SigmaCalc

Exponential Distribution Calculator

The distribution behind 'how long until the next event' — customer arrivals, equipment failures, radioactive decay.

Rate (λ, events per unit time)

Time (x)

P(X ≤ x)

0.632

F(x) = 1 − e^(−λx)

Density at x: 0.184

What you entered

Rate (λ, events per unit time): 0.5Time (x): 2

1
Probability density at x
0.5 × e^(−0.5 × 2)= 0.1839
2
P(X ≤ x) — cumulative probability
1 − e^(−0.5 × 2)= 0.6321

Result

P(X ≤ x): 0.632

At a rate of 0.5 events per unit time, there's a 63.21% chance of waiting 2 or less time units for the next event — the exponential distribution models memoryless waiting times.

Exponential distribution density and cumulative probability

The exponential distribution models waiting times between events occurring at a constant rate lambda. The CDF is the probability the wait is shorter than x.

Lambda (rate)	x	PDF at x	CDF - P(wait < x)
0.5	1	0.3033	0.3935
0.5	2	0.1839	0.6321
1	1	0.3679	0.6321
1	2	0.1353	0.8647
2	0.5	0.7358	0.6321
0.2	5	0.0736	0.6321

Four rows return a CDF of exactly 0.6321, and that is the distribution's signature: whenever x equals the mean waiting time of 1 over lambda, the probability of having waited less is always 1 minus 1 over e, about 63.21%, whatever the rate. So there is a 63% chance the wait is shorter than average, which is what a right-skewed distribution looks like. The exponential is memoryless - having already waited ten minutes tells you nothing about how much longer you will wait - which fits radioactive decay and call arrivals but fits equipment wear-out poorly, since machines do age.

The 'memoryless' property

The exponential distribution has a distinctive quirk: the probability of waiting an additional x time units doesn't depend on how long you've already waited. A machine that hasn't failed in the first hour is exactly as likely to fail in the next hour as a brand-new one.

Rate and units must match

λ (lambda) must be expressed in the same time units as x — if your average event rate is per hour but you're asking about a probability over 30 minutes, convert one of them to match the other first.

Frequently asked questions

A light bulb lasts an average of 1,000 hours. What's the probability it lasts at least 1,500 hours?

λ = 1/1000 per hour. P(X > 1500) = e^(−λ×1500) = e^(−1.5) = 0.223 = 22.3%. About 1 in 4 bulbs will last at least 1,500 hours — the exponential distribution has a long tail.

What does 'memoryless' mean for this distribution?

If a component has survived 500 hours, the probability of lasting another 500 hours is exactly the same as a brand-new component lasting 500 hours. Past survival gives you no information about future lifetime. This is unique to the exponential distribution.

How is exponential related to Poisson?

They're two sides of the same coin. Poisson models how many events happen in a fixed time interval. Exponential models how long you wait between events. If arrivals follow a Poisson process with rate λ, the time between arrivals is exponential with the same λ.