My notes give the following explanation of the Rayleigh distribution:
The Rayleigh distribution has CDF $F(x) = 1 - e^{-x^2/2}, x > 0$. To get the PDF, we differentiate the CDF, which gives
$f(x) = xe^{-x^2 / 2}, \ x > 0.$
For $x \le 0$ both the CDF and PDF are equal to $0$.
Let $X \sim$ Rayleigh. To find $P(X > 2)$:
$$P[X > 2] = \int_2^\infty x e ^{-x^2/2} \ dx = 1 - F(2) \approx 0.14$$
For the last part, my understanding is that the author used symmetry to show that $P[X > 2] =1 - P[X \le 2] = 1 - F[2]$, right?