About the Exponential Distribution Calculator
This exponential distribution calculator works out probabilities for waiting times and lifetimes that follow an exponential model. Give it either the rate λ (events per unit of time) or the mean time between events, and it returns the cumulative probability P(X ≤ x), the survival probability P(X > x), the probability of a value falling between two limits, the density at x, any percentile, and the mean, median, variance and standard deviation.
The exponential distribution describes the time until the next event in a Poisson process: the gap between customers arriving, calls reaching a help desk, radioactive decays or failures of components with a constant failure rate. It is a core topic in probability and queueing courses and a quick tool for reliability engineers estimating the chance a part survives a given time.
The model assumes a constant rate, which gives it the memoryless property: the chance of waiting another 5 minutes is the same no matter how long you have already waited. If failures become more likely with age, a Weibull or gamma model is more appropriate.
With the default inputs, the p(x ≤ x) is 0.77687. Change any value above to recalculate instantly.
How to use the exponential distribution calculator
- 1Choose whether you know the rate λ or the mean time between events.
- 2Enter the rate or the mean (in the same time unit as x).
- 3Enter x for “at most” and “more than” probabilities, and a and b for a range.
- 4Pick a percentile to find the time below which that share of waits fall.
- 5Read the results and use the chart to see how quickly probability accumulates.
Formula and method
An exponential random variable X with rate λ has density λe^(−λx) for x ≥ 0 and cumulative distribution F(x) = 1 − e^(−λx). The survival function P(X > x) = e^(−λx), and the probability between a and b is e^(−λa) − e^(−λb). Inverting the CDF gives the p-th percentile, −ln(1 − p)/λ; the median (p = 0.5) is ln 2/λ ≈ 0.693/λ.
The mean and standard deviation are both 1/λ and the variance is 1/λ². If events occur as a Poisson process at λ per unit time, the waiting time between them is exponential with that rate. Because P(X > s + t | X > s) = P(X > t), the distribution is memoryless — the only continuous distribution with this property.
- λ
- Rate parameter (events per unit time)
- 1/λ
- Mean time between events
- x
- Waiting time or lifetime
- p
- Percentile as a fraction (0.9 for the 90th)
Worked examples
Rate 0.5 per minute, wait of 3 minutes
With λ = 0.5 events per minute (mean 2 minutes), P(X ≤ 3) = 1 − e^(−1.5) = 0.7769. The chance of a wait between 2 and 5 minutes is e^(−1) − e^(−2.5) = 0.2858, the median wait is 1.39 minutes and 90% of waits finish within 4.61 minutes.
Support calls average 8 minutes apart
A mean of 8 minutes gives λ = 1/8 = 0.125 per minute. The chance of more than 10 minutes without a call is e^(−1.25) = 0.2865, and half of all gaps are shorter than 8 × ln 2 = 5.55 minutes.
Component with a 1,000-hour mean life
With a constant failure rate of 0.001 per hour, the part survives 500 hours with probability e^(−0.5) = 0.6065 and fails within 2,000 hours with probability 1 − e^(−2) = 0.8647. Ten percent of parts fail by 105.4 hours (the B10 life).
Frequently asked questions
What is the exponential distribution used for?+
It models the time until the next event when events happen independently at a constant average rate: arrivals at a queue, time between calls or website hits, radioactive decay and the lifetimes of electronic parts with a constant failure rate. It is the waiting-time partner of the Poisson distribution.
How do you find probabilities with an exponential distribution?+
Use P(X ≤ x) = 1 − e^(−λx) for “at most x”, P(X > x) = e^(−λx) for “more than x”, and e^(−λa) − e^(−λb) for a value between a and b. Make sure λ and x use the same time unit.
What is the relationship between λ and the mean?+
The mean is 1/λ. If customers arrive at λ = 4 per hour, the average gap is 1/4 hour or 15 minutes. The standard deviation also equals 1/λ, and the median is ln 2/λ, about 69% of the mean.
What does memoryless mean?+
The probability of waiting at least t more units does not depend on how long you have already waited: P(X > s + t | X > s) = P(X > t). A bus with exponential gaps is no more likely to arrive soon just because you have waited a long time.
What is the difference between the exponential and Poisson distributions?+
Both describe the same process. The Poisson distribution counts how many events occur in a fixed interval (a discrete count), while the exponential distribution measures the continuous time between consecutive events. A Poisson rate of λ per hour means exponential gaps with mean 1/λ hours.