Let $X$ be a random variable with a normal distribution such that $E(X) = µ$ and $Var(X) = \sigma^2$. Suppose that $μ=0$ and that $\sigma=1$, then my textbook posits the question to find $$P(X^2-X-1>0)$$
Thoughts: Usually with these types of problems I attempt to rewrite the left-hand side of the inequality for $Z= \frac {x -μ}{\sigma}$ but that doesn't seem to apply here.
My other thoughts is that this is equivalent to the integral $$\int_{0}^{\infty}\frac {e^{\frac {-(x^2-x-1)^2}{2}}}{\sqrt {2\pi}} dx$$ but I am not sure how this helps. Any tips or insights appreciated.
$f(x)=x^2-x-1$ has distinct real roots $r_1,r_2$ (which you can explicitly calculate). For clarity assume $r_1<r_2$. It factors to $(x-r_1)(x-r_2)$. Consequently "$X^2-X-1>0$" is equivalent to "$X-r_1<0$ or $X-r_2>0$". Since these events are disjoint you can compute their probabilities and add them up.