# 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.