Page 127 - DLIS401_METHODOLOGY_OF_RESEARCH_AND_STATISTICAL_TECHNIQUES

Methodology of Research and Statistical Techniques

Notes 11—13 9 17 interval and 17

8—10 7 8 Cases below 13.5
5—7 1 1
N = 34
If you examine the above table for computing median you come across N/2 = 34/2 = 17
against two class intervals showing cumulative frequencies. If one calculates median mechanically
he/she may arrive at a wrong answer. So it will not be so simple to apply the formula and
get the resullts. Also, if you calculate cumulative frequency from above, another class interval
(17 — 19) may also show 17 as cumulative frequency. We will solve this question in two
alternative ways (i) conceptually and (ii) empirically.
Solution 1 : On counting number of cases from top and bottom sides we find that there are
17 cases above 16.5 and 17 cases below 13.5. We are coming across two points (instead of one)
below and above which lie exactly fifty percent cases as the class interval 14-16 is void (without
any frequency). So extending the assumption to this class, (that frequencies are evenly distributed
within each class interval). We may add half of this void class interval to either side and find
median.
So Median = 13.5 + 3/2 = 13.5 + 1.5 = 15 (Answer)

Or Median = 16.5 – 3/2 = 16.5 – 1.5 = 15 (Answer)
Solution 2
C.I. f cf Modified C.I. f cf
26–28 2 34 26–28 2 34
23–25 2 32 23–25 2 32

20–22 7 30 20–22 7 30
17–19 6 23 15 to 19.5 6 23
14–16 0 17
11–13 9 17 10.5 to 15 9 (fm) 17
8–10 7 8 8–10 7 8 (fb)

5–7 1 1 5–7 1 1
N = 34
Alternatively, we modify the class interval where N/2 may fall. The class interval with zero
frequency which is affecting calculations is adjusted towards the adjoining class intervals on
either side and the size of the modified class interval is used while applying the formula. Half
of the Class interval 14-16 has been adjusted towards the class intervals 11-13 and 17-19
respectively. The modified class intervals are mentioned in terms of exact limits (modified size
of class interval being 3 + 1.5 = 4.5).
Median = L + {(N/2 – fb)/fm} × i
L = 10.5, fb = 8
N/2 = 34/2 = 17, fm = 9

Median = 10.5 + {(17 – 8)/9} × 4.5
= 10.5 + 4.5 = 15

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