Mode of grouped data
Mode of discrete frequency distribution
1. Calculate Mode from the following grouped data
Solution:
`x`
`(1)` | Frequency `(f)`
`(2)` |
| 0 | 1 |
| 1 | 5 |
| 2 | 10 |
| 3 | 6 |
| 4 | 3 |
| --- | --- |
| `n=25` |
Mode :
the frequency of observation `2` is maximum (`10`)
`:. Z = 2`
2. Calculate Mode from the following grouped data
| X | Frequency |
| 10 | 3 |
| 11 | 12 |
| 12 | 18 |
| 13 | 12 |
| 14 | 3 |
Solution:
`x`
`(1)` | Frequency `(f)`
`(2)` |
| 10 | 3 |
| 11 | 12 |
| 12 | 18 |
| 13 | 12 |
| 14 | 3 |
| --- | --- |
| `n=48` |
Mode :
the frequency of observation `12` is maximum (`18`)
`:. Z = 12`
Mode of continuous frequency distribution
In grouped frequency distribution, we have to find a class with the maximum frequency, and this class is called modal class.
Mode is a value inside modal class and it is found by formula
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
where
`:. L = `lower boundary point of mode class
`:. f_1 = ` frequency of the mode class
`:. f_0 = ` frequency of the preceding class
`:. f_2 = ` frequency of the succedding class
`:. c = ` class length of mode class
3. Calculate Mode from the following grouped data
| Class | Frequency |
| 2 - 4 | 3 |
| 4 - 6 | 4 |
| 6 - 8 | 2 |
| 8 - 10 | 1 |
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` |
| 2-4 | 3 |
| 4-6 | 4 |
| 6-8 | 2 |
| 8-10 | 1 |
| --- | --- |
| -- | `n = 10` |
To find Mode Class
Here, maximum frequency is `4`.
`:.` The mode class is `4 - 6`.
`:. L = `lower boundary point of mode class `=4`
`:. f_1 = ` frequency of the mode class `=4`
`:. f_0 = ` frequency of the preceding class `=3`
`:. f_2 = ` frequency of the succedding class `=2`
`:. c = ` class length of mode class `=2`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`=4 + ((4 - 3)/(2*4 - 3 - 2)) * 2`
`=4 + (1/3) * 2`
`=4 + 0.6667`
`=4.6667`
4. Calculate Mode 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
`(1)` | Frequency `(f)`
`(2)` |
| 0-2 | 5 |
| 2-4 | 16 |
| 4-6 | 13 |
| 6-8 | 7 |
| 8-10 | 5 |
| 10-12 | 4 |
| --- | --- |
| -- | `n = 50` |
To find Mode Class
Here, maximum frequency is `16`.
`:.` The mode class is `2 - 4`.
`:. L = `lower boundary point of mode class `=2`
`:. f_1 = ` frequency of the mode class `=16`
`:. f_0 = ` frequency of the preceding class `=5`
`:. f_2 = ` frequency of the succedding class `=13`
`:. c = ` class length of mode class `=2`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`=2 + ((16 - 5)/(2*16 - 5 - 13)) * 2`
`=2 + (11/14) * 2`
`=2 + 1.5714`
`=3.5714`
5. Calculate Mode 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
`(1)` | Frequency `(f)`
`(2)` |
| 10 - 20 | 15 |
| 20 - 30 | 25 |
| 30 - 40 | 20 |
| 40 - 50 | 12 |
| 50 - 60 | 8 |
| 60 - 70 | 5 |
| 70 - 80 | 3 |
| --- | --- |
| `n = 88` |
To find Mode Class
Here, maximum frequency is `25`.
`:.` The mode class is `20 - 30`.
`:. L = `lower boundary point of mode class `=20`
`:. f_1 = ` frequency of the mode class `=25`
`:. f_0 = ` frequency of the preceding class `=15`
`:. f_2 = ` frequency of the succedding class `=20`
`:. c = ` class length of mode class `=10`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`=20 + ((25 - 15)/(2*25 - 15 - 20)) * 10`
`=20 + (10/15) * 10`
`=20 + 6.6667`
`=26.6667`
6. Calculate Mode 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
`(1)` | Frequency `(f)`
`(2)` |
| 20 - 25 | 110 |
| 25 - 30 | 170 |
| 30 - 35 | 80 |
| 35 - 40 | 45 |
| 40 - 45 | 40 |
| 45 - 50 | 35 |
| --- | --- |
| `n = 480` |
To find Mode Class
Here, maximum frequency is `170`.
`:.` The mode class is `25 - 30`.
`:. L = `lower boundary point of mode class `=25`
`:. f_1 = ` frequency of the mode class `=170`
`:. f_0 = ` frequency of the preceding class `=110`
`:. f_2 = ` frequency of the succedding class `=80`
`:. c = ` class length of mode class `=5`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`=25 + ((170 - 110)/(2*170 - 110 - 80)) * 5`
`=25 + (60/150) * 5`
`=25 + 2`
`=27`
This material is intended as a summary. Use your textbook for detail explanation.
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