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Quartile for grouped data Example-3 ( Enter your problem )
  1. Formula & Example-1
  2. Example-2
  3. Example-3

3. Example-3





5. Calculate Quartile-3 from the following grouped data
ClassFrequency
10 - 2015
20 - 3025
30 - 4020
40 - 5012
50 - 608
60 - 705
70 - 803


Solution:
ClassFrequency
`f`
`cf`
10 - 201515
20 - 302540
30 - 402060
40 - 501272
50 - 60880
60 - 70585
70 - 80388
---------
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
ClassFrequency
20 - 25110
25 - 30170
30 - 3580
35 - 4045
40 - 4540
45 - 5035


Solution:
ClassFrequency
`f`
`cf`
20 - 25110110
25 - 30170280
30 - 3580360
35 - 4045405
40 - 4540445
45 - 5035480
---------
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.
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