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Lets assume I’m grading following a normal distribution but instead of having a standard grading scale from A to F I have a grading scale from A to J with no plus or minus. Whats the percentage for each grade knowing that J% = A% , I% = B% etc.. Thanks in advance :)

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  • $\begingroup$ About average, I would say $\endgroup$ – wolfies Dec 28 '17 at 17:51
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This depends on how far out into the tails of the distribution you want to go. For example if you use a z-score of 3 or above for A then 0.15% of the students would receive an A grade (and 0.15% would get a J), if you use a z-score of 2 (and -2) then about 2.5% would receive A's and the same J's. Choose some other Z cutoff for different values in the tails.

Once you have decided on the tail percentages/z-scores, then you can divide the remaining center portion into 8 equal segments (for B through I), for the 2.5% extremes you would use the z-score cut-offs of -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, and 2 and find the percentages between each pair.

But note that grading on the normal curve (or any preset curve) is like going to the tailor and buying a suit, finding the sleeves of the jacket are a little to short, so arranging for surgery to have your arms shortened to fit the suit jacket.

The proper use of the normal and other probability distributions is to find the appropriate distribution that fits the data. Choosing the distribution first, then forcing the data to fit it is completely backwards. Make/find the suit/distribution fit the person/data, not the other way around.

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