Exponential Distribution Calculator

The exponential distribution is the continuous probability distribution that models the time between events in a Poisson process: a process where events occur continuously and independently at a constant average rate. You encounter it in reliability engineering when modelling how long a component works before failing, in queuing theory when modelling customer arrival times, in telecommunications when modelling call durations, and in nuclear physics when modelling radioactive decay intervals. The key parameter is lambda, the rate at which events occur. If a server handles on average 0.5 customers per minute, lambda = 0.5 and the mean wait time is 1/0.5 = 2 minutes. The probability that the waiting time X is less than some value t is given by the cumulative distribution function: P(X less than or equal to t) = 1 minus e to the power negative lambda t. The survival function P(X greater than t) = e to the power negative lambda t gives the probability of waiting longer than t. A key feature of the exponential distribution is the memoryless property: the probability of waiting an additional time t is the same regardless of how long you have already waited. This calculator takes the rate parameter lambda and a time value t, and returns the CDF, survival probability, probability density at t, mean, median, variance and standard deviation.

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Rate parameter (lambda)

Time value (t)

0.7769

P(X ≤ 3): probability within time t

P(X > 3)0.2231

PDF at t0.1116

Mean (1/λ)2.0000

Median (ln2/λ)1.3863

Variance4.0000

Std dev2.0000

lambda must be positive. t must be non-negative. Mean = 1/lambda, Median = ln(2)/lambda, Variance = 1/lambda squared.

How it works

For an exponential distribution with rate parameter λ and time t ≥ 0:
PDF: f(t) = λ e^-λt
CDF: P(X ≤ t) = 1 - e^-λt
Survival: P(X > t) = e^-λt
Mean = 1/λ
Median = ln(2)/λ ≈ 0.6931/λ
Variance = 1/λ²
Std dev = 1/λ
Note that for the exponential distribution, mean = standard deviation = 1/λ. The coefficient of variation is always exactly 1.

Worked example

With lambda = 0.5 and t = 3: CDF = 1 - e^{-0.5 × 3} = 1 - e^-1.5 = 1 - 0.2231 = 0.7769. P(X > 3) = e^-1.5 = 0.2231. PDF at t = 0.5 × e^-1.5 = 0.1116. Mean = 1/0.5 = 2.0000. Median = ln(2)/0.5 = 1.3863. Variance = 1/0.25 = 4.0000. Std dev = 1/0.5 = 2.0000.

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Frequently asked questions

What is the exponential distribution?

The exponential distribution models the time between events in a Poisson process, where events occur continuously and independently at a constant average rate lambda. It is used to model waiting times, such as the time until the next customer arrives, the time until a component fails, or the time until a radioactive decay occurs. The probability density function is f(x) = lambda times e to the power minus lambda x, for x greater than or equal to zero.

What is the mean of the exponential distribution?

The mean of the exponential distribution is 1 divided by lambda. If events occur at rate lambda = 0.5 per unit time (meaning on average 0.5 events per unit), the mean time between events is 1/0.5 = 2 time units. The median is ln(2) divided by lambda, which is approximately 0.693 divided by lambda.

What is the memoryless property of the exponential distribution?

The memoryless property means that the probability of waiting an additional time t does not depend on how long you have already waited. Formally, P(X greater than s + t | X greater than s) = P(X greater than t) for all s, t greater than zero. The exponential distribution is the only continuous distribution with this property.