@Vadim's answer and @carlop's comment address the issue that there's very little you can formally say without further information. But it sounds like what is wanted are some ad hoc informal but concrete suggestions for distribution assumptions, so here are a few.
Normally your percentile should not depend on the number of test-takers, since, if you assume that scores arise independently from some fixed underlying probability distribution, then your percentile as a function of your score $S$ is best thought of as $\mathrm{Pr}[x. This is also the context of grading on a curve, the percentiles are defined independent of the number of scores to be assigned.
Idea 1: Assume the scores are uniform on an interval containing 100%. Then the mean $M$ is also the median, and your percentile is $(S-2M+100)/(200-2M)$ if your score is above the minimum $2M-100$ (and zero otherwise). This has the advantage that the calculation is very easy, but the disadvantage that it only works if M>50 and has no support below a minimum score.
Idea 2: Assume the scores are more-or-less normally distributed and that 100% is a specific kind of "extraordinary," say 3 standard deviations from the mean. I.e. the standard deviation is $(100-M)/3$. Again the calculation is easy but the choice of 3 std.dev. is entirely arbitrary.
Idea 3: Assume the scores are more-or-less normally distributed, the top score is 100, and that the test-takers' scores are uniform in the percentiles, i.e. the top score is at the $N/(N+1)$ percentile for $N$ takers. This looks more sophisticated because it uses the additional piece of information $N$, but as discussed above that's not necessarily appropriate. It also now requires the quantile function for the normal distribution to compute the standard deviation.
Here's a picture for the example with given mean 92%. 
Using Idea 1 we assume the scores are uniform from 84-100 (shown in blue) and 95 is at the 69th percentile.
Using Idea 2 we assume the std.dev. is 8/3 (shown in red) and 95 is at the 87th percentile.
Using Idea 3 with $N=14$ (shown in yellow) we assume 100 is at about the 93rd percentile, about 1.5 std.dev. from the mean, then 95 is at the 71st percentile. With $N=10^5$ (shown in green) we assume 100 is about 4.3 std.dev. from the mean, then 95 is at about the 95th percentile.
Finally, here's an approximate reconstruction of the shape of the distribution of scores from the 2006 SAT scores from the published percentiles (1600 scale). For what it's worth Idea 2 does a decent job. The percentile of the perfect score on the 1600 scale was 99.93 which would be about 3.2 std.dev. away for a normal distribution, on the 2400 scale was 99.98 which would correspond to about 3.5 std.dev. 