Use the tables to answer the question.
Table 1
Data Point 11 15 9 10 11 12 12 10 9
Distance from Mean 0 4 2 1 0 1 1 1 2
Table 2
Data Point 9 10 14 14 12 10 10 10 9
Distance from Mean 2 1 3 3 1 1 1 1 2
Compare the datasets in the tables. Which of the following correctly describes the relationship of their average deviations?
(1 point)
Responses
The second table has a larger average deviation.
The second table has a larger average deviation.
The average deviation is the same for both tables.
The average deviation is the same for both tables.
The first table has a larger average deviation.
The first table has a larger average deviation.
It is impossible to tell from this data.
It is impossible to tell from this data.
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Table 1
Data Point 11 15 9 10 11 12 12 10 9
Distance from Mean 0 4 2 1 0 1 1 1 2
Table 2
Data Point 9 10 14 14 12 10 10 10 9
Distance from Mean 2 1 3 3 1 1 1 1 2
Compare the datasets in the tables. Which of the following correctly describes the relationship of their average deviations?
(1 point)
Responses
The second table has a larger average deviation.
The second table has a larger average deviation.
The average deviation is the same for both tables.
The average deviation is the same for both tables.
The first table has a larger average deviation.
The first table has a larger average deviation.
It is impossible to tell from this data.
It is impossible to tell from this data.
Skip to navigation
Answers
Answered by
GPT-4o mini
To determine which dataset has a larger average deviation, we calculate the average deviation for both tables.
**Table 1:**
- Distance from Mean: 0, 4, 2, 1, 0, 1, 1, 1, 2
- To find the average deviation, we sum these distances and divide by the number of data points.
Sum of distances = \(0 + 4 + 2 + 1 + 0 + 1 + 1 + 1 + 2 = 12\)
Number of data points = 9
Average deviation for Table 1 = \(\frac{12}{9} \approx 1.33\)
**Table 2:**
- Distance from Mean: 2, 1, 3, 3, 1, 1, 1, 1, 2
- We again sum these distances and divide by the number of data points.
Sum of distances = \(2 + 1 + 3 + 3 + 1 + 1 + 1 + 1 + 2 = 15\)
Number of data points = 9
Average deviation for Table 2 = \(\frac{15}{9} \approx 1.67\)
Now, comparing the average deviations:
- Table 1 average deviation ≈ 1.33
- Table 2 average deviation ≈ 1.67
Thus, we can conclude:
**The second table has a larger average deviation.**
**Table 1:**
- Distance from Mean: 0, 4, 2, 1, 0, 1, 1, 1, 2
- To find the average deviation, we sum these distances and divide by the number of data points.
Sum of distances = \(0 + 4 + 2 + 1 + 0 + 1 + 1 + 1 + 2 = 12\)
Number of data points = 9
Average deviation for Table 1 = \(\frac{12}{9} \approx 1.33\)
**Table 2:**
- Distance from Mean: 2, 1, 3, 3, 1, 1, 1, 1, 2
- We again sum these distances and divide by the number of data points.
Sum of distances = \(2 + 1 + 3 + 3 + 1 + 1 + 1 + 1 + 2 = 15\)
Number of data points = 9
Average deviation for Table 2 = \(\frac{15}{9} \approx 1.67\)
Now, comparing the average deviations:
- Table 1 average deviation ≈ 1.33
- Table 2 average deviation ≈ 1.67
Thus, we can conclude:
**The second table has a larger average deviation.**
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