Since the standard deviation is not given, we assume the process is a Poisson process, with mean 12000 miles.
From Poisson distribution tables, we see that with a value of lambda(mean)=12000, the lower tail probability for x=12140 miles is 0.9000105, which means that the 90 percentile is 12140 miles.
The lower-tail probability of 9000 miles is 6.2*10^(-181) which is essentially zero.
NOTE: if the problem were modelled with a normal distribution, the probabilities would be completely different.
