Question

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.

Answer 1

To compare the variability of the datasets using the standard deviation, we first need to calculate the standard deviation of both datasets.

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

1. **Calculate the mean**:
\[
\text{Mean} = \frac{(11 + 7 + 12 + 8 + 9 + 5)}{6} = \frac{52}{6} \approx 8.67
\]

2. **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

3. **Calculate the variance**:
\[
\text{Variance} = \frac{(5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49)}{6} \approx \frac{33.37}{6} \approx 5.56
\]

4. **Calculate the standard deviation**:
\[
\text{Standard Deviation} = \sqrt{5.56} \approx 2.36
\]

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

1. **Calculate the mean**:
\[
\text{Mean} = \frac{(10 + 8 + 10 + 8 + 8 + 4)}{6} = \frac{58}{6} \approx 9.67
\]

2. **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

3. **Calculate the variance**:
\[
\text{Variance} = \frac{(0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.11)}{6} \approx \frac{40.67}{6} \approx 6.78
\]

4. **Calculate the standard deviation**:
\[
\text{Standard Deviation} = \sqrt{6.78} \approx 2.60
\]

### Summary:

- The standard deviation of **Dataset A** is approximately **2.36**.
- The standard deviation of **Dataset B** is approximately **2.60**.

Comparing the two standard deviations:

- Dataset A: 2.36
- Dataset B: 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, the standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.

Thus, 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.**