To compare the variability of the two datasets using standard deviation, we will first calculate the standard deviations of Dataset A and Dataset B.
Dataset A: 11, 7, 12, 8, 9, 5
-
Calculate the mean: \[ \text{Mean} = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67 \]
-
Calculate the squared differences from the mean: \[ (11 - 8.67)^2 = 5.44 \] \[ (7 - 8.67)^2 = 2.78 \] \[ (12 - 8.67)^2 = 11.11 \] \[ (8 - 8.67)^2 = 0.44 \] \[ (9 - 8.67)^2 = 0.11 \] \[ (5 - 8.67)^2 = 13.49 \]
-
Calculate the average of the squared differences: \[ \text{Variance} = \frac{5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49}{6} \approx \frac{33.37}{6} \approx 5.56 \]
-
Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{5.56} \approx 2.36 \]
Dataset B: 10, 8, 10, 8, 8, 4
-
Calculate the mean: \[ \text{Mean} = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67 \]
-
Calculate the squared differences from the mean: \[ (10 - 9.67)^2 = 0.11 \] \[ (8 - 9.67)^2 = 2.78 \] \[ (10 - 9.67)^2 = 0.11 \] \[ (8 - 9.67)^2 = 2.78 \] \[ (8 - 9.67)^2 = 2.78 \] \[ (4 - 9.67)^2 = 32.11 \]
-
Calculate the average of the squared differences: \[ \text{Variance} = \frac{0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.11}{6} \approx \frac{40.67}{6} \approx 6.78 \]
-
Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{6.78} \approx 2.60 \]
Final Standard Deviations:
- Dataset A: \( \approx 2.36 \)
- Dataset B: \( \approx 2.60 \)
Conclusion:
The standard deviation of Dataset A is approximately 2.36, and the standard deviation of Dataset B is approximately 2.60. Therefore, Dataset B has a larger standard deviation, indicating it has more variability.
From the given options, 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."
1 answer