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For data 0–20:17, 20–40:f₁, 40–60:32, 60–80:f₂, 80–100:19 with total frequency 120 and median 50, the median class is 40–60. Given cf before median class = 17 + f₁ and f = 32, l = 40, h = 20, and median = 50, find (17 + f₁).

Class 10StatisticsStatistics and Probability
  • A40
  • B48
  • C44
  • D88

Step-by-Step Solution

Median = l + h(n/2 − cf)/f → 50 = 40 + 20(60 − cf)/32 → 10 = 20(60 − cf)/32 → 16 = 60 − cf → cf = 44. So 17 + f₁ = 44.
Answer: C) 44

Key idea: Median for grouped data

**Step 1**: build cumulative frequency (cf) column.
**Step 2**: find n/2 (half the total frequency).
**Step 3**: locate the **median class** — the FIRST class whose cumulative frequency ≥ n/2.
**Step 4**: Apply formula.

**Median formula**:
**Median = l + h × (n/2 − cf) / f**

Where:
- l = lower limit of median class
- h = class width
- n = total frequency
- cf = cumulative frequency of class BEFORE median class
- f = frequency of median class

The formula uses linear interpolation within the median class to find where the n/2-th item falls.

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