# Applying an uncertainty to a prediction

I am estimating/predicting the numbers of students who will pass on an exam at a school this year. My method is very simple: Each student give me a guess of their own grade one month before the test. This way, the estimated mean of the passing percentage turns out to be 95 %.

Now, I would like to give a measure of uncertainty (like P10, P50, P90) in addition to this value. Last year the test was done at several schools in my area. It turns out that the percentages of students passing the test seem to be Weibull distributed with a mean of 92 % and a standard deviation of 3 %.

How wrong would it be to apply this standard deviation - and the distribution - to my estimate of 95 %? Is it completely, utterly wrong, or could this be okay to some pragmatic extent in the absence of other alternatives?

For an estimate of the mean on my school this year I would trust my own prediction (95 %) more than the one from several schools last year (92 %) because there are some parameters like size and quality of lectures that would vary. Could the standard deviation still be of some use? I guess it would not if we are strict, but I want to be pragmatic as well.

Any response is highly appreciated.

• I'm not quite sure how a percentage could follow a Weibull distribution (typically defined as a distribution on $(0, \infty)$ and not on $[0,1]$ or $[0,100]$). In any case, would you not think that students might be overoptimistic? Unless you have data on how students' guesses match up with final results, it may be hard to rely on these guesses and one might think that what happened at this school or other schools in previous years might be a better predictor. Jan 14, 2016 at 11:56
• Thanks for your comment. I use a flipped/reversed Weibull distribution so that it is on (-∞,100) instead of (0,∞). I could agree on your comment regarding overoptimism. However, my main point was whether or not to use the standard deviation of 3 % to my estimate of 95 % - given that I actually am supposed to find a distribution including uncertainty for this value. What would you think about that? Jan 14, 2016 at 12:37