# What distribution does my data follow?

Let us say that I have 1000 components and I have been collecting data on how many times these log a failure and each time they logged a failure, I am also keeping track of how long it took my team to fix the problem. In short, I have been recording the time to repair (in seconds) for each of these 1000 components. Data is given at the end of this question.

I took all these values and drew a Cullen and Frey graph in R using descdist from the fitdistrplus package. My hope was to understand if the time to repair follows a particular distribution. Here's the plot with boot=500 to get bootstrapped values:

I see that this plot is telling me that the observation falls into the beta distribution (or maybe not, in which case, what is it revealing?) Now, considering that I am a system architect and not a statistician, what is this plot revealing? (I am looking for a practical real-world intuition behind these results).

EDIT:

QQplot using the qqPlot function in package car. I first estimated the shape and scale parameters using the fitdistr function.

> fitdistr(Data$Duration, "weibull") shape scale 3.783365e-01 5.273310e+03 (6.657644e-03) (3.396456e+02)  Then, I did this: qqPlot(LB$Duration, distribution="weibull", shape=3.783365e-01, scale=5.273310e+03)


EDIT 2:

Updating with a lognormal QQplot.

Here's my data:

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• That diagram does not tell you your distribution is beta. It says the skewness and kurtosis are consistent with a beta - it could easily be lognormal, for example, but it probably isn't actually any of the distributions named on that diagram. Commented May 6, 2013 at 0:13
• @Glen_b: Thank you. I just included a qqplot for lognormal as well but even this does not seem to be a good fit. Is there anything else you recommend that I try out? I included my data in the question. Commented May 6, 2013 at 0:22
• I am curious why you call this a "Cullen Frey" plot, when it was introduced by Rhind in 1909 (and well known for generations afterwards), 90 years before Cullen and Frey wrote anything together! See the Wikipedia article on the Pearson system of distributions.
– whuber
Commented May 6, 2013 at 15:06
• We are seeing Stigler's Law of Eponymy in action. :-)
– whuber
Commented May 7, 2013 at 2:20
• @whuber It's a Cullen and Frey plot, not Rhind's visualization of the Pearson space. It has distinctly different features, such as the depiction of boostrapped values, the overlay of the uniform distribution, etc, etc. It builds on Rhind's graph, but everything in science builds on something before it (and we don't want to have to attribute everything to the original, unknown inventors of fire and the wheel...). Commented Nov 3, 2015 at 15:33

The thing is that real data doesn't necessarily follow any particular distribution you can name ... and indeed it would be surprising if it did.

So while I could name a dozen possibilities, the actual process generating these observations probably won't be anything that I could suggest either. As sample size increases, you will likely be able to reject any well-known distribution.

Parametric distributions are often a useful fiction, not a perfect description.

Let's at least look at the log-data, first in a normal qqplot and then as a kernel density estimate to see how it appears:

Note that in a Q-Q plot done this way around, the flattest sections of slope are where you tend to see peaks. This has a clear suggestion of a peak near 6 and another about 12.3. The kernel density estimate of the log shows the same thing:

In both cases, the indication is that the distribution of the log time is right skew, but it's not clearly unimodal. Clearly the main peak is somewhere around the 5 minute mark. It may be that there's a second small peak in the log-time density, that appears to be somewhere in the region of perhaps 60 hours. Perhaps there are two very qualitatively different "types" of repair, and your distribution is reflecting a mix of two types. Or just maybe once a repair hits a full day of work, it tends to just take a longer time (that is, rather than reflecting a peak at just over a week, it may reflect an anti-peak at just over a day - once you get longer than just under a day to repair, jobs tend to 'slow down').

Even the log of the log of the time is somewhat right skew. Let's look at a stronger transformation, where the second peak is quite clear - minus the inverse of the fourth root of time:

The marked lines are at 5 minutes (blue) and 60 hours (dashed green); as you see, there's a peak just below 5 minutes and another somewhere above 60 hours. Note that the upper "peak" is out at about the 95th percentile and won't necessarily be close to a peak in the untransformed distribution.

There's also a suggestion of another dip around 7.5 minutes with a broad peak between 10 and 20 minutes, which might suggest a very slight tendency to 'round up' in that region (not that there's necessarily anything untoward going on; even if there's no dip/peak in inherent job time there, it could even be something as simple as a function of human ability to focus in one unbroken period for more than a few minutes.)

It looks to me like a two-component (two peak) or maybe three component mixture of right-skew distributions would describe the process reasonably well but would not be a perfect description.

The package logspline seems to pick four peaks in log(time):

with peaks near 30, 270, 900 and 270K seconds (30s,4.5m,15m and 75h).

Using logspline with other transforms generally find 4 peaks but with slightly different centers (when translated to the original units); this is to be expected with transformations.

• +1 This is a gold mine of information of me. I am trying to digest everything you have written and so far this has taught me how to actually approach this type of problems. What is the point of the stronger transformation? May I ask how you came up with that? Is that with experience or is there a more formal way of choosing such a non-conventional transformation? Please pardon my ignorance if this is common wisdom in the stats community. But I would be thankful if you could point me to a good reference to learn this kind of "detective" work which feels awesome to me. Commented May 6, 2013 at 2:31
• Proper reference to EDA: Tukey, J. W. (1977). Exploratory Data Analysis. Addison-Wesley, Reading, MA. Commented May 6, 2013 at 3:04
• As mentioned in the above answer, you could try fitting a mixture distribution. Here's a paper that uses these hybrids for wind speed -- I think some of the distributions are combinations of 3 other distributions. journal-ijeee.com/content/3/1/27 Commented May 6, 2013 at 15:38
• For a mixture it's a matter of figuring out how many components you want, what distribution or distributions you're going to take a mixture of (which is what you originally posted about), and then how you'll identify the parameters of the components and the component proportions. There are a number of packages that can help with those tasks; here's a paper (pdf) on one of them. A few of the mixture-modelling packages are mentioned in the Cluster Analysis and Finite Mixture Modeling Task View ...(ctd) Commented May 6, 2013 at 23:15
• (ctd)... Another example package is rebmix. My own analysis above was based on simpler exploratory approaches but as it stands at present isn't yet a fully identified mixture model; it suggests that a 4-component mixture might be needed. The final part of my answer - the part with the log-spline is a different (nonparametric) approach to modelling complicated densities. Commented May 6, 2013 at 23:16

The descdist function has an option to bootstrap your distribution to get a sense of the precision associated with the estimate plotted. You might try that.

descdist(time_to_repair, boot=1000)


My guess is that your data are consistent with more than just the beta distribution.

In general, the beta distribution is the distribution of continuous proportions or probabilities. For example, the distribution of p-values from a t-test would be some specific case of a beta distribution depending on whether the null hypothesis is true and the amount of power your analysis has.

I find it extremely unlikely that the distribution of your times to repair would actually be beta. Note that that graph is only comparing the skew and kurtosis of your data to the specified distribution. The beta is bound by 0 and 1; I'll bet your data aren't, but that graph isn't checking that fact.

On the other hand, the Weibull distribution is common for lag times. From eyeballing the figure (without the bootsamples plotted to gauge the uncertainty), I suspect your data are consistent with a Weibull.

You could also check if you data are Weibull, I believe, using qqPlot from the car package to make a qq-plot.

• +1 Thank you. In the time that I am understanding your answer, I just updated my question with the bootstrap parameter set to 500 in the descdist function. And yes, you are right that my values are not in [0,1]. Is there a way I can show that fact (belonging to weibull) using this graph? I will try to update my question with a QQPlot shortly. Commented May 5, 2013 at 23:58
• Just updated my question with a qqPlot from the car package. Commented May 6, 2013 at 0:05
• Hmmm. Well, the qq-plot does not make it look like the Weibull distribution is a good fit. Commented May 6, 2013 at 0:10
• And one more for the lognormal distribution. Do you recommend any pre-processing that I should do with the data? Or is there a better way to estimate the best-fit? I'm still wondering how I can utilize the Cullen/Frey graph in my context. Commented May 6, 2013 at 0:20
• Also, updated my question with the data I'm using at the end in case it helps. Commented May 6, 2013 at 0:24

For what it is worth, using Mathematica's FindDistribution routine, the logarithms are very approximately a mixture of two normal distributions,

That is, $x=\ln(\text{data})$, and $$f(x)=0.0585522 e^{-0.33781 (x-11.7025)^2}+0.229776 e^{-0.245814 (x-6.66864)^2}$$

Using 3 distributions to make a mixture distribution this can be

$$f(x)=0.560456\text{ Laplace}(5.85532,0.59296)+0.312384\text{ LogNormal}(2.08338,0.122309)+0.12716\text{ Normal}(11.6327,1.02011) \,,$$ which numerically is $$\begin{array}{cc} \Bigg\{ & \begin{array}{ll} 0.472592 e^{-1.68646 (5.85532\, -x)}\, +0.0497292 e^{-0.480476 (x-11.6327)^2} & x\leq 0 \\ 0.472592 e^{-1.68646 (5.85532\, -x)}+0.0497292 e^{-0.480476 (x-11.6327)^2}+\frac{1.01893 }{x}e^{-33.4238 (\ln (x)-2.08338)^2} & 0<x<5.85532 \\ 0.472592 e^{-1.68646 (x-5.85532)}+0.0497292 e^{-0.480476 (x-11.6327)^2}+\frac{1.01893 }{x}e^{-33.4238 (\ln (x)-2.08338)^2} & \text{Otherwise} \\ \end{array} \\ \end{array}$$

There are many other possibilities. For example, fitting three normal distributions to the 1/10$^\text{th}$ power of the data. For Mathematica code, further methods are as per this link .