To compare the variability of the datasets using standard deviation, we first need to calculate the standard deviation for each dataset.
Calculation of Standard Deviation
Dataset A: 11, 7, 12, 8, 9, 5
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Mean: \[ \text{Mean} = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67 \]
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Variance: \[ \text{Variance} = \frac{(11 - 8.67)^2 + (7 - 8.67)^2 + (12 - 8.67)^2 + (8 - 8.67)^2 + (9 - 8.67)^2 + (5 - 8.67)^2}{6} \] \[ = \frac{(2.33)^2 + (-1.67)^2 + (3.33)^2 + (-0.67)^2 + (0.33)^2 + (-3.67)^2}{6} \] \[ = \frac{5.4289 + 2.7889 + 11.0889 + 0.4489 + 0.1089 + 13.4689}{6} \approx \frac{33.7734}{6} \approx 5.6289 \]
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Standard Deviation: \[ \sigma_A = \sqrt{5.6289} \approx 2.37 \]
Dataset B: 10, 8, 10, 8, 8, 4
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Mean: \[ \text{Mean} = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67 \]
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Variance: \[ \text{Variance} = \frac{(10 - 9.67)^2 + (8 - 9.67)^2 + (10 - 9.67)^2 + (8 - 9.67)^2 + (8 - 9.67)^2 + (4 - 9.67)^2}{6} \] \[ = \frac{(0.33)^2 + (-1.67)^2 + (0.33)^2 + (-1.67)^2 + (-1.67)^2 + (-5.67)^2}{6} \] \[ = \frac{0.1089 + 2.7889 + 0.1089 + 2.7889 + 2.7889 + 32.0489}{6} \approx \frac{40.6334}{6} \approx 6.7722 \]
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Standard Deviation: \[ \sigma_B = \sqrt{6.7722} \approx 2.60 \]
Summary of Results:
- The standard deviation of Dataset A is approximately 2.37.
- The standard deviation of Dataset B is approximately 2.60.
Conclusion
The standard deviation of Dataset A (2.37) is lower than that of Dataset B (2.60). This indicates that Dataset B has more variability than Dataset A.
This means the correct response based on the calculated values would be:
"The standard deviation of Dataset A is 2.37. 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."
However, you provided specific rounded values in your responses that do not match the calculations performed here; please ensure the correct calculations are shared. If we use the rounded values provided (2.58 for A and 2.19 for B), we still conclude that Dataset A has more variability. Thus, there’s an inconsistency.
In the options you provided, none match the calculations done above, so please check the original calculations again with the correct inputs.
1 answer