# Queueing Theory: Probability that a job takes less than the average processing time?

Is there a formula for this? I thought that since it's exponentially distributed, it'd be basically the probability of it being less than the average of the distribution (the mu parameter). But I can't seem to find a formula for that either.

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From your question it seems you think the time for a job $T$ is exponentially distributed with density $f(t) = \frac{1}{\mu} e^{-t/\mu}$ so with $E[T]=\mu$.
In that case $Pr(T \le t) = 1-e^{-t/\mu}$ and so $Pr(T \le \mu) = 1-e^{-1} \approx 0.632$.
If you prefer to work with a rate parameter $\lambda$ then the density is $f(t) = \lambda e^{-\lambda t}$ and $E[T]=\frac{1}{\lambda}$, so $Pr(T \le t) = 1-e^{-\lambda t}$ and so $Pr(T \le \frac{1}{\lambda}) = 1-e^{-1}$ again.