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Question
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distance.
|
Distance (km)
|
200 - 210 | 210 - 220 | 220 - 230 | 230 - 240 | 240 - 250 |
| No. of buses | 40 | 60 | 80 | 50 | 20 |
Solution
|
Class
(Distance in Kms) |
Frequency (Number of buses) fi |
Cumulaive frequency less than the upper limit |
| 200 - 210 | 40 | 40 |
| 210 - 220 | 60 | 100 |
| 220 - 230 (Median Class) |
80 | 180 |
| 230 - 240 | 50 | 230 |
| 240 - 250 | 20 | 250 |
| N = 250 |
From the above table, we get
L (Lower class limit of the median class) = 220
N (Sum of frequencies) = 250
h (Class interval of the median class) = 10
f (Frequency of the median class) = 80
cf (Cumulative frequency of the class preceding the median class) = 100
Now, Median = `L + ((N/2-cf))/f xx h`
= 220 + `((250/2 -100)/80)xx10`
= 220 + 3.13
= 223.13 km
Hence, the median of the distances is 223.13 km.
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