Take two independent, non-degenerate Normals, $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$.
Define $Z=aX + Y$ with $a>0$.
This implies $Z \sim N(a\mu_X+\mu_Y,a^2\sigma_X^2+\sigma_Y^2)$
If $Pr(Z\leq 0)$ decreases with $a$, what must be true about $\mu_X$ and $\mu_Y$?
If $Pr(Z\leq z)$ with $z>0$ decreases with $a$, what must be true about $(\mu_X, \sigma_X^2)$ and $(\mu_Y,\sigma_Y^2)$?
Increasing $a$ has two effects on $aX$.
- It scales up the mean of $aX$.
- It scales up the variance of $aX$.
If $\mu_X>0$, then the first effect increase the mean value of $aX$, suggesting a decrease in $\Pr(aX\leq0)$.
However, with $\mu_X>0$ the second effect makes $\Pr(aX\leq0)$ increase with $a$. The intuition is that for a Normal an increase in the variance means more mass is away from the mean, which means more mass falls in the region below $\mu_X$ and hence also below $0$.
We also know that $\Pr(aX\leq0) = \Pr(X\leq0)$, and hence $aX$ is not affected by $a$. This means that effects 1 and 2 must be exactly offsetting each other.
Coming back to $Z$, we also know that $Y$ is not affected by $a$.
So we still have the same two effects of $a$ now working on $Z$ instead of $aX$. Intuitively, it seems to me that the first effect is as strong on $Z$ as on $aX$, while the second effect is attenuated by having the additional variance $\sigma_Y^2$. This suggests that if $Pr(Z\leq 0)$ decreases with $a$, then $\mu_X>0$ and we learn nothing about $\mu_Y$.