How to predict top 1% grades in an exam with only pass mark and percentage who passed I am trying to figure out what the top 1% of marks are for a set of past exams, where I only have the passing mark of 58% and the percentage of people who passed. 45%.
I want to use the normal distribution to estimate the grade achieved by the top 1% of students who passed. So assuming the student's marks are normally distributed and 45% of people scored 58/100 or more.
Is there a way to do this?
 A: The information is not sufficient. You only know that the mark of the 55-th percentile is 58/100. But you do not know anything about the spread. Below you see the distribution of different normal distributions that have the same 58 score for the 55-th percentile (the score where 45% do better than that), but different scores for the 99-th percentile (the score of the best 1%).
The image below shows the cumulative distribution. It gives on the x-axis the score, and on the y-axis the percentage that has this score or lower.

Sidenote: even if you knew two points, the  an estimate based on the assumption that the distribution is a normal distribution can be very wrong. Normal distributions might be good models for an average, but distributions like scores on a test involve different processes that generate the distribution. E.g. you can have students that studied and students that did not study, splitting the distribution into two groups and making the distribution bimodal.
A: Of course, @SextusEmpiricus (+1) and @ChristianHenning are quite correct that this is
a silly, impossible problem---at least without making possibly unwarranted assumptions. But
assuming test scores to be roughly normally distributed is standard fare
in elementary probability problems.
Granted that it is reasonable to assume/guess that scores are normally distributed, you might also assume/guess that the top possible z-score (corresponding to 100%) is $3.$ [The Empirical Rule might suggest $3.]$ Then, together with the given 45th percentile, you can get two equations in two unknowns $\mu$ and $\sigma.$  Roughly, say $\mu = 59.7, \sigma = 13.44$ are their solution.
Then in R, you'd have:
pnorm(c(58,91,100), 59.7, 13.44)
[1] 0.4496728 0.9900670 0.9986435

Maybe it's not too much of a reach to say that the 99th percentile is about $91.$
You may want to make some refinements to the details, but I guess something like this might be what was expected by way of solution.
Note: Clearly my choice of $3,$ to which some commenters object with surprising vehemence, is
somewhat arbitrary, and I have clearly labeled it is a guess. But it's not completely arbitrary if you want the answer to make sense: for example, it wouldn't have been reasonable or productive to choose $2$ or $4.$
