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.