The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.
| Distance (km) | 200–210 | 210–220 | 220–230 | 230–240 | 240–250 |
|---|---|---|---|---|---|
| No. of buses | 40 | 60 | 80 | 50 | 20 |
Answer: Median ≈ 223.13 km
| Class | Frequency | Cumulative frequency (less than) |
|---|---|---|
| 200–210 | 40 | 40 |
| 210–220 | 60 | 100 |
| 220–230 | 80 | 180 |
| 230–240 | 50 | 230 |
| 240–250 | 20 | 250 |
\(N = 250\), so \(\frac{N}{2} = 125\). The first cumulative frequency greater than or equal to 125 is 180, so the median class is 220–230.
\(L = 220\), \(cf = 100\), \(f = 80\), \(h = 10\)
\[ \text{Median} = L + \left[\frac{\frac{N}{2} - cf}{f}\right] \times h \]
\[ = 220 + \frac{125 - 100}{80} \times 10 \]
\[ = 220 + 3.125 = 223.125 \approx 223.13 \]
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