Full disclosure: I'm a professor of electrical engineering and computer science at a reputable university, so I teach statistics as part of my living in the guise of performance measurement and evaluation. I get very good teaching ratings, so I'm probably good at it. Maybe the students just like me; it is not because I am an easy A.
You define the median correctly, but the median is not necessarily the mid-point of the range. I'll give you two hypothetical examples to illustrate the difference.
In the first example, imagine there are a total of 9 contracts in the Wyndham universe. One is the highest: $13/K. One is the lowest: $4.5/K. The others are roughly evenly distributed, so the sorted list is $4.5, $6, $7, $8, $8.75, $10, $11, $12, $13. In this population the median is $8.75, because there are four contracts that cost less ($4.5, $6, $7, $8) and four contracts that cost more ($10, $11, $12, $13). $8.75 also happens to be the midpoint of the range [($13-$4.5)/2]+$4.5. In this example, the median and the midpoint are identical.
In the second example, imagine there are a total of 9 contracts in the Wyndham universe. One is the highest: $13/K. One is the lowest: $4.5/K. The others are not evenly distributed, but instead occupy the bottom half of the range: $4.5, $5, $5.5, $6, $6.5, $7, $7.5, $8, $13. Now, the median is $6.5, because there are four contracts that cost less ($4.5, $5, $5.5, $6) and four that cost more ($7, 7.5, $8, $13). But the midpoint of the range is still $8.75, because the range is still $4.5 to $13. In this example, the median and the midpoint differ.
I believe that the universe of Wyndham deeds looks more like our second example than our first example, and the
MF history seems to back this up based on how I am reading it. So, while CWA's $/K cost is below the midpoint of the Wyndham $/K range, I suspect it is above the median of Wyndham deeds ordered by $/K.