Help with creating Pareto distribution given a limited set of data. I am trying to calculate/create a Pareto distribution with basically only knowing the $x$ and $y$-axis of the chart. The maximum set of data would be $200,000$ with the minimum being $1$. The amount of users would be $226,000$. In short, I am trying to, if possible, determine the sum of data in a vacuum where the top $0.000442\% (\frac{1}{226,000})$ has a value of $200,000$ in a world where the $\frac{80}{20}$ Pareto distribution holds true.
Is this possible and if so how?
 A: Pareto random variables have continuous distributions with support from a positive real number up to $+\infty$.  For $x \gt x_\min:$
$$\mathbb P(X\ge x) = \left(\frac{x_\min}{x}\right)^\alpha$$
Only Pareto distributions with shape parameter $\alpha =\frac{\log (0.2)}{\log(0.2/0.8)} \approx 1.160964$ precisely reflect the $80/20$ rule
Having $\alpha \approx 1.160964$ and $x_\min=1$ and $x=200000$ and $\mathbb P(X\ge x)=\frac 1 {226000}$ are not consistent with a Pareto distribution.  You could have:

*

*$\alpha \approx 1.160964$ and $x_\min=1$ and $x=200000$ and $\mathbb P(X\ge x)\approx\frac 1 {1426610}$

*$\alpha \approx 1.160964$ and $x_\min=1$ and $x\approx 40905$ and $\mathbb P(X\ge x)=\frac 1 {226000}$

*$\alpha \approx 1.160964$ and $x_\min\approx 4.889$ and $x=200000$ and $\mathbb P(X\ge x)=\frac 1 {226000}$

*$\alpha \approx 1.010013$ and $x_\min=1$ and $x=200000$ and $\mathbb P(X\ge x)=\frac 1 {226000}$
This last $\alpha \approx 1.01$ would correspond to about a $96.7/3.3$ rule rather than $80/20$
