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Please consider the following problem and my answer to it. The answer seems to low to me. Is my answer right?
Thanks,
Bob

Problem:
The median net worth of a certain population is $97K$. To be in the top $5\%$ you need to have a net worth of $2387K$. Assuming the population follows the Pareto distribution, what is the minimum net worth you need to be in top $1\%$?
Answer:
For the Pareto distribution we have: \begin{eqnarray*} P(X > x) &=& \Big( \frac{x}{x_m} \Big) ^ \alpha \\ \end{eqnarray*} The above formula assumes that $x$ is greater than $0$. Now, we need to find the two parameters $x_m$ and $\alpha$. \begin{eqnarray*} \Big( \frac{97}{x_m} \Big) ^ \alpha &=& 0.5 \\ \Big( \frac{2387}{x_m} \Big) ^ \alpha &=& 0.95 \\ \end{eqnarray*} Using software, I find: \begin{eqnarray*} x_m &=& 3083 \\ \alpha &=& 0.200386 \\ \end{eqnarray*} Now we have our model. \begin{eqnarray*} 0.99 &=& \Big( \frac{x}{x_m} \Big) ^ \alpha \\ 0.99 &=& \Big( \frac{x}{3083} \Big) ^ {0.200386}\\ 0.9510821 &=& \Big( \frac{x}{3083} \Big) \\ x &=& 3083( 0.9510821 ) \\ x &=& 2932.1861 \\ \end{eqnarray*}

Here is an updated solution:

For the Pareto distribution we have: \begin{eqnarray*} P(X > x) &=& \Big( \frac{x_m}{x} \Big) ^ \alpha \\ \end{eqnarray*} The above formula assumes that $x$ is greater than $0$. Now, we need to find the two parameters $x_m$ and $\alpha$. \begin{eqnarray*} \Big( \frac{x_m}{97} \Big) ^ \alpha &=& 0.5 \\ \Big( \frac{x_m}{2387} \Big) ^ \alpha &=& 0.95 \\ \end{eqnarray*} Using software, I find: \begin{eqnarray*} x_m &=& 3083.33 \\ \alpha &=& -0.200386 \\ \end{eqnarray*} Now we have our model. \begin{eqnarray*} 0.99 &=& \Big( \frac{x_m}{x} \Big) ^ \alpha \\ 0.99 &=& \Big( \frac{3083.33}{x} \Big) ^ { -0.200386 } \\ 1.010101 &=& \Big( \frac{3083.33}{x} \Big) ^ { 0.200386 } \\ 1.00514339 &=& \Big( \frac{3083.33}{x} \Big) \\ x &=& 2932.50 \\ \end{eqnarray*} Is this answer right? My feeling is that it is still too low? If so, what did I do wrong?

Bob
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    I edited your Question: Note how exponents work in JaX. If you type $y^12$ you get $y^12,$ but if you type $y^{12} you get $y^{12}.$ – BruceET Oct 14 '18 at 21:25

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Wikipedia says the CDF is $1 - (x_m/x)^\alpha,$ for $x > x_m,$ so you should check your formula for $P(X > x).$ Another reason to check is that your formula gives probabilities that exceed 1/2.

Suggest correcting your formula, re-estimating parameters, and trying again.

BruceET
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  • I believe my formula for $P(X > x)$ is correct. It is consisted with the CDF given in Wikipedia. Actually, I got the formula from Wikipedia. – Bob Oct 14 '18 at 21:26
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    In spite of finding $x = 2932.1861 < x_m =3038?$ – BruceET Oct 14 '18 at 21:29
  • I am not saying that my work is correct. In fact, I suspect it is wrong. What I am saying is that I believe my formula for $P(X > x)$ is correct. – Bob Oct 14 '18 at 22:49
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    Perhaps sooner or later -- after all else fails -- you will take another look at Wikipedia to fix your mistake. – BruceET Oct 15 '18 at 06:59
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    I have $\frac{x}{x_m}$ and it should be $\frac{x_m}{x}$. I will fix the post. – Bob Oct 15 '18 at 11:54