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Question
From the following data calculate the rent of 15th, 65th and 92nd house.
| House rent (in ₹) | 11000 | 12000 | 13000 | 15000 | 14000 | 16000 | 17000 | 18000 |
| No. of houses | 25 | 17 | 13 | 14 | 15 | 8 | 6 | 2 |
Solution
Arranging the given data in ascending order.
| House Rent (in ₹) | No. of houses (f) | Less than Cumulative frequency (c.f.) |
| 11000 | 25 | 25 ← P15 |
| 12000 | 17 | 42 |
| 13000 | 13 | 55 |
| 14000 | 15 | 70 ← P65 |
| 15000 | 14 | 84 |
| 16000 | 8 | 92 |
| 17000 | 6 | 98 ← P92 |
| 18000 | 2 | 100 |
| Total | 100 |
Here, n = 100
P15 = value of `15(("n" + 1)/100)^"th"` observation
= value of `15((100 + 1)/100)^"th"` observation
= value of (15 × 1.01)th observation
= value of (15.15)th observation
Cumulative frequency which is just greater than (or equal to) 15.15 is 25.
∴ P15 = 11000
P65 = value of `65(("n" + 1)/100)^"th"` observation
= value of `65((100 + 1)/100)^"th"` observation
= value of (65 × 1.01)th observation
= value of (65.65)th observation
Cumulative frequency which is just greater than (or equal to) 65.65 is 70.
∴ P65 = 14000
P92 = value of `92(("n" + 1)/100)^"th"` observation
= value of `92((100 + 1)/100)^"th"` observation
= value of (92 × 1.01)th observation
= value of (92.92)th observation
Cumulative frequency which is just greater than (or equal to) 92.92 is 98.
∴ P92 = 17000
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