problem-2.17

problem-2.17  First grab the data and check the units (minutes). The top 10% is actually the 0.10 quantile in this case, as shorter times are better.
> times = nym.2002$time
> range(times)                  # looks like minutes
[1] 147.3 566.8
> sum(times < 3*60)/length(times) * 100 # 2.6% beat 3 hours
[1] 2.6
> quantile(times,c(.10, .25))   # 3:28 to 3:53
  10%   25%
208.7 233.8
> quantile(times,c(.90))        # 5:31
  90%
331.8

It is doubtful that the data is symmetric. It is much easier to be relatively slow in a marathon, as it requires little talent and little training-just doggedness.