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I have this grouped data.

Income = below 5000, 5001 - 10000, 10001 - 20000, 20001 - 40000, 40001 - 100000, 100001 - 200000, above 200000

F(Income) = 59, 101, 153, 207, 223, 44, 6

How do I find the Median Income, especially the midpoint? I have problem with dealing with Below 5000 and Above 200000

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    $\begingroup$ Well, what have you tried? How many of your data points are $<10000.5$, say? How many are $<20000.5$? In this way you should at least find out which bucket contains the median. Now, beyond that you can't say much without making some assumption about how income within a bucket is distributed. $\endgroup$ – lulu Dec 19 '17 at 16:39
  • $\begingroup$ Thanks @lulu I have got it sorted out. Bruce actually gave me what I needed. $\endgroup$ – Instructius Dec 21 '17 at 16:10
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The median lies somewhere in the 4th interval (the one with frequency 207). There are various interpolation procedures for making assumptions and guessing where in the interval. If your book discusses one of these, I suggest you try to figure out how to use it. Often, it is good enough to say which interval is the 'median interval'. [The open-ended first and last intervals cause difficulty if you are guessing at the sample mean, but not for the median.]

In the table below cf stands for cumulative frequency and crf for cumulative relative frequency.

 i   f  cf    crf      
 1  59  59 0.0744
 2 101 160 0.2018
 3 153 313 0.3947
 4 207 520 0.6557  <-- This interval: First crf > .5
 5 223 743 0.9369
 6  44 787 0.9924
 7   6 793 1.0000
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  • $\begingroup$ A commonly used interpolation formula is used in this Q & A and other pages mentioned there. Maybe also see "Related" pages linked in the right margin here. $\endgroup$ – BruceET Dec 19 '17 at 23:09
  • $\begingroup$ Thanks so very much @BruceET, this is very much helpful. $\endgroup$ – Instructius Dec 20 '17 at 8:18

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