Acertain car battery brand can be driven an average of 12,000 miles before its first failure.
a. what is the 90th percentile of the battery brand’s lifetime in units of miles?
b. a car owner with this battery has driven 4,000 miles and is planning a 5,000 mile road trip. what is the probability that the owner will be able to complete the trip without having to replace the battery?

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.