Say I have a list of the letter grades of a class (meaning some number of As, Bs, Cs, Ds and Fs). Is there any way for me to take this discrete distribution and extrapolate it to the most likely continuous one? Especially given that the ranges for each letter grade usually vary (i.e. A might be from 90-100, a difference of 10% whereas F might be 0-50, a difference of 50%), I have no idea how I would even begin to go about this.

The final goal of this would be to be able to get an approximant for any percentile (i.e. if you are a nth percentile student, your grade will most likely be around x%). Is this even possible?

Side note: I'm sure there are multiple continuous distributions that will fit such a dataset, so I'm looking for the most likely/common one I suppose.

  • $\begingroup$ Without additional data, how would you determine which continuous distribution is the most likely one to have generated your grades? This seems impossible. $\endgroup$ Commented Aug 27, 2023 at 13:32
  • $\begingroup$ Well that's why I'm asking: I don't know if it's possible. I feel like there's some givens, i.e. it's a slim chance that all of the people who got an A got 100%, or that all of the people who got an F got 0%, etc. But I'm not sure what to do with that. $\endgroup$
    – Ghull
    Commented Aug 27, 2023 at 13:40
  • $\begingroup$ Yes there is a chance that this happened and you don't know how big or small it is. Thus I would argue that there is no solution. $\endgroup$ Commented Aug 27, 2023 at 14:56
  • $\begingroup$ @Ghull you seem to have answered your own question in comments. $\endgroup$
    – Glen_b
    Commented Aug 28, 2023 at 0:11

1 Answer 1


If I needed to do this and wouldn't have any more information, I'd probably replace every grade by its middle mark (e.g., 25 for "0-50", 95 for "90-100") in order to construct a numerical data set. Based on these you could estimate mean and variance of a normal distribution and use this.

Having done this, you could of course use normal distribution percentiles.

Note however that this is "quick and dirty". In fact there is usually no argument that the normal distribution is "true" or "more likely" than others. It's just that neither any other distribution can claim such a thing, and the normal is something of a standard choice in case we don't know anything better.

There is a certain possibility to check the normal against the data; if indeed a normal distribution were true, certain distribution patterns of A-F would be more expected than others, or rather, certain things could happen that would look distinctively non-normal (for example, to give an extreme example, if all grades were either A or F with nothing in between). The chi-squared test could be used to test normality, but it wouldn't suggest an alternative in case the normal is rejected, and there is no straightforward way to find an alternative. Also, if the data set is large, normality may be rejected even if it is still OK as approximation.

Furthermore, based on past numerical data, one could represent classes in ways different from the midpoint (for example, if percentages within the A-range are very skew, it may be better represented by 92 than by 95 depending on what available past data say; one could even introduce within-category variance, but this would make things more complicated).

Another thing is that the normal distribution is for data that can take any real value including negative ones or larger than 100. If most of your data are not concentrated in A or F this may not be much of a problem, because then the normal distribution probability for values smaller 0 or larger 100 will be very small anyway. Otherwise you could use a truncated normal distribution, but if results are very skewly distributed and/or concentrated at both ends of the scale, a normal isn't very good. One could consider Beta-distributions as alternative, although I'm not sure whether there may be issues with parameter estimation from very discrete data with categories of different range.

So this is an option, and it may be the one that will look most "standard" to many people, however keep in mind it cannot have any claim of "truth", "being likely", or "being optimal" in any sense. Whatever you do with it, interpret with caution.

  • $\begingroup$ This is what I was looking for. Thanks for the in-depth answer. $\endgroup$
    – Ghull
    Commented Aug 27, 2023 at 19:55
  • $\begingroup$ @Ghull From my own experience (and from numerous other sets of numerical grades I've seen), a normal distribution would hardly ever come close to being an adequate description of the numerical grades. Typically the results are skewed, multimodal, and clearly showing effects from at least one of the bounds and often with a number of other artifacts ... $\endgroup$
    – Glen_b
    Commented Aug 28, 2023 at 0:13
  • $\begingroup$ This is fitting a normal model; could be appropriate when grades are curved anyway. Otherwise, why not just interpolate? $\endgroup$
    – Ute
    Commented Aug 28, 2023 at 0:45
  • 1
    $\begingroup$ @Glen_b Interesting. I have often marked on a 0-100 scale and other than some mild skewness and maybe the odd outlier close to zero among otherwise solid marks, a normal approximation often looked rather good. Also keep in mind that the information the OP has is discrete at just six categories; many peculiarities cannot be properly modelled in this situation. (As I wrote, I agree that one shouldn't "believe" the normal, but coming up with anything more convincing is hard.) $\endgroup$ Commented Aug 28, 2023 at 11:14
  • $\begingroup$ @Ute If you can explain how, that would probably be worth your own answer. $\endgroup$ Commented Aug 28, 2023 at 11:15

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