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 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.
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 2.19. 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.
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 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.
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 2.19. 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.
Answers
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Answered by
GPT-4o mini
To compare the variability of the datasets using standard deviation, first, let's calculate the standard deviation for both Dataset A and Dataset B.
### 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 deviations from the mean:
- \( (11 - 8.67)^2 \approx 5.44 \)
- \( (7 - 8.67)^2 \approx 2.78 \)
- \( (12 - 8.67)^2 \approx 11.11 \)
- \( (8 - 8.67)^2 \approx 0.44 \)
- \( (9 - 8.67)^2 \approx 0.11 \)
- \( (5 - 8.67)^2 \approx 13.49 \)
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 deviations from the mean:
- \( (10 - 9.67)^2 \approx 0.11 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (10 - 9.67)^2 \approx 0.11 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (4 - 9.67)^2 \approx 32.11 \)
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:
\[
\text{Standard Deviation} = \sqrt{6.78} \approx 2.60
\]
### Comparison of Standard Deviations:
- **Standard Deviation of Dataset A:** approximately **2.36**
- **Standard Deviation of Dataset B:** approximately **2.60**
The rounded values could lead to:
- The final standard deviation calculations indicate that Dataset B has a higher standard deviation than Dataset A, implying that Dataset B has greater variability.
Therefore, 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.** However, this statement does not match the provided options. Therefore, please verify the details above to check for any discrepancies, as the calculations do suggest differing values than your listed responses.
### 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 deviations from the mean:
- \( (11 - 8.67)^2 \approx 5.44 \)
- \( (7 - 8.67)^2 \approx 2.78 \)
- \( (12 - 8.67)^2 \approx 11.11 \)
- \( (8 - 8.67)^2 \approx 0.44 \)
- \( (9 - 8.67)^2 \approx 0.11 \)
- \( (5 - 8.67)^2 \approx 13.49 \)
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 deviations from the mean:
- \( (10 - 9.67)^2 \approx 0.11 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (10 - 9.67)^2 \approx 0.11 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (8 - 9.67)^2 \approx 2.78 \)
- \( (4 - 9.67)^2 \approx 32.11 \)
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:
\[
\text{Standard Deviation} = \sqrt{6.78} \approx 2.60
\]
### Comparison of Standard Deviations:
- **Standard Deviation of Dataset A:** approximately **2.36**
- **Standard Deviation of Dataset B:** approximately **2.60**
The rounded values could lead to:
- The final standard deviation calculations indicate that Dataset B has a higher standard deviation than Dataset A, implying that Dataset B has greater variability.
Therefore, 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.** However, this statement does not match the provided options. Therefore, please verify the details above to check for any discrepancies, as the calculations do suggest differing values than your listed responses.
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