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प्रश्न
Following is the grouped data for duration of fixed deposits of 100 senior citizens from a certain bank:
| Fixed deposit (in days) | 0 – 180 | 180 – 360 | 360 – 540 | 540 – 720 | 720 – 900 |
| No. of senior citizens | 15 | 20 | 25 | 30 | 10 |
Calculate the limits of fixed deposits of central 50% senior citizens.
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उत्तर
We construct the less than cumulative frequency table as given below:
| Fixed deposit (in days) | No. of senior citizens (f) |
Less than Cumulative frequency (c.f.) |
| 0 – 180 | 15 | 15 |
| 180 – 360 | 20 | 35 ← Q1 |
| 360 – 540 | 25 | 60 |
| 540 – 720 | 30 | 90 ← Q3 |
| 720 – 900 | 10 | 100 |
| Total | 100 |
To find the limits of fixed deposits of central 50% senior citizens, we have to find Q1 and Q3.
Here, N = 100
Q1 class = class containing `(("N")/4)^"th"` observation
∴ `"N"/4=100/4` = 25
Cumulative frequency which is just greater than (or equal to) 25 is 35.
∴ Q1 lies in the class 180 – 360.
∴ L= 180, f = 20, c.f. = 15, h = 180
∴ Q1 = `"L"+"h"/"f"("N"/4 - "c.f.")`
= `180 + (180)/(20)(25 - 15)`
= 180 + 9(10)
= 180 + 90
∴ Q1 = 270
Q3 class = class containing `(("3N")/4)^"th"` observation
∴ `(3"N")/(4) = (3(100))/(4)` = 75
Cumulative frequency which is just greater than (or equal to) 75 is 90.
∴ Q3 lies in the class 540 – 720
∴ L= 540, f = 30, c.f. = 60, h = 180
∴ Q3 = `"L"+"h"/"f"("3N"/4 - "c.f.")`
= `540 + (180)/(30)(75 - 60)`
= 540 + 6(15)
= 540 + 90
∴ Q3 = 630
∴ Limits of the duration of fixed deposits of central 50% of senior citizens are from 270 to 630.
