Exponential distribution calculator
The exponential distribution models the waiting time between events that arrive at a steady random rate.
1 − e^(−0.25 × 2) = 0.3935
Modelling waiting times
If events arrive at random at an average rate λ (buses on a busy route, customers at a counter, radioactive decays), the time until the next one is exponentially distributed. The curve starts at its highest point and decays: short waits are the most common, long waits are rarer but possible.
f(x) = λe^(−λx) P(X ≤ x) = 1 − e^(−λx) mean = 1/λ
Worked example
Customers arrive on average every 4 minutes (λ = 0.25 per minute). The chance of waiting less than 2 minutes for the next one is 1 − e^(−0.5) ≈ 0.3935. The chance of waiting more than 10 minutes is e^(−2.5) ≈ 0.0821.
Median below the mean
The median wait is ln 2 / λ ≈ 0.693 × the mean. The long right tail pulls the mean up, just as with any right-skewed data. The count of events per interval is Poisson with the same λ.
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Questions students ask
What is the exponential distribution formula?
The density is f(x) = λe^(−λx) for x ≥ 0, and the cumulative probability is P(X ≤ x) = 1 − e^(−λx). The mean is 1/λ.
How is the exponential related to the Poisson?
If the number of events per unit time is Poisson with rate λ, the time between consecutive events is exponential with the same rate λ.
What does memoryless mean here?
P(X > s + t | X > s) = P(X > t). A bulb whose lifetime is exponential is as good as new at every moment, which is why real parts that wear out need other models.
Rate or mean: which should I enter?
Either. A mean wait of 4 minutes is a rate of 0.25 per minute. The calculator converts between them.