4. Calculate Five number summary 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:
Five number summary :
| 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 | -- |
Minimum value `=0`
Maximum value `=12`
First quartile `Q_1` :
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`
Median `Q_2` :
`Q_2` class :
Class with `((2n)/4)^(th)` value of the observation in `cf` column
`=((2*50)/4)^(th)` value of the observation in `cf` column
`=(25)^(th)` value of the observation in `cf` column
and it lies in the class `4 - 6`.
`:. Q_2` class : `4 - 6`
The lower boundary point of `4 - 6` is `4`.
`:. L = 4`
`Q_2 = L + ((2 n)/4 - cf)/f * c`
`=4 + (25 - 21)/13 * 2`
`=4 + (4)/13 * 2`
`=4 + 0.6154`
`=4.6154`
Third quartile `Q_3` :
`Q_3` class :
Class with `((3n)/4)^(th)` value of the observation in `cf` column
`=((3*50)/4)^(th)` value of the observation in `cf` column
`=(37.5)^(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 + (37.5 - 34)/7 * 2`
`=6 + (3.5)/7 * 2`
`=6 + 1`
`=7`
Thus Five number summary is
1. Minimum value `=0`
2. First quartile `Q_1=2.9375`
3. Median `Q_2=4.6154`
4. Third quartile `Q_3=7`
5. Maximum value `=12`
This material is intended as a summary. Use your textbook for detail explanation.
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