac{33.37}{6} \approx 5.56 \]
For Dataset B:
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Calculate each squared deviation 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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Sum the squared deviations. \[ \text{Sum} \approx 0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.49 \approx 41.05 \]
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Calculate the variance. \[ \text{Variance (B)} = \frac{41.05}{6} \approx 6.84 \]
Step 3: Calculate the Standard Deviation
For Dataset A: \[ \text{Standard Deviation (A)} = \sqrt{5.56} \approx 2.36 \]
For Dataset B: \[ \text{Standard Deviation (B)} = \sqrt{6.84} \approx 2.61 \]
Conclusion
- Standard Deviation of Dataset A: \( \approx 2.36 \)
- Standard Deviation of Dataset B: \( \approx 2.61 \)
After rounding to the nearest hundredths:
- Standard Deviation A: 2.36
- Standard Deviation B: 2.61
Comparing Variability
Dataset B has a slightly higher standard deviation than Dataset A, indicating that Dataset B has greater variability in its values compared to Dataset A.