Exponential Distribution Calculator

StatisticsLast updated: August 22, 2026

The exponential distribution is a continuous distribution that models the waiting time between events that occur independently and at a constant rate. The time between two calls arriving at a call center, the time left until a machine fails, the gap between requests hitting a web server, and the time until a radioactive nucleus decays are all described by it. It is directly linked to the Poisson process: if the number of events is Poisson distributed, the time between events is exponential.

This calculator computes P(X ≤ x), P(X > x), and the probability between two bounds from either λ (the event rate per unit of time) or the mean waiting time (μ = 1/λ), and reports the mean, variance, standard deviation, median, and selected quantiles alongside them. It also demonstrates memorylessness — the most distinctive property of the exponential distribution — with concrete numbers, and draws the probability area on a shaded density curve. It is one of the most frequently used calculations in reliability engineering, queueing theory, survival analysis, and operations research courses.

The two are reciprocals: 0.25 events per hour and a mean of 4 hours describe the same distribution.
Must be greater than zero. E.g. an average of 0.25 failures per hour.
The time or amount whose probability you are asking about. It cannot be negative.
If filled in, the interval probability P(x < X ≤ x₂) is computed as well.

Exponential Distribution Formulas

Density:             f(x) = λ·e^(−λx),  x ≥ 0
Cumulative:          F(x) = P(X ≤ x) = 1 − e^(−λx)
Survival:            P(X > x) = e^(−λx)
Interval:            P(x₁ < X ≤ x₂) = e^(−λx₁) − e^(−λx₂)
Mean:                μ = 1 / λ
Variance:            σ² = 1 / λ²
Standard deviation:  σ = 1 / λ = μ
Median:              m = ln2 / λ ≈ 0.6931 / λ
Quantile (inv. CDF): xq = −ln(1 − q) / λ
Memorylessness:      P(X > s + t | X > s) = P(X > t)

λ is the RATE parameter, not the scale: the mean is 1/λ. The exponential distribution is the only continuous distribution with a constant hazard rate; h(x) = f(x)/[1−F(x)] = λ for every x.

How to Calculate

  1. Choose the form in which you know the parameter: the event rate per unit of time (λ) or the mean waiting time (μ = 1/λ).
  2. Enter the value of the parameter you selected; because the two are reciprocals, the result is the same whichever you enter.
  3. Write the time or amount whose probability you are interested in into the x field (it cannot be negative).
  4. If you also want the probability between two bounds, fill in the upper bound (x₂); leave it empty if you do not.
  5. Read P(X ≤ x), P(X > x), the mean, the variance, the median, and the quantiles together in the results; the shaded area shows the probability that was computed.
  6. Check the memorylessness line: if the conditional probability equals P(X > x), the model has been set up correctly.

Worked Examples

Time between machine failures (λ = 0.25/hour)

Assuming an average of 0.25 failures per hour (one failure every 4 hours on average), the probability that the next failure occurs within 3 hours is 0.5276 (52.76%), and the probability of exceeding 3 hours is 0.4724. The probability that the failure occurs between 3 and 6 hours is 0.2492. The mean is 4 hours while the median is 2.77 hours; because the distribution is right-skewed, the half-life is shorter than the mean.

P(X ≤ 3): 0.5276 · P(X > 3): 0.4724 · P(3 < X ≤ 6): 0.2492

Entering the mean: an 8-minute call interval

If an average of 8 minutes passes between calls, λ = 0.125. The probability that the next call arrives within 5 minutes is 0.4647 (46.47%), and the probability that it takes longer than 5 minutes is 0.5353. The variance is 64 and the standard deviation is 8 minutes — in the exponential distribution the standard deviation always equals the mean.

P(X ≤ 5): 0.4647 · P(X > 5): 0.5353 · Density f(5): 0.066908

Bulb lifetime: the probability of exceeding the mean

A bulb with a mean lifetime of 1000 hours has only a 0.3679 probability of lasting more than 1000 hours (e⁻¹); that is, roughly 63.21% of the bulbs fail before the mean. The probability that the lifetime falls between 1000 and 2000 hours is 0.2325, and the median lifetime is 693.15 hours.

P(X ≤ 1,000): 0.6321 · P(X > 1,000): 0.3679 · P(1,000 < X ≤ 2,000): 0.2325

Frequently Asked Questions

What does the memoryless property of the exponential distribution mean?
Memorylessness is the identity P(X > s + t | X > s) = P(X > t): no matter how long the system has already been running, the probability of it lasting another t units does not change. If a bulb has burned for 500 hours, the distribution of its remaining life is the same as that of a brand-new bulb. This property is found only in the exponential (continuous) and geometric (discrete) distributions, and it is unrealistic wherever wear is involved.
Should I enter λ or the mean?
Both define the same distribution; enter whichever one you know. λ is the number of events per unit of time (0.25 failures per hour), while the mean is the expected time between events (4 hours). The relation between them is μ = 1/λ. The most common mistake is typing the mean into the λ field; that reverses the probabilities completely.
What is the relationship between the exponential and the Poisson distribution?
They are two faces of the same process. In a Poisson process the NUMBER of events per unit of time follows a Poisson(λ) distribution, while the TIME between two consecutive events follows an Exp(λ) distribution. If an average of 3 calls arrive per hour (Poisson, λ = 3), for example, the time between calls is exponential with a mean of 1/3 hour = 20 minutes.
How do I get the same result in Excel, SPSS, and R?
In Excel, `=EXPON.DIST(x, lambda, TRUE)` gives P(X ≤ x), while the FALSE argument returns the density. In R, use `pexp(x, rate = lambda)` and `qexp(q, rate = lambda)` — R's `rate` parameter is the same λ as here. In Python you write `scipy.stats.expon(scale = 1/lambda)`; because SciPy takes the SCALE parameter, you must enter 1/λ rather than λ.
Why is the median smaller than the mean?
The exponential distribution is markedly right-skewed (its skewness is always 2): most observations are small and a few are very large, and those large values pull the mean upward. Because the median is ln2/λ ≈ 0.693·μ, it corresponds to roughly 69% of the mean. This is why the median is often the more representative measure of center when summarizing waiting time data.
When is the exponential distribution not appropriate?
It is not appropriate when the event rate changes over time. The failure rate rises for wearing parts, and under an 'infant mortality' effect it starts high and then falls; the Weibull distribution is preferred in both cases. Durations that are the sum of several stages (the completion of k events in a row, say) follow a gamma/Erlang distribution. If the standard deviation of the data differs markedly from its mean, the exponential assumption is doubtful.
Why can the 'density' value in the results be greater than 1?
In continuous distributions f(x) is a density, not a probability; probability is obtained only from the AREA under the curve. If λ is large (λ = 5, say), the density near x = 0 can reach 5, and that is not an error. In a continuous distribution the probability of a single point is exactly zero, which is why we ask for P(X ≤ x) or an interval probability instead of P(X = x).