Page 151 - DMGT404 RESEARCH_METHODOLOGY

Unit 8: Descriptive Statistics

Solution: Notes
The median of a distribution is that value of the variate which divides the distribution into two
equal parts. In case of a grouped frequency distribution, this implies that the ordinate drawn at
the median divides the area under the histogram into two equal parts. Writing the given data in
a tabular form, we have:
Class Intervals (1) Frequency (f) (2) ‘Less than’ type c.f. (3) Frequency Density (4)
0 - 10 5 5 0.5
10 - 20 12 17 1.2
20 - 30 14 31 1.4
30 - 40 18 49 1.8
40 - 50 13 62 1.3
50 - 60 8 70 0.8

 Frequency of the class f 
  Note : requency density in a class  Width of the class  h  
F
For the location of median, we make a histogram with heights of different rectangles equal to
frequency density of the corresponding class. Such a histogram is shown below:
Figure 8.1: Histogram
2.0
Frequency Density 1.0
1.5

0.5

0
0 10 20 30M 40 50 60
d
Class Intervals

Since the ordinate at median divides the total area under the histogram into two equal parts,
therefore we have to find a point (like Md as shown in the Figure) on X-axis such that an ordinate
(AMd) drawn at it divides the total area under the histogram into two equal parts.
We may note here that area under each rectangle is equal to the frequency of the corresponding
class.
f
Since area = Length × Breadth = Frequency density × Width of class =  h  . f
h
Thus, the total area under the histogram is equal to total frequency N. In the given example
N
N = 70, therefore  35. We note that area of first three rectangles is 5 + 12 + 14 = 31 and the area
2
of first four rectangles is 5 + 12 + 14 + 18 = 49. Thus, median lies in the fourth class interval which
is also termed as median class. Let the point, in median class, at which median lies be denoted by
Md. The position of this point should be such that the ordinate AMd (in the above histogram)
divides the area of median rectangle so that there are only 35 – 31 = 4 observations to its left.
From the histogram, we can also say that the position of Md should be such that
M  30 4
d  ...(1)

40 30 18

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