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Asked: 2026-05-15T09:30:53+00:00

Ok, so here’s my problem. We are looking at purchasing a data set from

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Ok, so here’s my problem. We are looking at purchasing a data set from a company to augment our existing data set. For the purposes of this question, let’s say that this data set ranks places with an organic number (meaning that the number assigned to one place has no bearing on the number assigned to another). The technical range is 0 to infinity, but from sample sets that I’ve seen, it’s 0 to 70. Based on the sample, it’s most definitely not a uniform distribution (out of 10,000 there are maybe 5 places with a score over 40, 50 with a score over 10, and 1000 with a score over 1). Before we decide to purchase this set, we would like to simulate it so that we can see how useful it may be.

So, to simulate it, I’ve been thinking about generating a random number for each place (about 150,000 random numbers). But, I also want to keep to the spirit of the data, and keep the distribution relatively the same (or at least reasonably close). I’ve been racking my brain all day trying to think of a way to do it, and have come up empty.

One thought I had was to square the random number (between 0 and sqrt(70)). But that would favor both less than 1 and larger numbers.

I’m thinking that he real distribution should be hyperbolic in the first quadrant… I’m just blanking on how to turn a linear, even distribution of random numbers into a hyperbolic distribution (If hyperbolic is even what I want in the first place).

Any thoughts?

So, to sum, here’s the distribution I would like (approximately):

  • 40 – 70: 0.02% – 0.05%
  • 10 – 40: 0.5% – 1%
  • 1 – 10: 10% – 20%
  • 0 – 1 : Remainder (78.95% – 89.48%)
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1 Answer

  1. Editorial Team

    Look at distributions used in reliability analysis – they tend to have these long tails. A relatively simply possibility is the Weibull distribution with P(X>x)=exp[-(x/b)^a].

    Fitting your values as P(X>1)=0.1 and P(X>10)=0.005, I get a=0.36 and b=0.1. This would imply that P(X>40)*10000=1.6, which is a bit too low, but P(X>70)*10000=0.2 which is reasonable.

    EDIT
    Oh, and to generate a Weibull-distributed random variable from a uniform(0,1) value U, just calculate b*[-log(1-u)]^(1/a). This is the inverse function of 1-P(X>x) in case I miscalculated something.

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