Expert: Scott A Wilson - 1/30/2009

Question
QUESTION: There is wine in this question:

When it comes to classifying wine, a 50 to 100 point scale is used major magazine A; the 20 point scale by major magazine B.


- «A»'s scale can be resumed as follow:

96-100 - extraordinary
90-95 - outstanding
80-89 - good to very good
70-79 - average
60-69 - below average
50-59 - unacceptable. The worst possible wine will get 50 points; nothing below 70 is considered «approved».


- And «B»'s scale is something like:

0-9 - unacceptable («reproved»)
10-20 - average... good... very good... outstanding... extraordinary


In order to convert a 20-point based grade into a 50-100 point based one, then, one may get tempted to multiply it by 2,5 and add 50 to the result.

which will return something like...


0 to 20 ------- 50 to 100       |
0       ------- 50          |
...          |
5       ------- 62-63          |  reproved
...          |
9     ------- 72-73          |
9.5    ------- 74          |


10    ------- 75
10.5    ------- 76
11    ------- 77-78
11.5    ------- 79
12    ------- 80
12.5    ------- 81
13    ------- 82-83
13.5    ------- 84
14    ------- 85
14.5    ------- 86
15    ------- 87-88
15.5    ------- 89
16    ------- 90
16.5    ------- 91
17    ------- 92-93
17.5    ------- 94
18    ------- 95
18.5    ------- 96
19    ------- 97-98
19.5    ------- 99
20    ------- 100


However, this sort of translaction fails miserably...

Example: a 72-point wine (average, thus approved) gets a 9 (then, reproved) at the same time.

So, my first question is «what's wrong with this?»

And second...

Note, as an example, that a wine rated 89 is «very good». However, 15,5 is not «very good», but, at best... «fairly good».

Is there any «correct» way to convert the 100 point scale into the 20 point scale, minimizing this «distortions»? Where can I find more information about this?


ANSWER: It looks like the result is correct for the data given.
If I read this right, one scale goes from 0 to 20 and
the other scale goes from 50 to 100.
So to get from the first one (x) to the second one (y), y=50+2.5x is
the correct equation to do this.

A rating of x=9 turns into y = 50 + 2.5(6) = 65,
but it really gets a 72.

This would say the conversion is not linear.
I might get a sample of what a 1, 5, 10, 15, and 20 come out to be.
I would then be able to use approximation with a quadratic or  exponential and see which one fit better.

If you need this done, find a selection of several ratings between 1 and 20 that are also rated 50 to 100.  Send it to me and I could do a curve fit and send the results back to you.

If you wanted me to, tell me what curve is being used to approximate the function and some of the variables in the function, and I could tell you how to do it.

So hopefully you'll figure out if that really is a good wine you're looking at ...

---------- FOLLOW-UP ----------

QUESTION: thank you! but now here is a problem, perhaps THE problem...

x<70 @ the 0-100 scale, irrelevant;
x=70 @ "    " "  "   , y=10 @ the 20 pts scale;

x=98(?... let's say so)-100 @ "    " "  "   , y= 20 @ the 20 pts scale...

but what if «I» happen to consider uncertain all the values between [x=70,y=10] "or almost" and [x=98-100,y=20] "or almost"? will therea be a way to «generate» them or (at least) usable intervals?

Answer
x<70 @ the 0-100 scale, irrelevant;
x=70 @ the 0-100 scale, y=10 @ the 20 pts scale;

x=98-100 @ the 0-100 scale, y= 20 @ the 20 pts scale...

but what if I happen to consider uncertain all the values between [x=70,y=10] "or almost" and [x=98-100,y=20] "or almost"? will therea be a way to «generate» them or (at least) usable intervals?

Let's call the first scale I and the second scale II.
Scale I was between 50 and 100, scale II was between 1 and 20.

On I, is x<70, then it must be at least 50.  Let y = (x-50)/2.
This will set up y somewhere below 10.

When x=70, y=10 would be a good place to go,
and that's what that equation does.

Now there are 30 points in scale I between 70 and 100, and 10 points in scale II between 10 and 20.  Take 10 + (x-70)/3 as the y value if
x > 70.

When the whole thing is lumped together, we get
x≤70: y = Round[(x-50)/2]        or
x≥70: y = Round[10 + (x-70)/3].

If x = 50, y would be 0.
If x = 70, y would be 10 in either function.
If x = 100, y would be 20.

An even better approach would be to find a downward facing parabola.
Just a momeent while I use Excel to do that.

Aha!  I'm done now - that didn't take long.

Here is the data I generated
50   0
51   1
52   1
53   2
54   2
55   3
56   3
57   4
58   4
59   5
60   5
61   6
62   6
63   7
64   7
65   8
66   8
67   9
68   9
69   10
70   10
71   10
72   11
73   11
74   12
75   12
76   12
77   13
78   13
79   14
80   14
81   14
82   15
83   15
84   15
85   16
86   16
87   16
88   17
89   17
90   17
91   18
92   18
93   18
94   18
95   19
96   19
97   19
98   20
99   20
100   20

Here is the function that I used to do it:
y = Round(((270-x)x-1100)/300), or
y = Round((-x² + 270x - 1100)/300).