Find the median from the following data:
Class | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 | 26 - 30 | 31 - 35 | 36 - 40 | 41 - 45 |
Frequency | 7 | 10 | 16 | 32 | 24 | 16 | 11 | 5 | 2 |
To find median, Assume
Σfi = N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and Cf = cumulative frequency
Lets form a table and convert it into exclusive - type by adjusting from both ends of a class.
CLASS | FREQUENCY(fi) | Cf |
0.5 - 5.5 | 7 | 7 |
5.5 - 10.5 | 10 | 7 + 10 = 17 |
10.5 - 15.5 | 16 | 17 + 16 = 33 |
15.5 - 20.5 | 32 | 33 + 32 = 65 |
20.5 - 25.5 | 24 | 65 + 24 = 89 |
25.5 - 30.5 | 16 | 89 + 16 = 105 |
30.5 - 35.5 | 11 | 105 + 11 = 116 |
35.5 - 40.5 | 5 | 116 + 5 = 121 |
40.5 - 45.5 | 2 | 121 + 2 = 123 |
TOTAL | 123 |
So, N = 123
⇒ N/2 = 123/2 = 61.5
The cumulative frequency just greater than (N/2 = )61.5 is 65, so the corresponding median class is 15.5 - 20.5 and accordingly we get Cf = 33(cumulative frequency before the median class).
Now, since median class is 15.5 - 20.5.
∴ l = 15.5, h = 5, f = 32, N/2 = 61.5 and Cf = 33
Median is given by,
⇒
= 15.5 + 4.45
= 19.95
Thus, median is 19.95.