How to Calculate the Exponential Distribution

The distribution of the wait until the next event when events arrive at random. With a mean of μ, the chance of one arriving within a time t is 1 − e^(−t/μ). It models time to failure and the gaps between arriving customers.

This is the distribution of the wait until the next event when events arrive at random. It fits time to failure, the gaps between arriving customers and the time until a radioactive atom decays.

P(X \le t) = 1 - e^{-t/\mu}

$\mu$ is the mean gap, and its reciprocal $\lambda = 1 \div \mu$ is the rate. The chance of waiting longer than $t$ is $e^{-t/\mu}$ .

Example

Take events arriving on average every 10 minutes. The chance of one within 5 minutes is $1 - e^{-0.5} = 0.393$ , or 39.3%, and the chance of waiting longer is 60.7%. The median is $10 \times \ln 2 = 6.93$ minutes, well short of the mean of 10.

The mean sits above the median

Half of all events arrive within 6.93 minutes, yet the mean is 10. Occasionally the wait is very long indeed, and those cases drag the mean up. For waiting times, the mean alone misleads.

It has no memory

Having already waited 10 minutes, the chance of an event in the next 5 is the same 39.3% as it was at the start. Time already spent waiting adds nothing to the forecast, which is why nothing is ever due.

Notes

The distribution applies when events arrive independently and at random. It does not describe something on a timetable, such as a train.