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Exponential Distribution Calculator

Computes the density, the cumulative probability P(X ≤ x), the survival probability P(X > x) and optionally P(x < X ≤ x2) for an exponential distribution given its rate λ or its mean 1/λ, plus mean, median, variance and standard deviation.

When to use

You model the time between independent random events at a constant rate (time to failure, time between arrivals, radioactive decay) and need the probability of waiting at most or more than a given time.

Do not use when: You need the probability of a number of events in an interval (use poisson-distribution), the trials are discrete (use geometric-distribution), or the failure rate changes over time (Weibull).

Formula

f(x) = λ e^(−λx); P(X ≤ x) = 1 − e^(−λx); P(X > x) = e^(−λx); mean = 1/λ; median = ln(2)/λ; variance = 1/λ²

The exponential distribution is memoryless: P(X > s + t | X > s) = P(X > t). It is the continuous counterpart of the geometric distribution and the inter-arrival time of a Poisson process with rate λ.

Inputs

ParameterTypeUnitRequiredDescription
ratenumbernoEvents per unit time. Give rate or mean (mean = 1/rate). Range: > 0
meannumbernoAverage waiting time, alternative to rate. Range: > 0
xnumberyesTime (or distance) at which to evaluate the distribution, in the same units as 1/rate. Range: ≥ 0
x2numbernoOptional upper value; when given, P(x < X ≤ x2) is also returned. Range: ≥ 0

Outputs

OutputTypeUnitDescription
ratenumberRate used.
meannumber1/λ.
probability_densitynumberλ·e^(−λx) (not a probability).
probability_at_mostnumber1 − e^(−λx): probability the event has occurred by x.
probability_more_thannumbere^(−λx): probability of waiting longer than x (survival function).
probability_betweennumbere^(−λx) − e^(−λx2) (only when x2 is given).
mediannumberln 2 / λ: half of the waiting times are shorter than this.
variancenumber1/λ².
std_devnumber1/λ (equal to the mean).

Example

λ = 0.5 per hour, x = 2 hours: {"rate":0.5,"x":2}{"rate":0.5,"mean":2,"probability_density":0.18394,"probability_at_most":0.632121,"probability_more_than":0.367879,"median":1.386294,"variance":4,"std_dev":2}

Mean 10 years, x = 5 years, x2 = 20: {"mean":10,"x":5,"x2":20}{"rate":0.1,"probability_density":0.060653,"probability_at_most":0.393469,"probability_more_than":0.606531,"probability_between":0.471196,"median":6.931472,"variance":100}

GET https://tttkmbb.com/api/v1/calculate/exponential-distribution?rate=0.5&x=2

Machine access

Sources

FAQ

Rate or mean?

Either: a mean time between failures of 10 years is a rate of 0.1 per year. If both are given they must agree (rate × mean = 1).

Why is the median less than the mean?

The distribution is right-skewed: many short waits and a few very long ones. The median is ln 2 ≈ 0.693 times the mean.

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