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Exponential Waiting-Time Probabilities & Moments

Calculate exponential survival and cumulative probabilities, mean, median, variance and standard deviation from a constant event rate.

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Enter the known values in the units shown. Results update as you type.
Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
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At rate 2 events per hour, mean waiting time and standard deviation are 0.5 hours, variance is 0.25 hours squared, and median is ln(2)/2 hours.

P(T > t) = exp(−λt); P(T ≤ t) = 1 − exp(−λt); mean = SD = 1/λ

Enter these known values and leave the other values blank.

Use one consistent time unit: rate is per unit of time, durations are in that unit and variance is in its square. Probabilities are fractions from zero to one.
Rate is finite and positive; elapsed time is nonnegative. The model assumes a constant hazard or a homogeneous Poisson arrival process.
The distribution is continuous, so a particular exact waiting time has probability zero. This reports a tail and cumulative probability, not a point probability.
Rounded extreme probabilities can display zero or one. A finite rate has positive survival at finite time; survival zero or cumulative probability one cannot determine a finite inverse time.

They are reciprocals. Two events per hour gives mean waiting time half an hour.

No. A time-varying rate requires a different model or an integrated hazard.

Bookify permits elapsed time zero and requires positive mean, median, variance and standard deviation for a finite positive exponential rate.