3. Example-3
3. Calculate Five number summary from the following grouped data
| Class | Frequency | | 2 - 4 | 3 | | 4 - 6 | 4 | | 6 - 8 | 2 | | 8 - 10 | 1 |
Solution:
Five number summary :
| Class | Frequency
`f` | `cf` | | 2 - 4 | 3 | 3 | | 4 - 6 | 4 | 7 | | 6 - 8 | 2 | 9 | | 8 - 10 | 1 | 10 | | --- | --- | --- | | n = 10 | -- |
Minimum value `=2`
Maximum value `=10`
First quartile `Q_1` :
Here, `n = 10`
`Q_1` class :
Class with `(n/4)^(th)` value of the observation in `cf` column
`=(10/4)^(th)` value of the observation in `cf` column
`=(2.5)^(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 + (2.5 - 0)/3 * 2`
`=2 + (2.5)/3 * 2`
`=2 + 1.6667`
`=3.6667`
Median `Q_2` :
`Q_2` class :
Class with `((2n)/4)^(th)` value of the observation in `cf` column
`=((2*10)/4)^(th)` value of the observation in `cf` column
`=(5)^(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 + (5 - 3)/4 * 2`
`=4 + (2)/4 * 2`
`=4 + 1`
`=5`
Third quartile `Q_3` :
`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`
Thus Five number summary is
1. Minimum value `=2`
2. First quartile `Q_1=3.6667`
3. Median `Q_2=5`
4. Third quartile `Q_3=6.5`
5. Maximum value `=10`
This material is intended as a summary. Use your textbook for detail explanation.
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