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Statistics
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section D]
The following table gives the cumulative frequency distribution of 'more than' type for marks obtained by 80 students in a test:
Marks: 0 and above (80), 10 and above (77), 20 and above (72), 30 and above (65), 40 and above (55), 50 and above (43), 60 and above (28), 70 and above (16), 80 and above (10).
Convert the above distribution into a continuous frequency distribution and compute the Median.
The following table gives the cumulative frequency distribution of 'more than' type for marks obtained by 80 students in a test:
Marks: 0 and above (80), 10 and above (77), 20 and above (72), 30 and above (65), 40 and above (55), 50 and above (43), 60 and above (28), 70 and above (16), 80 and above (10).
Convert the above distribution into a continuous frequency distribution and compute the Median.
Step-by-Step Solution
Key Concept: Convert to class intervals (0-10, 10-20, ...), compute frequencies $f_i$, find median class where $cf \ge N/2 = 40$.
Continuous distribution table:
Class Intervals: 0-10 (3), 10-20 (5), 20-30 (7), 30-40 (10), 40-50 (12), 50-60 (15), 60-70 (12), 70-80 (6), 80-90 (10). $N = 80, N/2 = 40$.
Median Class: 50-60 ($L = 50, f = 15, cf = 37, h = 10$).
$\text{Median} = L + \left(\dfrac{N/2 - cf}{f}\right) \times h = 50 + \left(\dfrac{40 - 37}{15}\right) \times 10 = 50 + 2 = 52$. [5.0 Marks]
Class Intervals: 0-10 (3), 10-20 (5), 20-30 (7), 30-40 (10), 40-50 (12), 50-60 (15), 60-70 (12), 70-80 (6), 80-90 (10). $N = 80, N/2 = 40$.
Median Class: 50-60 ($L = 50, f = 15, cf = 37, h = 10$).
$\text{Median} = L + \left(\dfrac{N/2 - cf}{f}\right) \times h = 50 + \left(\dfrac{40 - 37}{15}\right) \times 10 = 50 + 2 = 52$. [5.0 Marks]
Correct Answer: Median = 52
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