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In powder metallurgy people often use particle distributions such as this one: particle distribution

I understand this distribution as a probability density function. I integrated this function to obtain the cumulative distribution function: CDF

The value of the CDF is supposed to be one for particle sizes over 50µm but this is not the case. Is there a coefficient to apply to the integration ?

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    $\begingroup$ Without more knowledge about what's on the y axis in your first plot, I'd think dividing by the sum might suffice. $\endgroup$ – N. Wouda May 31 '17 at 15:03
  • $\begingroup$ The yaxis is the percentage of volume for each size of particle. $\endgroup$ – snickers Jun 1 '17 at 8:48
  • $\begingroup$ No, the y axis doesn't show a percentage, because the area under the curve is not of the order of 100. It shows proportion given a certain bin width or equivalent. I don't have a better guess than you why the curve is presented as smoothed rather than binned. A rough average for the y variable is 0.05 (units something/$\mu$m) and the horizontal extent is about 50 $\mu$m. So the area under the curve is about 2.5 something. That suggests to me that the data are collected with a bin width of the order of 2.5 $\mu$m. What the machinery does should be documented somewhere! $\endgroup$ – Nick Cox Mar 19 at 8:41
  • $\begingroup$ The lack of a label on the vertical axis should make us cautious about interpreting the first graphic as a plot of a density: evidently it's something else that is only proportional to a density. $\endgroup$ – whuber Aug 23 at 14:29
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Well, you simply must scale your "probability density" so that it scales to one. Then integrate it the same way you did to get your second plot, and it will be right (having a horizontal asymptote at 1).

Or, as @Nick Cox says in a comment, it is much easier (and equivalent), to just divide the cumulative values with its maximum.

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    $\begingroup$ I think it's a lot easier to divide the distribution function by its maximum, the cumulative sum or integral under the entire curve. $\endgroup$ – Nick Cox Mar 19 at 8:31

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