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Worked Examples · Example 7
Q.

Calculate the mean deviation from the median for the following distribution.

Class interval20–3030–4040–6060–8080–90
Frequency (f)5102096
Kerala DhseTextbookSubjectiveImportance★★★★★est
23% · 6/26 Questions
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Median =50= 50, ∑f∣d∣=665\sum f|d| = 665, so M.D.=66550=13.3M.D. = \dfrac{665}{50} = 13.3.

Step 1 — Median. Here n=50n = 50, so the median is the size of the n2=25th\dfrac{n}{2} = 25\text{th} value. Cumulative frequencies are 5, 15, 35, 44, 50, so the 25th value lies in class 40–60. With L=40L = 40, c.f.=15c.f. = 15, f=20f = 20 and i=20i = 20: Median=40+25−1520×20=40+10=50Median = 40 + \dfrac{25 - 15}{20} \times 20 = 40 + 10 = 50.

Step 2 — Absolute deviations of mid-points from the median (50), weighted by frequency.

C.I.fm.p.|d| = |m − 50|f|d|
20–3052525125

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