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?
Options:
(A) mean > mode > median
(B) mode > median > mean
(C) mode > mean > median
(D) median > mode > mean
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.8)**
**1. Statistical Data & Definitions**:
- **Data Pairs (Marks $x$, Frequency $f$)**:
- $(3, 3), (4, 9), (5, 11), (6, 7), (7, 14), (8, 2), (9, 4)$
- **Total Frequency ($N$)**:
$N = 3 + 9 + 11 + 7 + 14 + 2 + 4 = 50$
**2. Step-by-Step Calculations**:
- **Mode**: The mark with the greatest frequency.
- The maximum frequency is $14$, which corresponds to Mark **$7$**.
- $\text{Mode} = 7$.
- **Median**: Value at position $\frac{N+1}{2} = \frac{51}{2} = 25.5$.
- Cumulative frequencies:
- Mark 3: $3$
- Mark 4: $3 + 9 = 12$
- Mark 5: $12 + 11 = 23$
- Mark 6: $23 + 7 = 30$
- Since positions $24$ through $30$ correspond to Mark $6$, the median is **$6$**.
- $\text{Median} = 6$.
- **Mean**: $\bar{x} = \frac{\sum f_i x_i}{N}$:
$\sum f_i x_i = (3\times 3) + (4\times 9) + (5\times 11) + (6\times 7) + (7\times 14) + (8\times 2) + (9\times 4)$
$\sum f_i x_i = 9 + 36 + 55 + 42 + 98 + 16 + 36 = 292$
$\bar{x} = \frac{292}{50} = 5.84$
- **Comparison**:
$\text{Mode } (7) > \text{Median } (6) > \text{Mean } (5.84)$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (B)** (`mode > median > mean`).
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