Exponential Distribution Calculator
Calculate probabilities for the exponential distribution — the continuous probability distribution that models the time between events in a Poisson process. Find CDF, PDF, percentiles, or range probabilities. See also our Poisson Distribution Calculator and Half-Life Calculator.
How to Use the Exponential Distribution Calculator
The exponential distribution is a continuous probability distribution that models the time between events in a Poisson process — events that occur continuously and independently at a constant average rate. It is the continuous analog of the geometric distribution and is characterized by the memoryless property: the probability of waiting an additional time t is independent of how long you have already waited.
Common applications include modeling the time between customer arrivals, the lifetime of electronic components, the time between radioactive decays, the time between phone calls at a call center, and the distance between defects on a wire. The distribution is parameterized by λ (rate parameter), where 1/λ is the mean time between events.
This calculator supports four modes: CDF calculates P(X ≤ x), PDF gives the density at a point, Inverse finds the x-value for a given probability (percentile), and Range calculates P(a ≤ X ≤ b). The exponential distribution is closely related to the Poisson distribution — if events follow a Poisson process with rate λ, then the inter-arrival times follow an Exponential(λ) distribution.
Exponential Distribution Formulas
Probability Density Function (PDF):
f(x) = λe^(-λx) for x ≥ 0
Cumulative Distribution Function (CDF):
F(x) = P(X ≤ x) = 1 - e^(-λx)
Survival Function:
P(X > x) = e^(-λx)
Inverse CDF (Quantile):
x = -ln(1-p) / λ
Properties:
Mean = 1/λ, Median = ln(2)/λ
Variance = 1/λ², Std Dev = 1/λ
Mode = 0, Skewness = 2
Memoryless Property:
P(X > s+t | X > s) = P(X > t)
Example Calculation
A light bulb has an average lifetime of 1000 hours (λ = 1/1000 = 0.001). What is the probability it lasts more than 1500 hours?
Given: λ = 0.001 per hour, x = 1500 hours
P(X > 1500) = e^(-λx) = e^(-0.001 × 1500)
= e^(-1.5) = 0.2231
There is a 22.31% chance the bulb lasts beyond 1500 hours
Mean lifetime: 1/λ = 1000 hours
Median lifetime: ln(2)/λ = 693 hours
Note: Due to the memoryless property, if the bulb has
already lasted 500 hours, P(lasting 1000 more) is still
e^(-0.001×1000) = 0.3679 — same as a brand new bulb.
Exponential Distribution Reference Table
| λ | Mean (1/λ) | Median | P(X≤1) | P(X≤2) |
|---|---|---|---|---|
| 0.5 | 2.000 | 1.386 | 0.3935 | 0.6321 |
| 1.0 | 1.000 | 0.693 | 0.6321 | 0.8647 |
| 2.0 | 0.500 | 0.347 | 0.8647 | 0.9817 |
| 3.0 | 0.333 | 0.231 | 0.9502 | 0.9975 |
| 5.0 | 0.200 | 0.139 | 0.9933 | 0.9999 |
| 10 | 0.100 | 0.069 | 0.9999 | 1.0000 |
When to Use the Exponential Distribution
Use when:
- Modeling the time between independent events that occur at a constant average rate
- The memoryless property applies (the probability of waiting t more units does not depend on how long you have already waited)
- Examples: time between customer arrivals, time until next equipment failure, radioactive decay intervals
- The corresponding count process follows a Poisson distribution
Do NOT use when:
- Failure rate changes over time (use Weibull for wear-out or burn-in)
- Events are not independent or rates are not constant
- The system has a "warm-up" period before reaching steady state
- You need to model count data (use Poisson instead)
Additional Solved Examples
Example: Server Reliability
A server has a mean time between failures (MTBF) of 500 hours. What is the probability it runs for at least 600 hours without failure?
Lambda = 1/500 = 0.002 per hour
P(T > 600) = e^(-lambda x t) = e^(-0.002 x 600)
= e^(-1.2) = 0.3012
Answer: There is a 30.12% probability the server survives beyond 600 hours. The exponential distribution predicts that 1 - e^(-1) = 63.2% of failures occur before the MTBF, which is often counterintuitive.
Example: Customer Service Wait Time
Customers arrive at a help desk at a rate of 6 per hour. What is the probability that the next customer arrives within 5 minutes?
Lambda = 6 per hour = 0.1 per minute
P(T < 5) = 1 - e^(-0.1 x 5)
= 1 - e^(-0.5) = 1 - 0.6065 = 0.3935
Answer: There is about a 39.35% chance the next customer arrives within 5 minutes. The mean waiting time is 1/lambda = 10 minutes.
Practice Questions
Question 1
Light bulbs have an exponential lifespan with mean 1000 hours. What fraction fail within the first 500 hours?
Answer: P(T < 500) = 1 - e^(-500/1000) = 1 - e^(-0.5) = 1 - 0.6065 = 0.3935. About 39.35% of bulbs fail within the first 500 hours.
Question 2
Earthquakes in a region occur at an average rate of 2 per year. What is the probability of waiting more than 1 year for the next earthquake?
Answer: Lambda = 2/year. P(T > 1) = e^(-2 x 1) = e^(-2) = 0.1353. About 13.53% probability of no earthquake in the next year.
Question 3
Explain the memoryless property: a component has survived 100 hours. What is its probability of lasting another 100 hours?
Answer: Due to the memoryless property, P(T > 200 | T > 100) = P(T > 100). The conditional probability of surviving another 100 hours is the same as the original probability of surviving 100 hours from new. Past survival provides no information about future lifespan.
Key Takeaways
- The exponential distribution models time between events, with P(T > t) = e^(-lambda*t) and mean = 1/lambda.
- The memoryless property means P(T > s+t | T > s) = P(T > t) - the future does not depend on the past.
- The exponential is the continuous analog of the geometric distribution and the interarrival time of a Poisson process.
- Unlike intuition suggests, the median (0.693/lambda) is less than the mean (1/lambda) because the distribution is right-skewed.
- In reliability engineering, the exponential models the "useful life" phase with constant failure rate (middle of bathtub curve).
Frequently Asked Questions
What is the exponential distribution?
The exponential distribution is a continuous probability distribution that models the time (or distance) between events in a Poisson process. It is defined by a single parameter λ (rate), where 1/λ is the mean time between events. It is the only continuous distribution with the memoryless property, making it fundamental in reliability engineering, queuing theory, and survival analysis.
What is the memoryless property?
The memoryless property means P(X > s+t | X > s) = P(X > t). In practical terms, if you have already waited s units of time, the probability of waiting an additional t units is the same as if you had just started waiting. This makes the exponential distribution suitable for modeling component lifetimes where wear does not accumulate (constant failure rate). It does NOT model aging or wear-out failures.
How is the exponential related to the Poisson distribution?
If events occur according to a Poisson process with rate λ (events per unit time), then the time between consecutive events follows an Exponential(λ) distribution. The Poisson counts events in a fixed interval (discrete); the exponential measures time between events (continuous). They share the same rate parameter λ and are two views of the same underlying process.
When should I use the exponential distribution?
Use the exponential distribution when: (1) events occur randomly and independently, (2) the rate is constant over time (no aging), (3) you are modeling time/distance between events. Examples: time between customer arrivals, time between radioactive decays, distance between potholes on a road. Do NOT use it for: component lifetimes with wear-out (use Weibull), or processes where the rate changes over time.
What is the hazard rate of the exponential distribution?
The hazard rate (failure rate) of the exponential distribution is constant and equal to λ. This means the probability of failure in the next instant is the same regardless of how long the item has been operating. This constant hazard rate corresponds to the "useful life" period in the bathtub curve of reliability engineering, between early failures (decreasing rate) and wear-out failures (increasing rate).
How do I estimate λ from data?
The maximum likelihood estimator of λ is the reciprocal of the sample mean: λ̂ = n / Σxᵢ = 1/x̄. For example, if you observe 10 inter-arrival times averaging 5 minutes, then λ̂ = 1/5 = 0.2 events per minute. The confidence interval for λ uses the chi-square distribution: 2nλ̂/χ²(α/2, 2n) to 2nλ̂/χ²(1-α/2, 2n).