Exponential Distribution Calculator

Compute P(X ≤ x), P(X > x) and P(a < X ≤ b) from either the rate λ or the mean 1/λ — with a shaded PDF diagram and full step-by-step working.

The exponential distribution explained, step by step

The exponential distribution models waiting time between events in a Poisson process — same rate λ as the Poisson Distribution Calculator, different question ("how long between events?" instead of "how many in an interval?"). Each card below covers one piece the calculator uses.

1. PDF and CDF — the shape and the cumulative area

The exponential PDF starts at height λ at x = 0 and decays exponentially. The CDF integrates that curve to give the probability of waiting at most x units of time — it climbs quickly at first and asymptotes to 1.

f(x) = λ · e−λx  ·  P(X ≤ x) = 1 − e−λx
λrate — events per unit time·xwaiting time, x ≥ 0
Example
Given
λ = 0.1/min
x = 5
Substitute
1 − e^(−0.5)
Answer
P ≈ 0.3935

2. Survival and interval probabilities

The survival function is the complement of the CDF — the chance you wait longer than x. Any interval probability is the difference of two survivals (or two CDFs), which is why the shaded area between a and b uses e^(−λa) − e^(−λb).

P(X > x) = e−λx  ·  P(a < X ≤ b) = e−λa − e−λb
Example
Given
λ = 0.1
a = 5
b = 15
Substitute
e^(−0.5) − e^(−1.5)
Answer
P ≈ 0.3834

3. Mean = SD = 1/λ (signature property)

Both the mean and the standard deviation equal 1/λ, and the variance is 1/λ². The median is smaller — ln(2)/λ — because the distribution is heavily right-skewed and the long tail pulls the mean above the median.

μ = 1/λ   σ²
= 1/λ²   σ
= 1/λ   median
= ln 2 / λ
Example
Given
λ = 0.1/min
Substitute
μ = 1/0.1
Answer
μ = σ
= 10 min

4. Memoryless — no aging in the distribution

Exponential is the only continuous distribution with the memoryless property: given you've already waited s units, the chance of waiting at least t more is the same as if you'd just started. This is why exponential fits Poisson-process waits (calls, decays) but not wear-out lifetimes — for those use a Weibull.

P(X > s + t | X > s) = P(X > t)
Poisson-process gaps → memoryless fits
wear-out lifetimes → use Weibull
discrete analogue → Geometric
Example
Given
λ = 0.2/yr
waited 3 yr
Substitute
P(X > 2) = e^(−0.4)
Answer
≈ 0.6703

Features of this calculator

  • Enter either the rate λ or the mean 1/λ — the other value is derived automatically so you don't have to convert by hand.
  • Supports three probability queries: P(X ≤ x), P(X > x), and P(a < X ≤ b) between two values.
  • Reports mean, variance and standard deviation with a reminder that mean = SD = 1/λ.
  • Shaded PDF curve shows exactly which region of the exponential distribution is being computed.
  • Show/hide step-by-step working writes out F(x) = 1 − e^(−λx) with your numbers substituted in.

Frequently asked questions

+How do I estimate λ from data?

The maximum-likelihood estimate of λ is 1 divided by the sample mean of the observed waiting times. So if the average time between calls in your data is 7 minutes, an estimate of λ is 1/7 ≈ 0.143 per minute.

+Why isn't the mean the same as the median?

The exponential distribution is heavily right-skewed. The median is ln(2)/λ ≈ 0.693/λ, always noticeably less than the mean 1/λ. Half of the wait times fall below the median, but the long tail on the right pulls the mean up.

+Is exponential the same as exponential decay in physics?

Very closely related. Radioactive decay is a canonical Poisson process, and the waiting time until any single atom decays is exponentially distributed. The macroscopic decay law N(t) = N₀·e^(−λt) is the survival function of a huge number of independent exponential lifetimes.

+How is exponential related to the geometric distribution?

The geometric distribution is the discrete analogue — waiting time (in whole trials) for the first success in a sequence of Bernoulli trials. Exponential is its continuous counterpart, and it inherits the memoryless property from geometric.

+What if my data has an obviously increasing failure rate?

Then exponential is the wrong model. Look at the Weibull distribution, which generalises exponential with a shape parameter that lets the hazard rate increase (wear-out), decrease (infant mortality) or stay constant (exponential is Weibull with shape = 1).

+Can I use this for the time between website visits?

As a first approximation during a stable traffic window, yes. Beware daily and weekly cycles — visits between 3am and 4am aren't drawn from the same distribution as visits between 2pm and 3pm, so fit exponential inside a homogeneous window rather than across a full day.

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