We found that if \(\lambda\) is the parameter for the Poisson process, then the Poisson distribution had mean \(\lambda T\) or if one presumed that T=1, then for that distribution, the mean is just \(\lambda\text{.}\) We will find out below that the mean for the exponential distribution will be \(\frac{1}{\lambda}\text{.}\) Therefore, we will eventually present this formula using the exponental mean \(\mu = \frac{1}{\lambda}\) rather than using \(\lambda\) in the actual formula.