Q.8 The above frequency chart shows the frequency distribution of marks obtained
by a set of students in an exam.
From the data presented above, which one of the following is CORRECT?
(A) mean > mode > median
(B) mode > median > mean
(C) mode > mean > median
(D) median > mode > mean
Introduction
In statistics, mean, median and mode are the three most widely used measures of central tendency.
This article explains how to determine the correct relationship between mean, median, and mode using a frequency chart of marks scored by students in an exam.
Understanding this comparison helps in analyzing the distribution pattern, especially whether it is skewed towards higher or lower values.
Given Frequency Distribution
| Marks | Frequency |
|---|---|
| 3 | 3 |
| 4 | 9 |
| 5 | 11 |
| 6 | 7 |
| 7 | 14 |
| 8 | 2 |
| 9 | 4 |
Total students = 3 + 9 + 11 + 7 + 14 + 2 + 4 = 50
Step 1: Find Mode
Mode = the value with highest frequency
Here, the highest frequency = 14 at marks = 7
👉 Mode = 7
Step 2: Find Median
Median = middle value when data are arranged
Since total = 50 (even), the median is the average of 25th and 26th observations.
Let’s accumulate frequencies:
| Marks | Freq | Cumulative Freq |
|---|---|---|
| 3 | 3 | 3 |
| 4 | 9 | 12 |
| 5 | 11 | 23 |
| 6 | 7 | 30 |
| 7 | 14 | 44 |
| 8 | 2 | 46 |
| 9 | 4 | 50 |
25th observation lies within mark 6
26th observation also lies within mark 6
👉 Median = 6
Step 3: Find Mean
Mean = Σ(f × x) ÷ Σf
Compute numerator:
- 3×3 = 9
- 4×9 = 36
- 5×11 = 55
- 6×7 = 42
- 7×14 = 98
- 8×2 = 16
- 9×4 = 36
Sum of fx = 9 + 36 + 55 + 42 + 98 + 16 + 36 = 292
Mean = 292 ÷ 50 = 5.84
👉 Mean ≈ 5.84
Final Comparison
- Mean ≈ 5.84
- Median = 6
- Mode = 7
Therefore: Mode > Median > Mean
Correct Answer
✅ (B) mode > median > mean
Explanation of Options
(A) mean > mode > median – Incorrect
Mean (~5.84) is less than both median and mode, not greater.
(B) mode > median > mean – Correct
- Mode (7)
- Median (6)
- Mean (5.84)
This matches the calculated values.
(C) mode > mean > median – Incorrect
Mean is not between mode and median; median exceeds mean.
(D) median > mode > mean – Incorrect
Median is not greater than mode; mode is highest.
Conclusion
For the given frequency distribution, the data is positively skewed, meaning most values are clustered on the lower side with a few higher marks pulling the tail.
This results in the correct relationship:
📌 mode > median > mean
