On the distribution of exam scores and the use of t-tests A while back, I did some educational research where I compared exam scores between two groups.  I remember reading somewhere that educational exams such as these tend to follow a normal distribution.  Of course, normal distributions must necessarily be infinite and educational exam scores are not.  Is it mathematically correct to say that these scores follow a "truncated normal distribution" rather than a traditional normal distribution?  Does this truncation preclude one from being able to use a t-test to compare two groups?
 A: 
Is it mathematically correct to say that these scores follow a "truncated normal distribution" rather than a traditional normal distribution? 

The scores are not truncated normal either.
The variable is bounded, but I'm not sure truncation is really the right term for this situation, since it's not like the observations that would otherwise score above the highest mark are lost. If anything you might envisage it as something nearer winsorization (but perhaps, given the possibility of someone capable of scoring very highly on a much harder test nevertheless making one mistake, even that's an imperfect description).
Even if we had exactly the right model for what happens at the bounds, they're still not <whateverhappensattheendpoints>-ized normal either. (I'll say "truncated" for this, to indicate that I'm talking about the thing you're calling truncation.) 
They're discrete, for example. But they're also not the same as a normal that has been "truncated" and discretized. 

Does this truncation preclude one from being able to use a t-test to compare two groups?

Not necessarily. In many cases, the t-test will have characteristics that are not badly affected by the non-normality.
If the data look approximately normal, you're probably fine. The less normal they look, the larger the sample size you need before the test has close to the anticipated properties.
