Page 126 - DLIS401_METHODOLOGY_OF_RESEARCH_AND_STATISTICAL_TECHNIQUES

Unit 8: Measurement of Central Tendency

30–34 8 (fm) 19 = number of cases (fb) below the Notes

25–29 7 interval containing Median
(N=40)
In the above example there are a total of 40 cases. We have to find a point below and above
which lie 20 cases. There are 13 cases in top 3 class intervals and 19 cases in the bottom four
class intervals. The point segregating the values into two halves may be found in class interval
30–34 which has 8 cases in it. It is thus called the Median class. Assuming that these 8 frequencies
are evenly distributed within the class interval 30 - 34 (exact limits 29.5 to 34.3, we may find
the median point which has to be 1 case above 29.5 (or 7 cases below 34.5).
There are 8 cases covering a space of 5 units so one case would take 5/8 spaces. Hence the
Median would be 29.5 + (1 × 5)/8 = 29.5 + 0.625
= 30.13 (taking approximation upto two decimal points)
This type of calculation gives rise to the formula :
Median = L+ {(N/2 – fb)/fm} × i

Where, L = Lower limit of Median class
N = Total number of cases
fb = Cumulative Frequency below the Median class
fm = Frequency in Median class
i = Size of the class interval
Using this formula for the previous example you can see that

Median = 29.5 + { (20 – 19)/8} × 5
Now we will calculate Median using this formula
= 29.5 + 5/8
= 29.5 + 0.625

= 30.13
Special Case for Calculation of Median
There may arise a special situation where there are no cases within the interval containing the
median, Let us take an example.
Example

Find the Median for the following frequency distribution :
Class Interval 5-7 8-10 11-13 14-16 17-19 20-22 23-25 26-28
Frequency 1 7 9 0 6 7 2 2
C.I. f cf

26—28 2 34
23—25 2 32
20—22 7 30
17—19 6 23 17 Cases above
14—16 0 17 16.5 Void Class

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