2. Quartile Example
1. Calculate Quartile-3 from the following grouped data
Solution:
`x` | Frequency `f` | `cf` | 0 | 1 | 1 | 1 | 5 | 6 | 2 | 10 | 16 | 3 | 6 | 22 | 4 | 3 | 25 | --- | --- | --- | | n = 25 | -- |
Here, `n = 25`
`Q_3 = ((3(n+1))/4)^(th)` value of the observation
`=((3*26)/4)^(th)` value of the observation
`=(19.5)^(th)` value of the observation
`=3`
2. Calculate Quartile-1 from the following grouped data
X | Frequency | 10 | 3 | 11 | 12 | 12 | 18 | 13 | 12 | 14 | 3 |
Solution:
`x` | Frequency `f` | `cf` | 10 | 3 | 3 | 11 | 12 | 15 | 12 | 18 | 33 | 13 | 12 | 45 | 14 | 3 | 48 | --- | --- | --- | | n = 48 | -- |
Here, `n = 48`
`Q_1 = ((n+1)/4)^(th)` value of the observation
`=(49/4)^(th)` value of the observation
`=(12.25)^(th)` value of the observation
`=11`
3. Calculate Quartile-3 from the following grouped data
Class | Frequency | 2 - 4 | 3 | 4 - 6 | 4 | 6 - 8 | 2 | 8 - 10 | 1 |
Solution:
Class | Frequency `f` | `cf` | 2 - 4 | 3 | 3 | 4 - 6 | 4 | 7 | 6 - 8 | 2 | 9 | 8 - 10 | 1 | 10 | --- | --- | --- | | n = 10 | -- |
Here, `n = 10`
`Q_3` class :
Class with `((3n)/4)^(th)` value of the observation in `cf` column
`=((3*10)/4)^(th)` value of the observation in `cf` column
`=(7.5)^(th)` value of the observation in `cf` column
and it lies in the class `6 - 8`.
`:. Q_3` class : `6 - 8`
The lower boundary point of `6 - 8` is `6`.
`:. L = 6`
`Q_3 = L + ((3 n)/4 - cf)/f * c`
`=6 + (7.5 - 7)/2 * 2`
`=6 + (0.5)/2 * 2`
`=6 + 0.5`
`=6.5`
4. Calculate Quartile-1 from the following grouped data
Class | Frequency | 0 - 2 | 5 | 2 - 4 | 16 | 4 - 6 | 13 | 6 - 8 | 7 | 8 - 10 | 5 | 10 - 12 | 4 |
Solution:
Class | Frequency `f` | `cf` | 0 - 2 | 5 | 5 | 2 - 4 | 16 | 21 | 4 - 6 | 13 | 34 | 6 - 8 | 7 | 41 | 8 - 10 | 5 | 46 | 10 - 12 | 4 | 50 | --- | --- | --- | | n = 50 | -- |
Here, `n = 50`
`Q_1` class :
Class with `(n/4)^(th)` value of the observation in `cf` column
`=(50/4)^(th)` value of the observation in `cf` column
`=(12.5)^(th)` value of the observation in `cf` column
and it lies in the class `2 - 4`.
`:. Q_1` class : `2 - 4`
The lower boundary point of `2 - 4` is `2`.
`:. L = 2`
`Q_1 = L + (( n)/4 - cf)/f * c`
`=2 + (12.5 - 5)/16 * 2`
`=2 + (7.5)/16 * 2`
`=2 + 0.9375`
`=2.9375`
5. Calculate Quartile-3 from the following grouped data
Class | Frequency | 10 - 20 | 15 | 20 - 30 | 25 | 30 - 40 | 20 | 40 - 50 | 12 | 50 - 60 | 8 | 60 - 70 | 5 | 70 - 80 | 3 |
Solution:
Class | Frequency `f` | `cf` | 10 - 20 | 15 | 15 | 20 - 30 | 25 | 40 | 30 - 40 | 20 | 60 | 40 - 50 | 12 | 72 | 50 - 60 | 8 | 80 | 60 - 70 | 5 | 85 | 70 - 80 | 3 | 88 | --- | --- | --- | | n = 88 | -- |
Here, `n = 88`
`Q_3` class :
Class with `((3n)/4)^(th)` value of the observation in `cf` column
`=((3*88)/4)^(th)` value of the observation in `cf` column
`=(66)^(th)` value of the observation in `cf` column
and it lies in the class `40 - 50`.
`:. Q_3` class : `40 - 50`
The lower boundary point of `40 - 50` is `40`.
`:. L = 40`
`Q_3 = L + ((3 n)/4 - cf)/f * c`
`=40 + (66 - 60)/12 * 10`
`=40 + (6)/12 * 10`
`=40 + 5`
`=45`
6. Calculate Quartile-1 from the following grouped data
Class | Frequency | 20 - 25 | 110 | 25 - 30 | 170 | 30 - 35 | 80 | 35 - 40 | 45 | 40 - 45 | 40 | 45 - 50 | 35 |
Solution:
Class | Frequency `f` | `cf` | 20 - 25 | 110 | 110 | 25 - 30 | 170 | 280 | 30 - 35 | 80 | 360 | 35 - 40 | 45 | 405 | 40 - 45 | 40 | 445 | 45 - 50 | 35 | 480 | --- | --- | --- | | n = 480 | -- |
Here, `n = 480`
`Q_1` class :
Class with `(n/4)^(th)` value of the observation in `cf` column
`=(480/4)^(th)` value of the observation in `cf` column
`=(120)^(th)` value of the observation in `cf` column
and it lies in the class `25 - 30`.
`:. Q_1` class : `25 - 30`
The lower boundary point of `25 - 30` is `25`.
`:. L = 25`
`Q_1 = L + (( n)/4 - cf)/f * c`
`=25 + (120 - 110)/170 * 5`
`=25 + (10)/170 * 5`
`=25 + 0.2941`
`=25.2941`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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