How to use this calculator
- Enter or choose i know the.
- Enter or choose probability that the time is.
- The probability updates as you type. Use Copy result or Share to keep it.
Method
P(X ≤ x) = 1 − e^(−λx); P(X > x) = e^(−λx); P(a < X ≤ b) = e^(−λa) − e^(−λb); mean = 1/λ, median = ln 2/λ, SD = 1/λ
- λ
- rate of events per unit time
- x, a, b
- times in the same unit as the rate
Worked example
- λ = 0.5 per minute, so the mean wait is 1 ÷ 0.5 = 2 minutes.
- P(X ≤ 2) = 1 − e^(−0.5 × 2) = 1 − e⁻¹ = 0.632121.
- Median = ln 2 ÷ 0.5 = 1.386 minutes.
With customers arriving at 0.5 per minute, there is a 63.2% chance the next one arrives within 2 minutes.
What the result means
The exponential distribution is memoryless: the chance of waiting another x minutes does not depend on how long you have already waited. The median is shorter than the mean, because a few long waits pull the mean up.
Assumptions and limits
- Events happen independently at a constant rate (a Poisson process), and time is measured in the same unit as the rate.
- Lifetimes of parts that wear out (rising failure rate) or that have early failures are better described by a Weibull distribution.
Questions
How is this linked to the Poisson distribution?
If counts per interval are Poisson with rate λ, the gaps between events are exponential with the same λ.
Why is the probability of waiting longer than the mean only 36.8%?
P(X > 1/λ) = e⁻¹ = 0.368 for any rate, because the distribution is skewed towards short waits.