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How to use
This calculator evaluates an exponential distribution with positive rate λ. It returns the density at x, CDF P(X≤x), survival probability P(X>x), interval probability, mean, variance, standard deviation and median. The model is commonly used for waiting times between independent events occurring at a constant average rate.
- Enter a positive event rate λ.
- Enter x and the interval endpoints a and b.
- Calculate the density, cumulative and survival probabilities and summary properties.
How it is calculated
Example
Example: with λ=0.5 and x=3, P(X>3)=e^-1.5≈0.22313, P(X≤3)≈0.77687 and the mean is 1/λ=2. If λ=0.5 events per minute, mean waiting time is 2 minutes. For the probability of waiting between 1 and 4 minutes, enter a=1 and b=4: e^−0.5−e^−2≈0.471195, or 47.1195%. All times must be in minutes because the rate is per minute.
Important notes
The exponential model assumes a constant rate and memorylessness. If the hazard changes with time, another model may be more appropriate. Some references parameterize the distribution by scale β=1/λ instead of rate λ.
Worked examples and interpreting results
Event rate 0.5 per time unit gives mean waiting time two units. Probability of waiting longer than two units is about 36.788% under the exponential distribution.
| Case | Calculation | Result |
|---|---|---|
| Probability of exceeding two units | e⁻⁽⁰·⁵ײ⁾ × 100 | 36.787944 % |
| Mean waiting time | 1 ÷ 0.5 | 2 |
| Probability of 1–4 minutes at rate 0.5/minute | exp(−0.5 × 1) − exp(−0.5 × 4) | 0.471195 |
| Median waiting time at the same rate | ln(2) ÷ 0.5 | 1.386294 min |
How to check the result
Enter rate 0.5, x=2 and interval bounds 0 and 2. Match time units to the rate unit; the model assumes constant rate and has the memoryless property.
Frequently asked questions
What is the difference between PDF and CDF?
The PDF is a density; the CDF is the probability that X is at most x.
Is P(X=x) equal to the PDF?
No. For a continuous distribution, a single exact point has probability zero.
What is the mean?
Mean and standard deviation are both 1/λ, while variance is 1/λ².
How do I calculate the probability between two times?
For 0≤a≤b, probability = e^(−λa)−e^(−λb). Enter both interval endpoints. Equal endpoints give zero probability because the distribution is continuous.
Should I enter the rate or mean waiting time?
The input expects rate λ, not the mean. If mean waiting time is 5 minutes, enter λ=1/5=0.2 per minute and use minutes for x and interval endpoints.