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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

λ = 1

λ (rate: events per unit of time)

Ask a question

x = 1

x (threshold)

Compare with

P(X ≤ 1) = 0.6321

P(X ≤ 1) = F(1) = 0.6321

X is a measurement, so a single value has probability 0 and the boundary makes no difference: P(X < x) = P(X ≤ x) = F(x).

Summary

Measure Exp(λ = 1)
mean 1.0000
variance 1.0000
standard deviation 1.0000
skewness 2.0000

The same in code

Python (scipy.stats)
expon.cdf(1, scale=1/1)
Excel
=EXPON.DIST(1, 1, TRUE)
scipy's expon takes the mean wait 1/λ as its scale; Excel's EXPON.DIST takes the rate λ.

Formulas

Try this

0 of 4 done
  1. 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.

  2. 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.

  3. 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.

  4. 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.

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