Work order completion time is exponentially distributed I have a work order system that serves the operational needs of a big organization. It has tickets and it gets completed in x days (0 < x < 1000) . I pulled out the data for completion time in days vs # of tickets. I plotted it and found out that this is exponentially distributed. I overlayed the exponential curve on top of the data and its a nice fit. I'm happy that it fits because now I can use the CDF & determine the probability of a new ticket getting completed within the deadline.
I have a hard time wrapping my head around it. My understanding of exponential distribution is "time between events in a poisson process". Why does this data fit in exponential distribution ? What would be the underlying poisson process ? Is there a rule of thumb to just look at the problem description and conclude its exponential ? 
 A: *

*The fact that your data look like an exponential and the the inter-event time in a Poisson process is exponential doesn't carry any logical implication that there's a Poisson process operating here. [It would be an argument analogous to  "snow is cold and frozen chicken is cold, therefore frozen chicken is snow".]
Certainly if you're certain that a Poisson process holds (or is so close to holding it will make little difference to the way the process behaves), you'd expect an exponential inter-event time (or nearly so). But there's nothing inherent in "time to process a work order" that makes it specifically exponential. It does happen that many such sets of time-data are reasonably well modelled by an exponential, but on the other hand, many are not. [e.g. It's not hard to think of reasons why some things would not be exponential -- consider if inherent in every work order there was an extensive - but constant - amount of paperwork, even for short jobs; then the distribution could not be exponential, but might be close enough to a shifted-exponential. Or imagine if every job had two consecutive parts, both of which were close to exponential in their completion time; then the overall completion time would not be exponential.]

*It sounds like your recorded data may not be close to continuous. You've got a discrete number of days, and exponential isn't discrete. (The underlying time may be continuous but from my understanding you only have days recorded. You might consider whether you are better off using the geometric - the discrete "equivalent" - to deal with the data at hand.)

*Even the underlying time won't be exponential; however, it might well be close enough for your purposes (which is to say, even though I could fairly safely deny that your distribution is exactly either exponential or geometric, I would probably just go ahead and use it too, since it will probably serve quite well).

*However, I want to raise some potential censoring issues
a. You mention $x<1000$. Is this a hard recording limit (i.e. censoring), or is it that you just haven't seen a job that takes so long (in which case it's of a different kind than the "$0<x$" part? After all, a job that takes "-3 days" isn't simply a value "not yet seen" but more "not ever possible for real completion times")
b. Incidentally, can you have a work order that is "just given up on" because no solution is available, where you're recording the time until the problem is shelved?
A: It helps to think whether there is a Poisson process behind.
By Poisson process we usually mean a pure birth process $X(t)$ that satisfies:


*

*$X(0)=0$

*$X(t)-X(s)$ has Poisson distribution with parameter proportional to $t-s$

*$X(t)$ has independent increments.


In your case, let's denote the number of ticket completed at time $t$ by $X(t)$, it may be considered as the "underlying Poisson process" if the following can be satisfied (search for Poisson postulates for why we need these assumptions):


*

*Time recorded in continuous fashion, so record hours, minutes, even seconds to make it "more continuous" than just days.

*No two or more tickets can be completed simultaneously, and the chance of getting one more ticket done in a short period is proportional to the length of that period.

*Each ticket is independent of others, so completion of one ticket does not depend/affect other tickets.


If these assumptions seems reasonable, then we can model that the inter-arrival time of $X(t)$, that is, the time it take until next ticket is done, with exponential distribution.
