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
use one of the provided responses
(1 point)
Responses

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 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 2.19. 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.
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.

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.

Answer 1

To calculate the standard deviation for both datasets, let's go through the steps:

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

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

2. **Calculate the squared deviations from the mean**:
\[
(11 - 8.67)^2 \approx 5.56
\]
\[
(7 - 8.67)^2 \approx 2.78
\]
\[
(12 - 8.67)^2 \approx 11.11
\]
\[
(8 - 8.67)^2 \approx 0.45
\]
\[
(9 - 8.67)^2 \approx 0.11
\]
\[
(5 - 8.67)^2 \approx 13.49
\]

3. **Sum of squared deviations**:
\[
5.56 + 2.78 + 11.11 + 0.45 + 0.11 + 13.49 \approx 33.50
\]

4. **Calculate the variance** (since this is a sample, we use \(n - 1\) for the denominator):
\[
\text{Variance} = \frac{33.50}{6 - 1} \approx 6.70
\]

5. **Standard Deviation**:
\[
\text{Standard Deviation} = \sqrt{6.70} \approx 2.58
\]

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

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

2. **Calculate the squared deviations from the mean**:
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(4 - 9.67)^2 \approx 32.49
\]

3. **Sum of squared deviations**:
\[
0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.49 \approx 41.05
\]

4. **Calculate the variance** (using \(n - 1\) for the denominator):
\[
\text{Variance} = \frac{41.05}{6 - 1} \approx 8.21
\]

5. **Standard Deviation**:
\[
\text{Standard Deviation} = \sqrt{8.21} \approx 2.86
\]

Now we summarize our results:

- Standard Deviation of Dataset A: **2.58**
- Standard Deviation of Dataset B: **2.86**

### Conclusion
The correct response is:
**The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.86. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.**