What's the best approach for results of a running race? I am a student in a good statistics program, but I'm not always the best at picking the tools/process to apply to a problem. To be clear, this is NOT homework, I am asking for a project that I have in mind for myself.
Let's say there is a runner in 5K races. The runner has data on their placing in the event in relation to the population of runners in the race. So they know they finished in X place out of N runners. The data would look like:
Placing   Field of Runners
10          120
2           135
23          131   
560         810
1           10

Now the runner wants to answer - "I am running in a race with 118 people today, what is my likely placing in the race?". The runner would be interested in either a placing or a percentage placing in the field, like the top 5%, the top 10%, etc. Like, "what is my likelihood of placing in the top 10% of the field today?".
At first I thought of doing something with a normal distribution and a z-score. But I'm not sure if that would make sense because of the disparity in the size of the fields; some fields may have 10 runners, other times it will have thousands. 
Then I was thinking about normalizing the data with a percentage placing - like top 12%, top 80%, etc - but then I'm a little hazy on how to go on from there. 
Or maybe it's a linear regression issue???
Can anyone please help me with this?
 A: You are looking for the distribution of your place in the race. It's normal, see this note.
So, the simplest estimator is to get the percentiles from your data set for each race and average them. That would be your estimate of your place in the next race. You can weight the ranks by $\sqrt{n_i}$, where $n_i$ is the number of runners in races, this will wight the big races heavier. 
Here's MATLAB example for both simple and weighted means, and your place in next race with 100 runners:
a=[10          120
2           135
23          131   
560         810
1           10]

% unweighted mean
m = mean(a(:,1)./a(:,2))

%wighted mean

mw = mean((a(:,1)./a(:,2) .* sqrt(a(:,2))) / sum (sqrt(a(:,2))))

% place in 100 runners race

round(mw*100,0)

Output:
m =

   0.213015738384695


mw =

   0.070342921102228


ans =

     7

This is assuming that you got to popular 5K races with significant number of runners, hundreds. In small races it's hard to argue that the runners would be similar race to race. In big races such as Kormen's pink run, the runners body doesn't change much. 
