Average waiting times in line A local store has an average waiting time of 3 minutes.
I need to find the percentage of customers that wait less than 2 minutes and more than 4 minutes. 
The store is running a promotion where the customer's purchases are free if they wait longer than a certain time. If the store doesn't want to give away products to more than 3% of the customers. What should be advertised then?
 A: I think it is a good bet that you are supposed to use an exponential distribution with mean 3 (rate 1/3) for the waiting time $X.$ Then it is an easy integration problem to find the answers to the first two questions, and not difficult to solve the final one. You can
find the pdf of this exponential distribution in a probability text or in the Wikipedia article on 'exponential distribution.' (An exponential distribution might be reasonably accurate unless
waiting lines for service are involved.)
Here are answers from R statistical software (in which the cdf is denoted pexp), against which you can
check your answers from calculus.
 pexp(2, rate=1/3)
 ## 0.4865829        # P(X < 2)
 1 - pexp(4, rate=1/3)
 ## 0.2635971        # P(X > 4)
 qexp(.97, rate=1/3)
 ## 10.51967         # P(X > 10.52) = 0.03

Here is a graph of the density function of $Exp(rate = 1/3)$ with vertical
dotted lines separating relevant areas.

Note: In case you are expected to use a uniform distribution, the
relevant choice would be $Unif(0, 6)$, and you can get probabilities
without calculus. But that would be an extremely unrealistic
distribution for waiting times.
