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Generated on: July 20, 2026 at 09:33 PM

Exponential Distribution Calculator

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Rate0.5
Value3
Lower bound1
Upper bound4

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f(t)
t f(t)

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Mathematics

Exponential Distribution Calculator

Calculate PDF f(x), CDF P(X≤x), tail probability P(X>x), interval probability P(a≤X≤b), mean, variance, and median for an exponential distribution from the rate λ.

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f(t)
t f(t)

Steps

\begin{aligned} f(x) &= \lambda e^{-\lambda x} \\ &= 0.5 \cdot e^{-0.5 \cdot 3} \\ &= ? \end{aligned}
\begin{aligned} P(X \le x) &= 1 - e^{-\lambda x} \\ &= 1 - e^{-0.5 \cdot 3} \\ &= ? \end{aligned}
\begin{aligned} P(X > x) &= e^{-\lambda x} \\ &= e^{-0.5 \cdot 3} \\ &= ? \end{aligned}
\begin{aligned} P(a \le X \le b) &= e^{-\lambda a} - e^{-\lambda b} \\ &= e^{-0.5 \cdot 1} - e^{-0.5 \cdot 4} \\ &= ? \end{aligned}
\begin{aligned} \mu &= \dfrac{1}{\lambda} \\ &= \dfrac{1}{0.5} \\ &= ? \end{aligned}
\begin{aligned} \sigma^2 &= \dfrac{1}{\lambda^2} \\ &= \dfrac{1}{0.5^2} \\ &= ? \end{aligned}
\begin{aligned} \text{median} &= \dfrac{\ln 2}{\lambda} \\ &= \dfrac{\ln 2}{0.5} \\ &= ? \end{aligned}

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Exponential Distribution

The exponential distribution is a continuous probability distribution that models the waiting time between independent events occurring at a constant average rate. It is the continuous partner of the Poisson distribution: where Poisson counts how many events happen in a fixed interval, the exponential distribution describes how long you wait between them. It appears throughout reliability engineering, queueing theory, and survival analysis.

Formula

Given a rate parameter λ > 0 and a value x ≥ 0, the probability density function (PDF) is:

f(x)=λe−λx,x≥0f(x) = \lambda e^{-\lambda x}, \quad x \geq 0

The cumulative distribution function (CDF) gives the probability of a waiting time of at most x:

P(X≤x)=1−e−λxP(X \leq x) = 1 - e^{-\lambda x}

The survival function (tail probability) is its complement — the probability of still waiting after x:

P(X>x)=e−λxP(X > x) = e^{-\lambda x}

The probability that the waiting time falls in an interval [a, b] is the difference of the survival function at the endpoints:

P(a≤X≤b)=e−λa−e−λbP(a \leq X \leq b) = e^{-\lambda a} - e^{-\lambda b}

The mean, variance, and median are compact functions of the rate:

μ=1λ,σ2=1λ2,median=ln⁡2λ\mu = \frac{1}{\lambda}, \qquad \sigma^2 = \frac{1}{\lambda^2}, \qquad \text{median} = \frac{\ln 2}{\lambda}

Worked example

A support line receives calls at an average rate of λ = 0.5 calls per minute, so the mean gap between calls is 1/0.5 = 2 minutes. What is the probability that the next call arrives within 3 minutes?

Using the CDF with x = 3:

P(X≤3)=1−e−0.5×3=1−e−1.5≈0.7769P(X \leq 3) = 1 - e^{-0.5 \times 3} = 1 - e^{-1.5} \approx 0.7769

There is a 77.7 % chance that the next call arrives within 3 minutes. The chance of waiting longer than 3 minutes is the complement:

P(X>3)=e−1.5≈0.2231P(X > 3) = e^{-1.5} \approx 0.2231

The probability that the wait lands between 1 and 4 minutes is:

P(1≤X≤4)=e−0.5−e−2≈0.6065−0.1353=0.4712P(1 \leq X \leq 4) = e^{-0.5} - e^{-2} \approx 0.6065 - 0.1353 = 0.4712

The density at x = 3 is:

f(3)=0.5 e−1.5≈0.1116f(3) = 0.5 \, e^{-1.5} \approx 0.1116

Formula derivation

The CDF follows directly from integrating the density from 0 to x:

P(X≤x)=∫0xλe−λt dt=[−e−λt]0x=1−e−λxP(X \leq x) = \int_0^x \lambda e^{-\lambda t}\, dt = \left[ -e^{-\lambda t} \right]_0^x = 1 - e^{-\lambda x}

The mean is found by integrating x against the density:

μ=∫0∞x λe−λx dx=1λ\mu = \int_0^\infty x\,\lambda e^{-\lambda x}\, dx = \frac{1}{\lambda}

The median solves P(X ≤ m) = 0.5:

1−e−λm=12  ⟹  e−λm=12  ⟹  m=ln⁡2λ1 - e^{-\lambda m} = \tfrac{1}{2} \;\Longrightarrow\; e^{-\lambda m} = \tfrac{1}{2} \;\Longrightarrow\; m = \frac{\ln 2}{\lambda}

Because ln 2 ≈ 0.693 < 1, the median is always smaller than the mean 1/λ, reflecting the distribution's right skew.

The memoryless property

The exponential distribution is the only continuous distribution that is memoryless:

P(X>s+t∣X>s)=P(X>t)P(X > s + t \mid X > s) = P(X > t)

Having already waited s units of time gives no information about the remaining wait. This is why the exponential model suits components with a constant failure rate — parts that do not wear out or "age" in a statistical sense — and why it underpins the Markov property of continuous-time processes.

Relationship to other distributions

  • Poisson: If events follow a Poisson process with rate λ, the count in a fixed interval is Poisson(λ) while the time between successive events is Exponential(λ). Counting uses Poisson; timing the gaps uses the exponential.
  • Geometric: The geometric distribution models the number of discrete trials until the first success. The exponential distribution is its continuous-time analogue, describing the waiting time until the first event.
  • Gamma: The sum of k independent Exponential(λ) waiting times follows a Gamma(k, λ) distribution — the time until the k-th event in a Poisson process.
  • Weibull: The exponential distribution is the special case of the Weibull distribution with shape parameter 1, corresponding to a constant hazard rate.

Practical applications

  • Reliability engineering: modelling the time to failure of components with a constant hazard rate, the basis of the "useful life" phase of the bathtub curve.
  • Queueing theory: representing inter-arrival and service times in M/M/1 and related queue models.
  • Survival analysis: providing a baseline hazard model for time-to-event data in medicine and actuarial work.
  • Physics: describing the time between radioactive decays, where each nucleus decays independently at a fixed rate.
  • Telecommunications: modelling the idle time between packet arrivals in network traffic analysis.

Frequently Asked Questions (FAQ)

When is the exponential distribution appropriate?

Use the exponential distribution to model the waiting time between independent events that occur at a constant average rate. Typical examples include the time until the next customer arrives, the lifetime of a component with a constant failure rate, and the interval between radioactive decays. It is the continuous counterpart of the geometric distribution and the waiting-time partner of the Poisson distribution.

What is the memoryless property?

The exponential distribution is the only continuous distribution that is memoryless: P(X > s + t | X > s) = P(X > t). In words, having already waited s units of time gives no information about how much longer you must wait — the remaining wait has the same distribution as a fresh start. This is why it models systems without wear-out, such as a component whose failure rate does not increase with age.

What is the difference between the rate λ and the mean?

The rate λ counts how many events occur per unit of time, while the mean 1/λ measures the average time between events. They are reciprocals: if events arrive at λ = 2 per hour, the mean gap is 1/2 = 0.5 hours. Some references parameterise the exponential distribution by the mean (often called the scale) instead of the rate; this calculator uses the rate λ.

How is the exponential distribution related to the Poisson distribution?

They describe the same process from two viewpoints. If events occur according to a Poisson process with rate λ, then the number of events in a fixed interval follows a Poisson distribution, and the time between consecutive events follows an exponential distribution with the same rate λ. Counting events uses Poisson; timing the gaps between them uses the exponential.

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