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Question
Following data gives the age distribution of 240 employees of a firm. Calculate Q.D. of the distribution:
| Age (in years) |
20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 |
| No. of employees | 30 | 40 | 60 | 50 | 46 | 14 |
Solution
We construct the less than cumulative frequency table as follows:
| Age (In years) |
No. of employees (f) |
Less than cumulative frequency (c.f.) |
| 20 – 25 | 30 | 30 |
| 25 – 30 | 40 | 70 ← Q1 |
| 30 – 35 | 60 | 130 |
| 35 – 40 | 50 | 180 ← Q3 |
| 40 – 50 | 46 | 226 |
| 45 – 50 | 14 | 240 |
| Total | N = 240 |
Here, N = 240
For Q1, class = class containing `("N"/4)^"th"` observation
∴ `"N"/(4) = (240)/(4)` = 60
Cumulative frequency which is just greater than (or equal to) 60 is 70.
∴ Q1 Lies in the class 25 – 30.
∴ L = 25, f = 40, c.f. = 30, h = 5
Q1 = `"L"+"h"/"f"("N"/4-"c.f.")`
∴ Q1 = `25 + (5)/(40)(60 - 30)`
∴ Q1 = `25 + (1)/(8)(30)`
∴ Q1 = 25 + 3.75
∴ Q1 = 28.75
For Q3 class = class containing `("3N"/4)^"th"` observation
∴ `(3"N")/4=(3xx240)/4` = 180
Cumulative frequency which is just greater than (or equal to) 180 is 180.
∴ Q3 Lies in the class 35 – 40
∴ L = 35, f = 50, c.f. = 130, h = 5
Q3 = `"L"+"h"/"f"((3"N")/4-"c.f.")`
∴ Q3 = `35 + (5)/(50)(180 - 130)`
∴ Q3 = `35 + (1)/(10)xx50`
∴ Q3 = 35 + 5
∴ Q3 = 40
Q.D. = `("Q"_3 - "Q"_1)/(2)`
∴ Q.D. = `(40 - 28.75)/(2)`
∴ Q.D. = `(11.25)/(2)`
∴ Q.D. = 5.625
Notes
The question has been modified.
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