The frequency distribution above summarizes a set of data, where a is a positive integer. How much greater is the...
GMAT Problem-Solving and Data Analysis : (PS_DA) Questions
The frequency distribution above summarizes a set of data, where \(\mathrm{a}\) is a positive integer. How much greater is the mean of the set of data than the median?
| Value | Frequency |
|---|---|
| 1 | \(\mathrm{a}\) |
| 2 | \(\mathrm{2a}\) |
| 3 | \(\mathrm{3a}\) |
| 4 | \(\mathrm{2a}\) |
| 5 | \(\mathrm{a}\) |
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1. TRANSLATE the frequency table information
- Given information:
- Values: 1, 2, 3, 4, 5
- Frequencies: \(\mathrm{a, 2a, 3a, 2a, a}\) (where \(\mathrm{a}\) is a positive integer)
- Need to find: how much greater the mean is than the median
2. INFER the pattern in the data
- Key insight: Look at the frequency pattern: \(\mathrm{a, 2a, 3a, 2a, a}\)
- This creates a symmetric distribution around the middle value 3
- In symmetric distributions, the mean equals the median
- Therefore: \(\mathrm{mean - median = 0}\)
3. Verify with explicit calculations (optional)
Finding the median:
- Total frequency = \(\mathrm{9a}\)
- The median is the middle value with equal data above and below
- Values below 3: \(\mathrm{a + 2a = 3a}\) items
- Values above 3: \(\mathrm{2a + a = 3a}\) items
- Median = 3
Finding the mean:
- SIMPLIFY: Mean = \(\mathrm{(1×a + 2×2a + 3×3a + 4×2a + 5×a) ÷ 9a}\)
\(\mathrm{= 27a ÷ 9a}\)
\(\mathrm{= 3}\)
Answer: A. 0
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students miss the symmetric pattern and try to calculate mean and median separately, making computational mistakes along the way.
They might incorrectly find the median position or make arithmetic errors in the mean calculation (like getting \(\mathrm{27a ÷ 9a = 3a}\) instead of 3). This leads to confusion and guessing among the answer choices.
Second Most Common Error:
Missing conceptual knowledge: Students don't recognize that in a symmetric distribution, mean equals median.
Instead, they spend time on lengthy calculations, increasing the chance of arithmetic errors. Even if they calculate correctly, they don't see the elegant shortcut that makes this problem much simpler.
The Bottom Line:
This problem tests whether students can recognize patterns in data distributions. The key insight is spotting the symmetry, which immediately tells you the answer without tedious calculations.
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