Problem Statement
The median of the following grouped frequency distribution of 100 observations is 525. Find the missing frequencies $x$ and $y$:
| Class Interval | Frequency |
| :--- | :--- |
| 0 - 100 | 2 |
| 100 - 200 | 5 |
| 200 - 300 | $x$ |
| 300 - 400 | 12 |
| 400 - 500 | 17 |
| 500 - 600 | 20 |
| 600 - 700 | $y$ |
| 700 - 800 | 9 |
| 800 - 900 | 7 |
| 900 - 1000 | 4 |
Verified Solution & Marking Scheme
Construct Cumulative Frequency Table and Sum Relation
Cumulative frequencies (cf):
- 0-100: 2
- 100-200: 7
- 200-300: $7 + x$
- 300-400: $19 + x$
- 400-500: $36 + x$
- 500-600: $56 + x$
- 600-700: $56 + x + y$
- 700-800: $65 + x + y$
- 800-900: $72 + x + y$
- 900-1000: $76 + x + y$
Total frequency $N = 100$:
$76 + x + y = 100 \implies x + y = 24 \quad \text{--- (1)}$
Identify Median Class
Given Median $= 525$, which lies in the class interval $500 - 600$.
Therefore:
- Lower limit of median class $l = 500$
- Frequency of median class $f = 20$
- Cumulative frequency of preceding class $cf = 36 + x$
- Class width $h = 100$
- $N/2 = 50$
Apply Median formula to find x
$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 14 - 5 = 9$
Find y from equation (1)
Substitute $x = 9$ into (1):
$9 + y = 24 \implies y = 15$
Check: $x = 9, y = 15 \implies x + y = 24$. All frequencies are non-negative integers.