2 3/4-(-1 1/2)+(-5/6)-(-3/7)-(+4 2/3)

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aaron2113

07.03.2021

Mathematics

$F_{T}(t)=\frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})$

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:

$P(X=x)=\lambda e^{-\lambda x}$

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 $t\geq 0$ the CDF for the waiting time is given by:

$F_{T}(t) =P(T\leq t)$

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:

$F_{T}(t)=P(T\leq t|0 customers)P(0 customers)+ P(T\leq t|1 customers)P(1 customers)$

On this case the probability that we need to wait if we have o customers ahead is $P(T\leq t|0 customers)=1$ 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 $P(0 customers)=P(1 customers)=\frac{1}{2}$ . And replacing we have:

$F_{T}(t)=\frac{1}{2} +\frac{1}{2}\int_{0}^{t} \lambda e^{-\lambda \tau} d\tau = \frac{1}{2}+\frac{1}{2}[-e^{-\lambda t}+e^{-\lambda 0}]= \frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})$

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