Ages of 20 Students Enrolled in a College ClassAgeFrequency186195204212221231301The table above shows the distribution of ages of the 20 students...
GMAT Problem-Solving and Data Analysis : (PS_DA) Questions
Ages of 20 Students Enrolled in a College Class
| Age | Frequency |
|---|---|
| 18 | 6 |
| 19 | 5 |
| 20 | 4 |
| 21 | 2 |
| 22 | 1 |
| 23 | 1 |
| 30 | 1 |
The table above shows the distribution of ages of the 20 students enrolled in a college class. Which of the following gives the correct order of the mean, median, and mode of the ages?
1. TRANSLATE the frequency table information
- Given information:
- 20 students total with ages ranging from 18 to 30
- Age 18: 6 students, Age 19: 5 students, Age 20: 4 students
- Age 21: 2 students, Age 22: 1 student, Age 23: 1 student, Age 30: 1 student
- What this tells us: We need to find mode, median, and mean from this frequency data, then order them.
2. TRANSLATE each statistic definition to this context
- Mode: The age that appears most frequently = age with highest frequency count
- Median: The middle age when all 20 ages are listed in order
- Mean: The average age = total of all ages ÷ 20
3. Calculate the mode
- TRANSLATE frequency data: Age 18 appears 6 times (highest frequency)
- Mode = 18
4. Calculate the median
- INFER the approach: With 20 students (even number), median is average of 10th and 11th values in ordered list
- VISUALIZE ordered data: 18,18,18,18,18,18,19,19,19,19,19,20,20,20,20,21,21,22,23,30
- The 10th and 11th positions are both 19
- Median = 19
5. Calculate the mean
- SIMPLIFY using weighted sum: \((18×6) + (19×5) + (20×4) + (21×2) + (22×1) + (23×1) + (30×1)\)
- SIMPLIFY: \(108 + 95 + 80 + 42 + 22 + 23 + 30 = 400\)
- Mean = \(400 ÷ 20 = 20\)
6. INFER the correct ordering
- Mode = 18, Median = 19, Mean = 20
- Therefore: \(\mathrm{mode} \lt \mathrm{median} \lt \mathrm{mean}\)
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students treat the frequency table like a simple list of ages rather than understanding that each age appears multiple times according to its frequency.
They might calculate mean as \((18+19+20+21+22+23+30) ÷ 7 = 21.9\), completely ignoring the frequency weights. Or they might think the median is simply the middle age value (20) rather than finding the actual middle position in the full dataset of 20 students.
This leads to incorrect calculations and may cause them to select Choice B or C depending on which mistakes they make.
Second Most Common Error:
Insufficient INFER reasoning: Students correctly calculate individual statistics but struggle with the median calculation for even-numbered datasets.
They might think the median is 19.5 (averaging positions 10 and 11) without realizing both the 10th and 11th actual data points are the same value (19). This misconception doesn't change the final answer but shows incomplete understanding of median calculation.
The Bottom Line:
Success on this problem requires recognizing that frequency tables represent repeated data values, not just unique values. The key insight is properly weighting calculations by frequency and understanding positional concepts for median in the full expanded dataset.