Question 3.12: A bus is supposed to arrive at a given bus stop at 10:00 a.m......
A bus is supposed to arrive at a given bus stop at 10:00 a.m., but the actual time of arrival is a r.v. X which is uniformly distributed over the 16-minute interval from 9:52 to 10:08. If a passenger arrives at the bus stop at exactly 9:50, what is the probability that the passenger will board the bus no later than 10 minutes from the time of his/her arrival?
The p.d.f. of X is~f(x)=1/16 for x ranging between 9:52 and 10:08, and 0 otherwise. The passenger will board the bus no later than 10 minutes from the time of his/her arrival at the bus stop if the bus arrives at the bus stop between 9:52 and 10:00 (as the passenger will necessarily have to wait for 2 minutes, between 9:50 and 9:52). The probability for the bus to arrive between 9:52 and 10:00 is 8/16 = 0.5. This is the required probability.
REMARK 4 It has been stated (see comments after relation (7)) that sometimes a handful of moments of a r.v. X completely determine the distribution of X. Actually, this has been the case in all seven distributions examined in this section. In the Binomial distribution, knowledge of the mean amounts to knowledge of p, and hence of f . The same is true in the Geometric distribution, as well as in the Poisson distribution. In the Hypergeometric distribution, knowledge of the first two moments, or equivalently, of the mean and variance of X (see expressions for the expectation and variance), determine mand n and hence the distribution itself. The same is true of the Gamma distribution, as well as the Normal and Uniform distributions.