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The dot plots provided show two samples, labeled "Sample A" and "Sample B," representing the number of lunches purchased by seventh-grade students.

### Step 1: Calculate the means of both samples.
- **Sample A** consists of the following values: 12, 13, 14, 14, 15, 15, 18.
  - Mean of Sample A = \( \frac{12 + 13 + 14 + 14 + 15 + 15 + 18}{7} = \frac{101}{7} = 14.43 \) (rounded to two decimal places)
  
- **Sample B** consists of the following values: 11, 13, 13, 15, 16, 17, 20.
  - Mean of Sample B = \( \frac{11 + 13 + 13 + 15 + 16 + 17 + 20}{7} = \frac{105}{7} = 15 \).

### Step 2: Find the difference in the means.
- Difference between means = \( 15 - 14.43 = 0.57 \).

### Step 3: Calculate the mean absolute deviations (MAD).
MAD is calculated as the average of the absolute differences between each data point and the mean.

- **MAD for Sample A:**
  - \( \text{MAD} = \frac{|12 - 14.43| + |13 - 14.43| + |14 - 14.43| + |14 - 14.43| + |15 - 14.43| + |15 - 14.43| + |18 - 14.43|}{7} \)
  - \( \text{MAD for A} \approx \frac{2.43 + 1.43 + 0.43 + 0.43 + 0.57 + 0.57 + 3.57}{7} \approx \frac{9.43}{7} \approx 1.35 \).

- **MAD for Sample B:**
  - \( \text{MAD} = \frac{|11 - 15| + |13 - 15| + |13 - 15| + |15 - 15| + |16 - 15| + |17 - 15| + |20 - 15|}{7} \)
  - \( \text{MAD for B} = \frac{4 + 2 + 2 + 0 + 1 + 2 + 5}{7} = \frac{16}{7} \approx 2.29 \).

### Step 4: Compare the difference in means to the MADs.
- The difference in the means is approximately 0.57.
- Comparing this to the MAD of Sample A (\( 1.35 \)):
  - \( \frac{0.57}{1.35} \approx 0.42 \).
- Comparing this to the MAD of Sample B (\( 2.29 \)):
  - \( \frac{0.57}{2.29} \approx 0.25 \).

### Conclusion:
The difference in means is best approximated as 0.5 times the mean absolute deviation of Sample A, based on the options given.

Thus, the best-supported statement is:
- **"The difference in the means is about 0.5 times the mean absolute deviation of Sample A."**

Would you like more details or have any questions?

### Related Questions:
1. How is the mean absolute deviation (MAD) different from the standard deviation?
2. How does the distribution shape affect the mean and MAD values?
3. Can the mean be misleading if the data has outliers?
4. How would the result change if we calculated the variance instead of MAD?
5. Why is it important to compare the difference in means relative to the MAD?

### Tip:
Always check the spread of data when comparing samples; the mean alone might not provide the full picture!

The dot plots show two different random samples of the number of lunches purchased in a month by some seventh grade students. Which statement is BEST supported by the data?

Analyze dot plots showing the number of lunches purchased by seventh grade students. Learn to calculate the mean and mean absolute deviation (MAD), and understand how to compare the difference in means relative to the MAD. Suitable for students in Grades 6-8.

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