The exponential distribution
The waiting time until the next event, when events happen at an average rate λ. The mean wait is 1/λ, and the distribution forgets how long you have already waited. Move λ, then ask a question: the area it counts is shaded.
Parameters
Ask a question
Compare with
P(X > 10) = 0.1353
P(X > 10) = 1 − F(10) = 1 − 0.8647 = 0.1353
Summary
| Measure | Exp(λ = 0.2) |
|---|---|
| mean | 5.0000 |
| variance | 25.0000 |
| standard deviation | 5.0000 |
| skewness | 2.0000 |
The same in code
- Python (scipy.stats)
expon.sf(10, scale=1/0.2)- Excel
=1-EXPON.DIST(10, 0.2, TRUE)
Formulas
Try this
Mean and median
Open the preset for the median wait.
Look for: The median wait is 3.4657 minutes, well below the mean of 5: the long right tail pulls the mean up.
It forgets
With λ = 0.2, note P(X > 10), then ask for P(X > 20).
Look for: P(X > 10) = 0.1353 and P(X > 20) = 0.0183 = 0.1353 × 0.1353. Having waited 10 minutes, the chance of waiting 10 more is the same as at the start.
One shape
Move λ anywhere and watch the skewness.
Look for: It stays at 2. The rate only stretches the time axis; it never changes the shape.
Not a bell
Turn on the normal with the same mean and σ.
Look for: The normal puts about 16 % of its probability below 0, a negative waiting time, which is impossible. Waiting times need a skewed model.