Exponential Distribution Calculator
Enter the rate λ or the mean of an exponential distribution and a value x, and get the density, the cumulative and survival probabilities, a range probability, the median and any percentile — with the working shown, a memorylessness check and a shaded curve. Free, no sign-up.
Adds P(x ≤ X ≤ this value) to the results.
Shows that the wait still to come does not depend on s.
Finds the x with P(X ≤ x) = p, such as 0.9 for the 90th percentile.
Results
λ = 1/mean = 1/4 = 0.25
| Probability | Meaning | Value |
|---|---|---|
| P(X ≤ 6) | at most 6 (also: less than 6) | 0.7769 |
| P(X > 6) | more than 6 (also: at least 6; the survival function) | 0.2231 |
| f(6) | Density at x: f(x) = λe^(−λx). A density, not a probability: it can exceed 1. | 0.055783 |
4
Mean (1/λ)
16
Variance (1/λ²)
4
Standard deviation (1/λ)
2.7726
Median (ln 2 / λ)
0.25
Hazard rate (λ)
Memorylessness
P(X > s + x | X > s) = P(X > 9) / P(X > 3) = 0.1054 / 0.4724 = 0.2231
P(X > x) = P(X > 6) = 0.2231. The two are equal: having waited 3 already changes nothing about the wait still to come.
Quantile
x = −ln(1 − p) / λ = −ln(1 − 0.9) / 0.25 = 9.2103. So P(X ≤ 9.2103) = 0.9.
Density curve
Exponential density with λ = 0.25. The shaded area under the curve is P(X ≤ x).
How to Use the Exponential Distribution Calculator
- 1. Say which parameter you know. Choose the rate λ (events per unit of time) or the mean 1/λ (units of time per event), then type it in. The calculator converts the other one and shows the conversion.
- 2. Enter x. The time, distance or other amount you are asking about. Add an upper end if you want the probability of a range.
- 3. Optional extras. Enter a time already waited, s, to see the memoryless property at work, or a probability p to find the x that cuts it off (p = 0.9 gives the 90th percentile).
- 4. Read the results. P(X ≤ x), P(X > x), the density f(x), the mean, variance, median and hazard rate, and a curve with your probability shaded.
What the Exponential Distribution Models
The exponential distribution describes the waiting time until the next event when events occur independently at a constant average rate: the gap between calls to a help line, the time until a radioactive atom decays, the distance between flaws along a cable. It is the continuous counterpart of the geometric distribution, and it is tied to the Poisson distribution: if the number of events in a period is Poisson with rate λ, the waits between events are exponential with the same λ.
| Quantity | Formula |
|---|---|
| Density | f(x) = λe−λx |
| Cumulative probability | P(X ≤ x) = 1 − e−λx |
| Survival function | P(X > x) = e−λx |
| Quantile for probability p | x = −ln(1 − p) / λ |
| Mean and standard deviation | 1/λ each |
| Variance | 1/λ² |
| Median | ln 2 / λ ≈ 0.6931 / λ |
Rate or mean? Check which one your source uses
Textbooks split on the parameter. Some write the rate λ; others write the mean (or scale) θ = β = 1/λ. Software splits the same way: R's dexp, pexp and qexp take the rate, and so does Excel's EXPON.DIST, while Python's scipy.stats.expon takes scale = 1/λ. Mixing them up is the commonest error with this distribution — a mean of 4 minutes is a rate of 0.25 per minute, not 4. Pick the matching option above and the calculator does the conversion.
Worked Example: Waiting Time Between Calls
Calls to a help line arrive on average every 4 minutes, so the mean is 4 and λ = 1/4 = 0.25 per minute. The chance that the next call takes at most 6 minutes is P(X ≤ 6) = 1 − e−1.5 = 0.7769, and the chance it takes longer is P(X > 6) = e−1.5 = 0.2231. The chance the wait falls between 2 and 6 minutes is e−0.5 − e−1.5 = 0.3834.
The median wait is ln 2 / 0.25 = 2.7726 minutes — shorter than the mean of 4, because the distribution is right-skewed. The 90th percentile is −ln(0.1) / 0.25 = 9.2103 minutes: nine calls in ten arrive within about 9.2 minutes of the last.
Memorylessness
Suppose 3 minutes have passed with no call. The chance of waiting at least 6 more is P(X > 9) / P(X > 3) = 0.1054 / 0.4724 = 0.2231 — exactly P(X > 6), the figure for a fresh start. The exponential is the only continuous distribution with this property, and it follows from the constant hazard rate: at every moment the chance of an event in the next instant is λ, however long you have waited.
When the exponential is the wrong model
Memorylessness is a strong assumption. A battery or a person ages, so a lifetime model with a rising hazard (a Weibull or gamma) fits better than an exponential; the exponential suits events that strike without wear, such as random arrivals.
Frequently Asked Questions
- How do I find λ from the mean?
- Take the reciprocal: λ = 1/mean. A mean wait of 4 minutes gives λ = 0.25 per minute, and a rate of 3 events per hour gives a mean of 1/3 hour, or 20 minutes. Keep the units of x consistent with the units of λ, or the probabilities will be wrong.
- What does the memoryless property mean?
- It means P(X > s + t | X > s) = P(X > t): the time you have already waited does not change the distribution of the time still to come. If a bus arrives at random with mean gap 10 minutes, after waiting 20 minutes your expected remaining wait is still 10 minutes.
- How are the exponential and Poisson distributions related?
- They describe the same process from two sides. The Poisson counts events in a fixed period; the exponential measures the time between events. If calls arrive at 3 per hour (Poisson, λ = 3), the gap between calls is exponential with mean 20 minutes, and the chance of waiting over 5 minutes is e−3 × 5/60 = 0.7788.
- How do I calculate P(X > x) for an exponential distribution?
- Use the survival function e−λx. With λ = 0.25 and x = 6 that is e−1.5 = 0.2231. There is no need to subtract from 1 — which is why the calculator computes the upper tail directly and keeps its precision far out in the tail.
- Why is the median smaller than the mean?
- The distribution is strongly right-skewed: many short waits and a few very long ones pull the mean up. The median is ln 2 / λ, about 69% of the mean 1/λ. Half of all waits are shorter than 0.69 of the average wait.
- Is the exponential distribution a good model for lifetimes?
- Only when the failure rate is constant, as for random shocks or electronic parts in their useful life. Wear-out, infant mortality and ageing all make the hazard rate change over time, and the exponential then misjudges both tails. Check the fit before relying on it.