To calculate the standard deviation for both datasets, we follow these steps:
- Calculate the mean of each dataset.
- Calculate the variance.
- Take the square root of the variance to find the standard deviation.
Dataset A: 4, 5, 6, 12, 13
-
Mean: \[ \text{Mean} = \frac{4 + 5 + 6 + 12 + 13}{5} = \frac{40}{5} = 8 \]
-
Variance: \[ \text{Variance} = \frac{(4-8)^2 + (5-8)^2 + (6-8)^2 + (12-8)^2 + (13-8)^2}{5} \] \[ = \frac{(-4)^2 + (-3)^2 + (-2)^2 + (4)^2 + (5)^2}{5} = \frac{16 + 9 + 4 + 16 + 25}{5} = \frac{70}{5} = 14 \]
-
Standard Deviation: \[ \text{Standard Deviation} = \sqrt{14} \approx 3.74 \]
Dataset B: 4, 5, 7, 9, 10
-
Mean: \[ \text{Mean} = \frac{4 + 5 + 7 + 9 + 10}{5} = \frac{35}{5} = 7 \]
-
Variance: \[ \text{Variance} = \frac{(4-7)^2 + (5-7)^2 + (7-7)^2 + (9-7)^2 + (10-7)^2}{5} \] \[ = \frac{(-3)^2 + (-2)^2 + (0)^2 + (2)^2 + (3)^2}{5} = \frac{9 + 4 + 0 + 4 + 9}{5} = \frac{26}{5} = 5.2 \]
-
Standard Deviation: \[ \text{Standard Deviation} = \sqrt{5.2} \approx 2.28 \]
Summary of Results
- Standard Deviation of Dataset A: approximately 3.74
- Standard Deviation of Dataset B: approximately 2.28
Conclusion
The standard deviation of Dataset A (3.74) is larger than that of Dataset B (2.28). Therefore, Dataset A has more variability.
So the correct response is: The standard deviation of Dataset A is 3.74. The standard deviation of Dataset B is 2.28. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability.