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प्रश्न
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.
|
Average Speed of
Vehicles (Km/hr) |
60 - 64 | 64 - 69 | 70 - 74 | 75 - 79 | 79 - 84 | 84 - 89 |
| No. of vehicles | 10 | 34 | 55 | 85 | 10 | 6 |
बेरीज
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उत्तर
|
Class
(Average speed of vehicles (km/hr)) |
Continuous
classes |
Frequency (Number of workers) |
Cumulative frequency less than the upper limit |
| 60 - 64 | 59.5 - 64.5 | 10 | 10 |
| 64 - 69 | 64.5 - 69.5 | 34 | 44 |
| 70 - 74 | 69.5 - 74.5 | 55 | 99 → cf |
| 75 - 79 | 74.5 - 79.5 | 85 → f | 184 |
| 79 - 84 | 79.5 - 84.5 | 10 | 194 |
| 84 - 89 | 84.5 - 89.5 | 6 | 200 |
| Total | N = 200 |
Here, total frequency = ∑fi = N = 200
∴ `"N"/2 = 200/2 = 100`
Cumulative frequency, which is just greater than (or equal) to 100, is 184.
∴ The median class is 74.5 – 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5
Now,
∴ Median = `"L" + (("N"/2 - "cf")/"f") × "h"`
= `74.5 + ((100 - 99)/85) xx 5`
= `74.5 + 1/85 xx 5`
= `74.5 + 1/17`
= 74.5 + 0.0588
= 74.558 ≈ 75
∴ The median of the given data is 75 km/hr.
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