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Exponential Distribution Calculator

Calculate density, cumulative probabilities and percentiles of waiting times in an exponential distribution.

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Explanation

The exponential distribution models the waiting time until the next event when those events occur independently and at a constant rate \( \lambda \). It is the continuous counterpart of the geometric distribution and the direct complement of the Poisson distribution: if the number of events per unit of time follows a Poisson(\( \lambda \)), the time between consecutive events follows an Exponential(\( \lambda \)). Its mean is \( 1/\lambda \) and its median \( \ln(2)/\lambda \), always smaller than the mean due to the distribution's positive skew.

The memoryless property is the characteristic that defines it mathematically: \( P(X > s + t \mid X > s) = P(X > t) \). This means that a component that has already been running for \( s \) hours has the same probability of failing in the next \( t \) hours as a brand-new one. Use it when the process is genuinely memoryless: intervals between phone calls, time until the next accident on a stretch of road, lifetime of electronic components with a constant failure rate, or service times in queues.

Formula

\( f(x;\lambda)=\lambda e^{-\lambda x},\quad x\geq 0 \)

Parameters

  • λ (lambda): mean rate of occurrence per unit of time (or space).
  • x: waiting time until the event occurs.

Worked example

Situation: An electronic component fails at a mean rate of \(\lambda = 0.5\) failures per hour. The time until the first failure follows an exponential distribution with parameter \(\lambda = 0.5\), that is, with a mean lifetime of \(1/\lambda = 2\) hours.

Question 1: What is the probability that the component fails before 3 hours, \(P(X \leq 3)\)?

Solution: We apply the exponential CDF: \[ P(X \leq 3) = 1 - e^{-\lambda \cdot 3} = 1 - e^{-0.5 \times 3} = 1 - e^{-1.5} \approx 1 - 0.2231 \approx 0.7769 \] There is approximately a 77.69% probability that the component fails within the first 3 hours.

Question 2: What is the median lifetime, that is, the time \(t\) such that \(P(X \leq t) = 0.5\)?

Solution: We invert the CDF by setting it equal to 0.5: \[ 1 - e^{-0.5\,t} = 0.5 \implies e^{-0.5\,t} = 0.5 \implies t = \frac{\ln 2}{0.5} = 2\ln 2 \approx 1.386 \text{ h} \] The median lifetime is approximately 1.386 hours, lower than the mean of 2 hours, which reflects the right skew of the exponential distribution.

Interpretation: Half of the components fail before 1.386 hours, but the mean rises to 2 hours due to the distribution's long tail. In preventive maintenance, this indicates that planning replacements around the median, rather than the mean, guarantees covering the 50% earliest failures.

How to interpret the result

The calculator offers three types of output. The PDF, \( f(x) = \lambda e^{-\lambda x} \), gives the probability density at the point \( x \): it is maximal at \( x = 0 \) and decreases exponentially. A high PDF value at a point indicates that waiting times near that value are relatively frequent. In the chart, the curve starts at \( \lambda \) at the origin and falls toward zero; the green area represents the cumulative probability for the range you selected on the horizontal axis (time or duration).

The CDF, \( P(X \leq x) = 1 - e^{-\lambda x} \), gives the probability that the waiting time is less than or equal to \( x \). For example, if \( \lambda = 0.2 \) events per minute (equivalent to a mean time of 5 minutes), then \( P(X \leq 3) = 1 - e^{-0.6} \approx 0.451 \): there is a 45.1% probability that the waiting time does not exceed 3 minutes. The right tail \( P(X > x) = e^{-\lambda x} \) is the survival probability, widely used in reliability analysis and queueing theory.

The quantile result inverts the CDF: given a probability level \( p \), it returns the time \( x = -\ln(1-p)/\lambda \) such that \( P(X \leq x) = p \). The 90th percentile indicates the time exceeded by only 10% of events; the 50th percentile (median) is \( \ln(2)/\lambda \approx 0.693/\lambda \), always smaller than the mean \( 1/\lambda \), which reflects the distribution's skew.

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Frequently asked questions

  • What does the exponential distribution model? The waiting time until the next event when events occur at a constant rate λ: time to the next call, failure of a non-ageing component, arrival of the next customer.
  • What is the memoryless property? It means P(X > s + t | X > s) = P(X > t): the time already elapsed does not change the probability of waiting longer. It is the only continuous distribution with this property.
  • How does it relate to the Poisson and the gamma? If events follow a Poisson process with rate λ, the times between events are exponential, and the sum of k independent exponential times follows a gamma (Erlang) with shape k.

Reference: Exponential distribution — Wikipedia