I am trying to model particles generated by a mechanical crusher. We put a big chunk into crusher and we get small particles. These particles are categorised into 3 categories Size <2 mm, 2-50 mm and >50 mm. Assume these particles are spherical in shape.

Target is to find the particle size distribution such that.

Size <2 mm particles weight corresponds to 1.5% of input weight

Size 2-50 mm particles weight corresponds to 89% of input weight

Size >50 mm particles weight corresponds to 9.5% of input weight

I have assumed normal distribution for particle sizes and written a small code in python for varying mean and std, generate 1000 particles, categorise them and calculate their weights and check if the condition is satisfied or not.

I could not find a solution.

I did the same exercise for skewed normal distribution but no luck.

Note that normal distribution generates negative number, I wrote an if condition when negative number is observed take a number between 0.01 mm to 2 mm following uniform distribution

Is my approach correct or is their any simpler approach.

  • $\begingroup$ There are infinitely many solutions -- but to be realistic, they should assume a distributional shape that has zero (or at least vanishlingly small) chance of being negative. A standard choice for particle size distributions is the family of lognormal distributions. $\endgroup$
    – whuber
    Mar 2 at 15:58
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    $\begingroup$ "I have assumed normal distribution for particle sizes" -- this doesn't seem likely to be realistic and I expect it will yield unsatisfactory results. Common choices for particle size distributions include lognormal and Weibull (sometimes called Rosin-Rammler in particle-size analysis), though a few other possibilities do occur. The Weibull seems a bit more common for analyses involving grinding/milling/ crushing of material but I expect the lognormal will likely do quite well. $\endgroup$
    – Glen_b
    Mar 2 at 16:09
  • $\begingroup$ @Glen FWIW, a nice property of the lognormal family is that when any of the diameters, volumes, or surface areas of the particles have a lognormal distribution, so do the other two properties. $\endgroup$
    – whuber
    Mar 3 at 16:30
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    $\begingroup$ Indeed. This handy property is related to the part of my recent answer discussing lognormality of $Y^p$ here: stats.stackexchange.com/a/641246/805 ... (nevertheless Weibull seems somwwhat more common in analyses with crushed media). However, I think this property is also true of the Weibull $\endgroup$
    – Glen_b
    Mar 3 at 21:10
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    $\begingroup$ Specifically, since Weibulls are powers of exponentials, they're also powers of each other. Indeed, that neat property should hold for the distribution of $Y= \exp(X)$ with $X$ being from a location-scale family (normal for lognormal, negative of a Gumbel for Weibull, etc) $\endgroup$
    – Glen_b
    Mar 3 at 22:37

1 Answer 1


This is what I have done:

For lognormal distribution, $\ln(x)\sim N(\mu,\sigma)$

$$ \int_0^2 x^3P(x)=0.015E[x^3] $$ After some math $$ \operatorname{erf}\left(\frac{3\sigma^2+\mu-\ln 2}{\sqrt{2}\sigma}\right)=0.97 $$

$$ \int_2^{50} x^3P(x)=0.89E[x^3] $$ After some math $$ \operatorname{erf}\left(\frac{3\sigma^2+\mu-\ln 2}{\sqrt{2}\sigma}\right)-\operatorname{erf}\left(\frac{3\sigma^2+\mu-ln50}{\sqrt{2}\sigma}\right) =1.78 $$

For Weibull distribution $f(x)=\frac{k}{\lambda}(\frac{x}{\lambda})^{k-1}\exp(-(\frac{x}{\lambda})^{k}),x\geqslant 0$

Applying same conditions I got

$$ 1-\frac{\Gamma(1+\frac{3}{k},\frac{2^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})}=0.015 $$

$$ \frac{\Gamma(1+\frac{3}{k},\frac{2^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})}-\frac{\Gamma(1+\frac{3}{k},\frac{50^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})} =0.89 $$

Then calculated the variables.


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