The exponential distribution describes the time between events that happen independently at a constant average rate – customers arriving at a counter, calls reaching a help desk, radioactive decays, or failures of a component that does not wear out. If events follow a Poisson process with rate λ per unit of time, the waiting time to the next event is exponential with the same λ.
Its key feature is memorylessness: the chance of waiting another t minutes does not depend on how long you have already waited. That is realistic for random arrivals but not for parts that age, where the Weibull distribution is the better model.
Enter the rate λ in events per unit of time (if you know the average waiting time instead, λ = 1/average). The calculator gives P(X ≤ x), P(X > x), probabilities for ranges, the time by which a given share of waits are over, and simulated waiting times. The mean waiting time is 1/λ but the median is only ln 2/λ ≈ 0.693/λ, because a few long waits pull the average up.
How to use the Exponential Distribution Calculator
- Rate λ (events per unit of time) — If you know the mean time between events, λ = 1 ÷ mean.
- What to calculate
- Time x (x₁ for a range)
- Upper value x₂
- Cumulative probability p
- How many random values
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Probabilities come from the closed-form CDF 1 − e^(−λx); quantiles from x = −ln(1 − p)/λ, which is also how the random sample is generated. Choose what to calculate: a cumulative probability, an upper tail, the probability between or outside two values, a quantile (inverse CDF) or a random sample generated by inverse-transform sampling. The chart shades exactly the area you asked for.
Worked example
A school cafeteria serves 15 clients per hour, so λ = 15/60 = 0.25 per minute and the mean gap between clients is 4 minutes. The probability that the next client arrives within 3 minutes is 1 − e^(−0.25 × 3) = 52.76%; the probability of a gap longer than 10 minutes is e^(−2.5) = 8.21%. The median gap is ln 2/0.25 = 2.77 minutes and 90% of gaps are shorter than −ln(0.1)/0.25 = 9.21 minutes.
Key properties
| Measure | Formula | For λ = 0.25 |
|---|---|---|
| Mean | 1/λ | 4 |
| Median | ln 2/λ | 2.7726 |
| Mode | 0 | 0 |
| Variance | 1/λ² | 16 |
| Standard deviation | 1/λ | 4 |
| Skewness | 2 | 2 |
| Excess kurtosis | 6 | 6 |
Formula
f(x) = λe^(−λx) · P(X ≤ x) = 1 − e^(−λx) · P(X > x) = e^(−λx) · mean = 1/λ · median = ln 2/λ · variance = 1/λ²
Worked example
Inputs — Rate λ (events per unit of time): 0.25; What to calculate: P(X ≤ x) – probability at or below x; Time x (x₁ for a range): 3.
Results — Result = P(X ≤ 3) = 0.527633 (52.7633%); Probability density f(x) = 0.118092; P(X ≤ x) – cumulative (CDF) = 52.7633 %; P(X > x) – survival = 47.2367 %; Mean (expected value) = 4; Median = 2.7726.
Key terms
- Rate (λ)
- The average number of events per unit of time; its reciprocal is the mean waiting time.
- Memoryless property
- The remaining waiting time does not depend on the time already waited.
- Survival function
- P(X > x) = e^(−λx), the chance the event has not yet happened by time x.
- Poisson process
- Events occurring independently at a constant average rate.
Frequently asked questions
What is the exponential distribution used for?
Waiting times between independent random events: arrivals, calls, decays, failures of components that do not age.
How do I get λ from an average?
λ = 1/mean. If clients arrive every 4 minutes on average (15 per hour), λ = 0.25 per minute.
What is the chance of waiting less than 3 minutes when λ = 0.25 per minute?
1 − e^(−0.75) = 52.76%.
What does memoryless mean?
P(X > s + t | X > s) = P(X > t): having waited s minutes does not change the distribution of the remaining wait.
Why is the median smaller than the mean?
The distribution is skewed right (skewness 2). The median is 0.693/λ while the mean is 1/λ.
How is it related to the Poisson distribution?
If the number of events per unit of time is Poisson(λ), the times between them are exponential with rate λ.
What is the half-life in terms of λ?
The median, ln 2/λ – the time by which half of the waits (or decays) are over.