Question
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 2.58. 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 2.19. 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.
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 A is larger than that of Dataset B, so Dataset A has more variability.
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 2.58. 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 2.19. 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.
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 A is larger than that of Dataset B, so Dataset A has more variability.
Answers
Answered by
GPT-4o mini
To compare the variability of the two datasets, we need to calculate the standard deviation for each. Let's calculate them.
### Dataset A: 11, 7, 12, 8, 9, 5
1. **Mean**:
\[
\text{Mean}_A = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67
\]
2. **Calculate the variance**:
\[
\text{Variance}_A = \frac{(11 - 8.67)^2 + (7 - 8.67)^2 + (12 - 8.67)^2 + (8 - 8.67)^2 + (9 - 8.67)^2 + (5 - 8.67)^2}{6}
\]
Calculating each squared difference:
\[
(11 - 8.67)^2 \approx 5.44
\]
\[
(7 - 8.67)^2 \approx 2.78
\]
\[
(12 - 8.67)^2 \approx 11.11
\]
\[
(8 - 8.67)^2 \approx 0.44
\]
\[
(9 - 8.67)^2 \approx 0.11
\]
\[
(5 - 8.67)^2 \approx 13.49
\]
Adding these up:
\[
5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49 \approx 33.37
\]
Dividing by the number of values (4 for sample standard deviation):
\[
\text{Variance}_A = \frac{33.37}{6} \approx 5.56
\]
3. **Standard deviation**:
\[
\text{SD}_A = \sqrt{5.56} \approx 2.36
\]
Now let's calculate for Dataset B.
### Dataset B: 10, 8, 10, 8, 8, 4
1. **Mean**:
\[
\text{Mean}_B = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67
\]
2. **Calculate the variance**:
\[
\text{Variance}_B = \frac{(10 - 9.67)^2 + (8 - 9.67)^2 + (10 - 9.67)^2 + (8 - 9.67)^2 + (8 - 9.67)^2 + (4 - 9.67)^2}{6}
\]
Calculating each squared difference:
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(4 - 9.67)^2 \approx 32.11
\]
Adding these up:
\[
0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.11 \approx 40.67
\]
Dividing by the number of values (4 for sample standard deviation):
\[
\text{Variance}_B = \frac{40.67}{6} \approx 6.78
\]
3. **Standard deviation**:
\[
\text{SD}_B = \sqrt{6.78} \approx 2.60
\]
### Conclusion
Now comparing the two results:
- **Dataset A SD**: 2.36
- **Dataset B SD**: 2.60
The standard deviation of Dataset B (2.60) is larger than that of Dataset A (2.36), indicating that Dataset B has more variability.
**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.**
### Dataset A: 11, 7, 12, 8, 9, 5
1. **Mean**:
\[
\text{Mean}_A = \frac{11 + 7 + 12 + 8 + 9 + 5}{6} = \frac{52}{6} \approx 8.67
\]
2. **Calculate the variance**:
\[
\text{Variance}_A = \frac{(11 - 8.67)^2 + (7 - 8.67)^2 + (12 - 8.67)^2 + (8 - 8.67)^2 + (9 - 8.67)^2 + (5 - 8.67)^2}{6}
\]
Calculating each squared difference:
\[
(11 - 8.67)^2 \approx 5.44
\]
\[
(7 - 8.67)^2 \approx 2.78
\]
\[
(12 - 8.67)^2 \approx 11.11
\]
\[
(8 - 8.67)^2 \approx 0.44
\]
\[
(9 - 8.67)^2 \approx 0.11
\]
\[
(5 - 8.67)^2 \approx 13.49
\]
Adding these up:
\[
5.44 + 2.78 + 11.11 + 0.44 + 0.11 + 13.49 \approx 33.37
\]
Dividing by the number of values (4 for sample standard deviation):
\[
\text{Variance}_A = \frac{33.37}{6} \approx 5.56
\]
3. **Standard deviation**:
\[
\text{SD}_A = \sqrt{5.56} \approx 2.36
\]
Now let's calculate for Dataset B.
### Dataset B: 10, 8, 10, 8, 8, 4
1. **Mean**:
\[
\text{Mean}_B = \frac{10 + 8 + 10 + 8 + 8 + 4}{6} = \frac{58}{6} \approx 9.67
\]
2. **Calculate the variance**:
\[
\text{Variance}_B = \frac{(10 - 9.67)^2 + (8 - 9.67)^2 + (10 - 9.67)^2 + (8 - 9.67)^2 + (8 - 9.67)^2 + (4 - 9.67)^2}{6}
\]
Calculating each squared difference:
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(10 - 9.67)^2 \approx 0.11
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(8 - 9.67)^2 \approx 2.78
\]
\[
(4 - 9.67)^2 \approx 32.11
\]
Adding these up:
\[
0.11 + 2.78 + 0.11 + 2.78 + 2.78 + 32.11 \approx 40.67
\]
Dividing by the number of values (4 for sample standard deviation):
\[
\text{Variance}_B = \frac{40.67}{6} \approx 6.78
\]
3. **Standard deviation**:
\[
\text{SD}_B = \sqrt{6.78} \approx 2.60
\]
### Conclusion
Now comparing the two results:
- **Dataset A SD**: 2.36
- **Dataset B SD**: 2.60
The standard deviation of Dataset B (2.60) is larger than that of Dataset A (2.36), indicating that Dataset B has more variability.
**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.**