To find the standard deviation for both datasets, we can follow these steps:
- Calculate the mean (average) of each dataset.
- Calculate the variance by finding the average of the squared differences from the mean.
- Take the square root of the variance to find the standard deviation.
Let's calculate the standard deviations for both datasets.
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: \[ \begin{align*} \text{Variance} & = \frac{(11 - 8.67)^2 + (7 - 8.67)^2 + (12 - 8.67)^2 + (8 - 8.67)^2 + (9 - 8.67)^2 + (5 - 8.67)^2}{6} \ & = \frac{(2.33)^2 + (-1.67)^2 + (3.33)^2 + (-0.67)^2 + (0.33)^2 + (-3.67)^2}{6} \ & = \frac{5.4289 + 2.7889 + 11.0889 + 0.4489 + 0.1089 + 13.4689}{6} \ & = \frac{33.4134}{6} \approx 5.57 \end{align*} \]
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Standard Deviation: \[ \text{Standard Deviation} = \sqrt{5.57} \approx 2.36 \]
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: \[ \begin{align*} \text{Variance} & = \frac{(10 - 9.67)^2 + (8 - 9.67)^2 + (10 - 9.67)^2 + (8 - 9.67)^2 + (8 - 9.67)^2 + (4 - 9.67)^2}{6} \ & = \frac{(0.33)^2 + (-1.67)^2 + (0.33)^2 + (-1.67)^2 + (-1.67)^2 + (-5.67)^2}{6} \ & = \frac{0.1089 + 2.7889 + 0.1089 + 2.7889 + 2.7889 + 32.1489}{6} \ & = \frac{41.7334}{6} \approx 6.95 \end{align*} \]
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Standard Deviation: \[ \text{Standard Deviation} = \sqrt{6.95} \approx 2.64 \]
Summary of Results
- Standard deviation of Dataset A: 2.36
- Standard deviation of Dataset B: 2.64
Comparing both, the standard deviation of Dataset B is larger than that of Dataset A, which indicates that Dataset B has more variability.
Therefore, the correct answer is: 2. The standard deviation of Dataset A is 2.36. The standard deviation of Dataset B is 2.64. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.