First, sorry for, sending on so many groups, by I tried to send it
only to groups connected with math, statistics and algorithms.
Second, sorry for my mathematical language, which is not perfect :)
I hope, that I stated my problem quite clearly. I'm not sure, but the
sought formula may be something like standardization formula, but with
1 instead of 0 for average, 0 instead of -infinity for minimum, and
the value of a (an element of A) taken into consideration. The last I
can cope with by multiplying by a, but the first to changes are
difficult.
Anyway, that is my problem:
I need to have a formula to solve a problem.
I've a set A = {a: a in R}. I want to assign to every number being an
element of the set A another number meeting these criteria:
1. For the lowest number in A f(min)=0
2. For the average number in A f(av)=1
3. For the lowest number in A f(max)=an finite positive number.
4. For every a and b ( f(a) < f(b) ) <=> ( a < b ).
5. Value of f(a) depends on:
a.) the value of a (p. 4)
b.) difference (imprecise term!) from a to the minimum of the A set,
or the "difficulty" (dif) of getting an a value in A. By the
"difficulty" I understand a value dependent on the number of values in
A lower or equal a and on the number of values in A lower of a. So for
sets X = {0,1,2,3} and Y = {1,2,2,3} difX(0) = difY(1) < difX(1) <
difY(2) [!] < difX(2) < difX(3) = difY(3). Difference of value 2 in
the Y set is lower because there is more 2s in Y than in X. So as if
the difference was a relative result in a sports rally, while X and Y
sets of results in these rallies.
I thought of a formula like this:
f(a) = a * ( a - min ) / ( average - min )
but it doesn't comply with point 5b
--
Ludwik
.
.
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