<< The picture was supposed to show a rectangle of width 1 and height 2 whose bottom is centered at x=1, and to the right of it, a rectangle of width 1 and height 1 whose bottom is centered at x=2. The base of the first rectangle goes from x=1/2 to x=3/2, and the base of the second rectangle goes from x=3/2 to x=5/2. The total area under the histogram is (1)(2)+(1)(1) = 3. The area to the left of the line x=5/4 is (5/4-1/2)(2) = 3/2, which is half of the total area. So x=5/4 is the "median".
After getting an explanatory e-mail from John Robertson, I thought I posted a "now I get it" post to math-fun, but it apparently didn't get through. (Your original message was perfectly clear, and my attempt to reproduce your calculation neglected to take account of the +1/2 poisitioning of the bars, even though I implied that positioning myself.) But: For very small samples, the median doesn't generally convey much information. For a large sample, however, this "bar-graph" median can be unduly influenced by outlying data -- much as an average can. But the idea of the median, as I see it, is to characterize the middle of the sorted data without much influence on the size of data points not close to the middle. Or am I still misunderstanding something? --Dan