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Use the standard deviation to compare the variability of the datasets. Round to the nearest hundredths. Dataset A: 11, 7, 12, 8, 9, 5 Dataset B: 10, 8, 10, 8, 8, 4 (1 point) Responses The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability. The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability. The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.58. Dataset A and Dataset B have the same standard deviation, so the variability is the same for both datasets. The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.58. Dataset A and Dataset B have the same standard deviation, so the variability is the same for both datasets. The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability. The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability. The standard deviation of Dataset A is 2.19. The standard deviation of Dataset B is 2.58. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.

1 answer

To compare the variability of the datasets using standard deviation, let's calculate the standard deviation for both datasets.

Dataset A: 11, 7, 12, 8, 9, 5

  1. Mean: \[ \text{Mean} = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} = 8.67 \]

  2. Variance:

    • Calculate the squared differences from the mean:

      1. \( (11 - 8.67)^2 = (2.33)^2 = 5.4289 \)
      2. \( (7 - 8.67)^2 = (-1.67)^2 = 2.7889 \)
      3. \( (12 - 8.67)^2 = (3.33)^2 = 11.0889 \)
      4. \( (8 - 8.67)^2 = (-0.67)^2 = 0.4489 \)
      5. \( (9 - 8.67)^2 = (0.33)^2 = 0.1089 \)
      6. \( (5 - 8.67)^2 = (-3.67)^2 = 13.4689 \)

    • Sum of squared differences: \[ 5.4289 + 2.7889 + 11.0889 + 0.4489 + 0.1089 + 13.4689 = 33.8334 \]

    • Variance: \[ \text{Variance} = \frac{33.8334}{6} = 5.6389 \]

  3. Standard Deviation: \[ \text{Standard Deviation} = \sqrt{5.6389} \approx 2.38 \]

Dataset B: 10, 8, 10, 8, 8, 4

  1. Mean: \[ \text{Mean} = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} = 9.67 \]

  2. Variance:

    • Calculate the squared differences from the mean:

      1. \( (10 - 9.67)^2 = (0.33)^2 = 0.1089 \)
      2. \( (8 - 9.67)^2 = (-1.67)^2 = 2.7889 \)
      3. \( (10 - 9.67)^2 = (0.33)^2 = 0.1089 \)
      4. \( (8 - 9.67)^2 = (-1.67)^2 = 2.7889 \)
      5. \( (8 - 9.67)^2 = (-1.67)^2 = 2.7889 \)
      6. \( (4 - 9.67)^2 = (-5.67)^2 = 32.1489 \)

    • Sum of squared differences: \[ 0.1089 + 2.7889 + 0.1089 + 2.7889 + 2.7889 + 32.1489 = 40.7334 \]

    • Variance: \[ \text{Variance} = \frac{40.7334}{6} = 6.7889 \]

  3. Standard Deviation: \[ \text{Standard Deviation} = \sqrt{6.7889} \approx 2.60 \]

Results:

  • The standard deviation of Dataset A is approximately 2.38.
  • The standard deviation of Dataset B is approximately 2.60.

Therefore, the correct choice based on the calculated standard deviations should reflect that Dataset B has a larger standard deviation than Dataset A, indicating that Dataset B has more variability.

In the responses provided, none of the calculations match the results we derived here, which indicates that the values given in the original statement were likely miscalculated or incorrect.

So, to summarize:

  • The standard deviation of Dataset A is approximately 2.38.
  • The standard deviation of Dataset B is approximately 2.60.
  • Dataset B has more variability than Dataset A.