No problem. I'm always glad to help out fellow NBA Live players.
As you've said, there are occasional times where the grade is completely off from the rating. So to find out what the expected rating for each grade level are, I have determine the standard deviation by punching in every single number into my graphing calcuator. Then the SD should be a good way to figure out the expected range of ratings (Average rating-SD to average rating+SD). If there are any math experts out there that knows a better way, can you please tell me how?
I'll update it again when I'm done.
Update:
The number of times each letter grade showed up in my experiment:
A+ : 0
A : 0
A- : 9
B+ : 4
B : 12
B- :16
C+ : 0
C : 25
C- : 19
D+ : 0
D : 21
D- : 21
E : 6
As asgsjb pointed out, there were no A+'s, A's, C+'s, nor D+'s in my experiment (although that doesn't necessarily mean that there will be none that shows up in the game).
The average will give you a preliminary guess at what the rating will be for each letter grade. I put a star next to some of the averages not as accurate as they have too little information. Here are the averages:
A- : 83.89
B+ : 73.25*
B : 72.25
B- : 66.81
C : 58.72
C- : 51.67
D : 46.05
D+ : 37
E : 17.33
I added and subtracted the standard deviation from the averages to get a more advanced range to predict where the rating will be. The standard deviation and the expected range are as follows:
A- : 3.66, 80 to 88
B+ : 3.44, 70 to 77
B : 5.54, 67 to 78
B- : 3.57, 63 to 70
C : 4.92, 54 to 64
C- : 4.63, 47 to 56
D : 5.03, 41 to 51
D- : 6.26, 31 to 43
E : 6.5, 11 to 24
A chart of the above information:
The standard deviation can also be used to calcuate how evenly spread out the values are (or not). Note that this formula is not mathematically proven. It's just a method I like to use to find the "spreadoutness" of the values.
First, I take find out the SD of the number of values in a given set (say, the A- set). This is really hard to explain so I will use an example. So we have the A- set of values. The SD for that is 3.66. There are 9 numbers in that set so I find out the SD for any 9 consecutive numbers (1, 2, 3, 4, 5, 6, 7, 8, 9) and it truns out to be 2.58. Since the SD for the A-'s is more than the SD for the 9 consecutive numbers, the values are farther from the average than any 9 given numbers with the same average.
So its like:
1, 2, 3, 4, 5, 6 has average of 3.5
SD of that is 1.707
1, 1, 1, 6, 6, 6 also has average of 3.5
But SD is 2.5
3, 3, 3, 4, 4, 4 again has average of 3.5
But its SD is 0.5
Sort of get what I mean now? Sorry if this is all flying over your head but its the best I can do.