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?

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 $4.$
