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Question
Calculate Q1, D6, and P15 for the following data:
| Mid value | 25 | 75 | 125 | 175 | 225 | 275 |
| Frequency | 10 | 70 | 80 | 100 | 150 | 90 |
Solution
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:
| Class interval | Frequency (f) |
Less than cumulative frequency (c.f.) |
| 0 – 50 | 10 | 10 |
| 50 – 100 | 70 | 80 ← P15 |
| 100 – 150 | 80 | 160 ← Q1 |
| 150 – 200 | 100 | 260 |
| 200 – 250 | 150 | 410 ← D6 |
| 250 – 300 | 90 | 500 |
| Total | 500 |
Here, N = 500
Q1 class = class containing `("N"/4)^"th"` observation
∴ `"N"/4 = 500/4` = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
∴ Q1 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q1 = `"L"+"h"/"f"("N"/4-"c.f.")`
= `100 + 50/80 (125 - 80)`
= `100 + 5/8 (45)`
= 100 + 28.125
= 128.125
D6 class = class containing `((6"N")/10)^"th"` observation
∴ `(6"N")/10=(6xx500)/10` = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D6 lies in the class 200 – 250
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D6 = `"L"+"h"/"f"((6"N")/10-"c.f.")`
= `200 + 50/150 (300 - 260)`
= `200 + 1/3 (40)`
= 200 + 13.33
= 213.33
∴ P15 class = class containing `((15"N")/100)^"th"` observation
∴ `(15"N")/100 =(15 xx 500)/100` = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P15 = `"L"+"h"/"f"((15"N")/100-"c.f.")`
= `50 + 50/70 (75 - 10)`
= `50 + 5/7 (65)`
= `50 + 325/7`
= 50 + 46.4286
= 96.4286
∴ Q1 = 128.125, D6 = 213.33, P15 = 96.4286
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