Median of grouped data
Median of discrete frequency distribution
If
n is odd, then
M= value of
(n+12)th observation
If
n is even, then
M=Value of (n2)th observation+Value of (n2+1)th observation2
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+12)th observation
= value of
(262)th observation
= value of
13th observation
From the column of cumulative frequency
cf, we find that the
13th 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+12)th observation
= value of
(492)th observation
= value of
24.5th observation
From the column of cumulative frequency
cf, we find that the
24.5th 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
n2.
The class whose cumulative frequency is
≥n2 is called Median class
Median
M=L+n2-cff⋅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
=2Median
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
=2Median
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
=10Median
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
=5Median
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