Calamity Jane goes to the bank to make a withdrawal, and is equally likely to find 0 or 1 customers ahead of her. The service time of the customer ahead, if present, is exponentially distributed with parameter λ. What is the CDF of Jane’s waiting time?

Answer:[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]Step-by-step explanation:The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]We are interested on the cumulative distribution function (CDF) of Jane’s waiting time. Let T the random variable that represent the Jane’s waiting time, and t possible value for the random variable T.For [tex]t\geq 0[/tex] the CDF for the waiting time is given by:[tex]F_{T}(t) =P(T\leq t)[/tex]For this case we can use the total rule of probability and the conidtional probability. We have two options, find 0 or 1 customer ahead of Jane, and for each option we have one possibility in order to find the CDF, like this:[tex]F_{T}(t)=P(T\leq t|0 customers)P(0 customers)+ P(T\leq t|1 customers)P(1 customers)[/tex] On this case the probability that we need to wait if we have o customers ahead is [tex]P(T\leq t|0 customers)=1[/tex] since if we don't have a customer ahead, so on this case Jane will not wait. And assuming that the probability of find 0 or 1 customers ahead of Jane is equal we have [tex]P(0 customers)=P(1 customers)=\frac{1}{2}[/tex]. And replacing we have:[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}\int_{0}^{t} \lambda e^{-\lambda \tau} d\tau = \frac{1}{2}+\frac{1}{2}[-e^{-\lambda t}+e^{-\lambda 0}]= \frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]