Exponential Distribution Calculator - Waiting Time and MTTF

Exponential distribution: how long until the next event

A request arrives every 15 minutes on average. Planning around "15 minutes" goes wrong in both directions: in 48.66% of cases the next one shows up within 10 minutes, and in 13.53% of cases there is a silence of more than half an hour. This exponential distribution calculator gives the chance that the event happens before time t, after it or between a and b, and works backward too: how long you have to allow to be X% sure the event has happened.

The problem with the average. The waiting time is skewed: many short gaps, a few very long ones. The median is only 0.693 of the mean, so with a mean of 15 minutes half of all gaps end before 10.4 minutes, while a buffer that covers 90% of cases needs 34.54 minutes.

Solution: four questions, one formula

What you know - the mean time between events (for example 15 minutes) or the rate λ (for example 4 per hour). One is the reciprocal of the other.
Value and unit - minutes, hours, days, weeks, months, years, or miles and kilometers when the events are scattered along a road, a cable or a pipe.
Your question - the event happens within t, the wait lasts longer than t, the event falls between a and b, or "how long until it has happened with X% certainty".
Time t (and the end b of an interval) or the certainty as a percentage - in the same unit as the mean.
Read the result - the main probability or time, tiles with the mean, the median and λ, and a table of how long the wait lasts in 10%, 50%, 90% and 99% of cases.

Behind every answer is the same cumulative distribution function from the NIST/SEMATECH e-Handbook of Statistical Methods: P(T ≤ t) = 1 - e^-λt. Its reliability chapter lists the mean 1/λ, the median ln 2/λ ≈ 0.693/λ and the variance 1/λ², and checks the formula with λ = 0.01: at 100 hours the CDF is 0.6321. The calculator returns 63.21% for the same inputs.

Picking the right question for your plan

You want to know	Choose	Example with a mean of 15 minutes
Will it happen soon?	Within t, P(T ≤ t)	within 10 minutes: 48.66%
How likely is a long silence?	Longer than t, P(T > t)	more than 30 minutes: 13.53%, about one gap in seven
Will it land in a window?	Between a and b	between 5 and 15 minutes: 34.87%
How much time to allow?	Time for X% certainty	90%: 34.54 minutes; 95%: 44.94 minutes

Four situations, four answers

An interval in days. An event every 10 days on average, between day 5 and day 15. Result: e^-0.5 - e^-1.5 = 38.34%.

Time to failure. NIST calls 1/λ the Mean Time To Fail (MTTF). With an MTTF of 5 years: failure in the first year 18.13%, still running after 10 years 13.53%.

Distance instead of time. 0.2 defects per mile along a cable, the first 7 miles. Result: 1 - e^-1.4 = 75.34% that a defect turns up before mile 7.

No memory. Mean 15 minutes, you have already waited 25. The chance of waiting more than another 10 minutes is 51.34%, exactly the same as P(T > 10) from the start. Time already spent waiting does not bring the event closer.

Waiting times as a multiple of the mean

The NIST percent point function is -ln(1 - q) × mean, so one column of multiples works for any unit. The last two columns show it for a mean of 10 minutes and of 2 days:

The event happens in	Multiple of the mean	Mean 10 minutes	Mean 2 days
10% of cases	0.105	1.05 min	0.21 days
25% of cases	0.288	2.88 min	0.58 days
50% (median)	0.693	6.93 min	1.39 days
75% of cases	1.386	13.86 min	2.77 days
90% of cases	2.303	23.03 min	4.61 days
95% of cases	2.996	29.96 min	5.99 days
99% of cases	4.605	46.05 min	9.21 days

When the exponential model fits, and when it misleads

It fits events that arrive at random, each independent of the last, at an average pace that does not change over the period you look at. NIST describes it as the model with a constant failure rate and notes that it works well for inter-arrival times, while the Poisson distribution describes the total number of events in a given period. Its reliability chapter ties it to the flat middle of the "bathtub curve".

It misleads when the rate changes, as on a help line that is busier at 9 a.m. than at 2 p.m. (calculate each period separately), or for parts like tires and bearings that fail more often with age. A quick check on your own data: the standard deviation of the gaps should be close to their mean, since for this distribution the two are equal.

Exponential distribution questions

Why is the median shorter than the mean?

The distribution has a long right tail: most gaps are short and a few very long ones pull the mean up. The median is 0.693 of the mean, so with a 15-minute mean half of the gaps end before 10.4 minutes.

What is the memoryless property?

P(T > s + t | T > s) = P(T > t). If you have already waited s minutes, the chance of waiting another t minutes is the same as at the start. Together with the geometric distribution, the exponential is the only distribution with this property.

How is it related to the Poisson distribution?

When events happen independently at a constant average rate λ, their count in a period follows a Poisson distribution and the gaps between them follow an exponential one. An average of 4 calls an hour means an average gap of 15 minutes.

Why does "within the mean time" always give 63.21%?

Because P(T ≤ 1/λ) = 1 - e^-1 = 0.6321 for every λ. It is a handy check: if t equal to the mean gives anything else, the units are mixed up.

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