Median of grouped data
Median of discrete frequency distribution
If `n` is odd, then
`M=` value of `((n+1)/2)^(th)` observation
If `n` is even, then
`M=(text{Value of } (n/2)^(th) text{ observation} + text{Value of } (n/2 + 1)^(th) text{ observation})/2`
1. Calculate Median from the following grouped data
Solution:`x` `(1)` | Frequency `(f)` `(2)` | `cf` `(5)` |
0 | 1 | 1 |
1 | 5 | 6 |
2 | 10 | 16 |
3 | 6 | 22 |
4 | 3 | 25 |
--- | --- | --- |
| `n=25` | -- |
Median :M = value of `((n+1)/2)^(th)` observation
= value of `(26/2)^(th)` observation
= value of `13^(th)` observation
From the column of cumulative frequency `cf`, we find that the `13^(th)` observation is `2`.
Hence, the median of the data is `2`.
2. Calculate Median from the following grouped data
X | Frequency |
10 | 3 |
11 | 12 |
12 | 18 |
13 | 12 |
14 | 3 |
Solution:`x` `(1)` | Frequency `(f)` `(2)` | `cf` `(5)` |
10 | 3 | 3 |
11 | 12 | 15 |
12 | 18 | 33 |
13 | 12 | 45 |
14 | 3 | 48 |
--- | --- | --- |
| `n=48` | -- |
Median :M = value of `((n+1)/2)^(th)` observation
= value of `(49/2)^(th)` observation
= value of `24.5^(th)` observation
From the column of cumulative frequency `cf`, we find that the `24.5^(th)` observation is `12`.
Hence, the median of the data is `12`.
Median of continuous frequency distribution
To find Median class, we find cumulative frequencies of all classes and then find `n/2`.
The class whose cumulative frequency is `>= n/2` is called Median class
Median `M = L + (n/2 - cf)/f * c`
where
`:. L = `lower boundary point of median class
`:. n = `Total frequency
`:. cf = `Cumulative frequency of the class preceding the median class
`:. f = `Frequency of the median class
`:. c = `class length of median class
3. Calculate Median 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)` | `cf` `(6)` |
2-4 | 3 | 3 |
4-6 | 4 | 7 |
6-8 | 2 | 9 |
8-10 | 1 | 10 |
--- | --- | --- |
-- | `n = 10` | -- |
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(10/2)^(th)` observation
= value of `5^(th)` observation
From the column of cumulative frequency `cf`, we find that the `5^(th)` observation lies in the class `4 - 6`.
`:.` The median class is `4 - 6`.
Now,
`:. L = `lower boundary point of median class `=4`
`:. n = `Total frequency `=10`
`:. cf = `Cumulative frequency of the class preceding the median class `=3`
`:. f = `Frequency of the median class `=4`
`:. c = `class length of median class `=2`
Median `M = L + (n/2 - cf)/f * c`
`=4 + (5 - 3)/4 * 2`
`=4 + (2)/4 * 2`
`=4 + 1`
`=5`
4. Calculate Median 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)` | `cf` `(6)` |
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` | -- |
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(50/2)^(th)` observation
= value of `25^(th)` observation
From the column of cumulative frequency `cf`, we find that the `25^(th)` observation lies in the class `4 - 6`.
`:.` The median class is `4 - 6`.
Now,
`:. L = `lower boundary point of median class `=4`
`:. n = `Total frequency `=50`
`:. cf = `Cumulative frequency of the class preceding the median class `=21`
`:. f = `Frequency of the median class `=13`
`:. c = `class length of median class `=2`
Median `M = L + (n/2 - cf)/f * c`
`=4 + (25 - 21)/13 * 2`
`=4 + (4)/13 * 2`
`=4 + 0.6154`
`=4.6154`
5. Calculate Median 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)` | `cf` `(7)` |
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` | ----- |
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(88/2)^(th)` observation
= value of `44^(th)` observation
From the column of cumulative frequency `cf`, we find that the `44^(th)` observation lies in the class `30 - 40`.
`:.` The median class is `30 - 40`.
Now,
`:. L = `lower boundary point of median class `=30`
`:. n = `Total frequency `=88`
`:. cf = `Cumulative frequency of the class preceding the median class `=40`
`:. f = `Frequency of the median class `=20`
`:. c = `class length of median class `=10`
Median `M = L + (n/2 - cf)/f * c`
`=30 + (44 - 40)/20 * 10`
`=30 + (4)/20 * 10`
`=30 + 2`
`=32`
6. Calculate Median 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)` | `cf` `(7)` |
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` | ----- |
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(480/2)^(th)` observation
= value of `240^(th)` observation
From the column of cumulative frequency `cf`, we find that the `240^(th)` observation lies in the class `25 - 30`.
`:.` The median class is `25 - 30`.
Now,
`:. L = `lower boundary point of median class `=25`
`:. n = `Total frequency `=480`
`:. cf = `Cumulative frequency of the class preceding the median class `=110`
`:. f = `Frequency of the median class `=170`
`:. c = `class length of median class `=5`
Median `M = L + (n/2 - cf)/f * c`
`=25 + (240 - 110)/170 * 5`
`=25 + (130)/170 * 5`
`=25 + 3.8235`
`=28.8235`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then