Simple question about difference in percentiles -- what's wrong with my solution?

Question

An insurer's annual weather-related loss, X, is a random variable with density function $$f(x) = \frac{2.5(200)^{2.5}}{x^{3.5}} \qquad \text{for } x > 200$$ and $f(x) = 0$ otherwise. Calculate the difference between the $25$th and $75$th percentile.

My approach: $$F(X) = \int_{200}^x2.5(200)^{2.5}(t^{-3.5}) \; dt = 1-\left(\frac{200}{x}\right)^{2.5}$$

So, the desired difference is $$F(0.75)-F(0.25) = 1-\left(\frac{200}{0.75}\right)^{2.5} - 1+\left(\frac{200}{0.25}\right)^{2.5} = \left(\frac{200}{0.25}\right)^{2.5} - \left(\frac{200}{0.75}\right)^{2.5}$$ which gives $16940694$ (approximated) while the correct answer is $124$. Where have I gone wrong?

Answer 1

If $$F(x) = 1-\left(\frac{200}{x}\right)^{2.5}$$ then $$F^{-1}(y) = \frac{200}{(1-y)^{0.4}}$$ and $$F^{-1}(0.75)-F^{-1}(0.25) \approx 348.22-224.39 = 123.83$$

Answer 2

Short answer: Your $F(0.75) - F(0.25)$ is computed incorrectly and doesn't mean what you think it means.

Long answer: recall that $$F(x) = 1 - \left(\dfrac{200}{x} \right)^{2.5}$$ for $x > 200$ (this is key) and $F(x) = 0$ for $x \leq 200$.

Thus, $F(0.75) - F(0.5) = 0 - 0 = 0$ (because both $0.75 \leq 200$ and $0.5 \leq 200$). This would be the probability $$F(0.75) - F(0.5) = \mathbb{P}(X \leq 0.75) - \mathbb{P}(X \leq 0.5) = \mathbb{P}(0.5 < X \leq 0.75) = 0\text{.} $$ Obviously, this isn't what you're looking for. What you're actually looking for are the percentiles $p_{0.25}$ and $p_{0.75}$. These are the values satisfying $$F(p_{0.25}) = 0.25 = 1 - \left(\dfrac{200}{p_{0.25}} \right)^{2.5}$$ $$F(p_{0.75}) = 0.75 = 1 - \left(\dfrac{200}{p_{0.75}} \right)^{2.5}$$ Use algebra to solve for $p_{0.25}$ and $p_{0.75}$, and then compute the difference $p_{0.75} - p_{0.25}$.