[LMB] OT: Adventures in grading II
Mark A. Mandel
mamandel at Filker.Org
Sun, 1 Dec 2002 00:54:46 -0500 (EST)
On Mon, 25 Nov 2002, Jim Parish wrote:
#This semester, I'm teaching a general education course in statistics.
#On the most recent midterm, I asked the following question: "You are
#told that a certain distribution <has thus-and-such properties>. Can
#you tell which is most likely to be larger, the median or the mean? The
#median or the mode? The mode or the mean?" One student answered, "The
#median is probably larger than the mean. The mode is probably larger
#than the median. The mean is probably larger than the mode."
#My initial reaction was a visible start at the contradiction. (Though I
#had said nothing, another teacher who was in the room glanced up and
#said, "Yeah, I get some of those too.") Later, though, I thought about
#it some more and realized that the three statements the student had
#made were not, in fact, contradictory. So, a challenge for the (rest of
#the) listmind: why not? Brownie points for the first acceptable answer.
#(Hints: you don't have to know what the words "distribution", "mean",
#"median", or "mode" mean; if you have any acquaintance with psephology,
#you've seen this before in a different guise.)
To simplify in accordance with your hint, let's make this:
"Probably A > B.
Probably B > C.
Probably C > A."
Why is this set of statements not contradictory?
Suppose that A > B with a probability of 2/3, or about 67%: that is, given
as many cases as you want, A > B in about 2/3 of them. Since 2/3 is
significantly greater than 1/2, "probably" is an accurate verbal
description of this probability. And suppose similarly that B > C with a
probability of 2/3, and that the relationship of A to B is independent of
the relationship of B to C. (Forget about C vs. A for the moment.)
Oh, and let's say that on whatever scale we're using, whenever A > B or
B > C the difference is exactly 1.
What about that other 1/3 of the cases? Well, suppose, just suppose, that
whenever B > A the difference is exactly 3, and likewise whenever C > B.
Now consider the possible values of A - C, worked out in the table below.
In the row and column headers and each cell there are three lines, the
"relationship" (> or <), the "difference" (x = y +/- z), and the
"probability" of the cell. For the headers, the values of these lines come
from the definitions above. For the cells, the probabilities are the
product of the row and column probabilities, 'cause that comes from the
definition of "and" for two independent events. (Tsk, tsk, Jim, you said,
or implied, that we wouldn't have to understand probability theory.) The
differences are computed from the row and column differences, and the
relationships are read off the differences.
(I broke my own rule of tabs here. I hope nobody's tab stops are set < 7.)
rel A>B A<B
diff A=B+1 A=B-3
prob 2/3 1/3
B>C : A>C A<C
B=C+1: A=C+2 A=C-2
2/3 : 4/9 2/9
B<C : A<C A<C
B=C-3: A=C-2 A=C-6
1/3 : 2/9 1/9
Now look at our results! In 4/9 of the cases A > C, but in the remaining
5/9 C > A: a probability of 55%. Not as much as 67%, but still enough to
say "probably" without uttering a falsehood.
QED. I wanna brownie; you can keep yer points.
-- Mark A. Mandel