Maneuvering normal distribution inequality

Question

I know $\Pr[N(\mu, \sigma^2) \geq \mu + k \sigma] = \Pr[N(0, 1) \geq k]$.

Say I am given, $\Pr[N(\mu, \sigma^2) \geq \mu + q \sigma] = \Pr[N(0, 1) \leq k]$.

How can I find a relation between $q$ and $k$?

Answer 1

$k=-q$ as @Tunococ implied. Following your notation we have $\Pr[N(\mu, \sigma^2) \geq \mu + q \sigma] = \Pr[N(0, 1) \ge q]$.

We are given that $\Pr[N(\mu, \sigma^2) \geq \mu + q \sigma] = \Pr[N(0, 1) \leq k]$.

Equating RHS we get ($\Phi()$ = standard normal cdf)

$\Pr[N(0, 1) \ge q] = \Pr[N(0, 1) \leq k] \Rightarrow 1-\Pr[N(0, 1) \le q] = \Pr[N(0, 1) \leq k] $

$\Rightarrow 1- \Phi(q) = \Phi(k) \Rightarrow \Phi(-q)=\Phi(k) \Rightarrow -q=k $.