Simplify $\Pr(Z\ge P(X))$ with $Z\sim N(0,1)$ Consider a discrete rv $X$ with probability function $p(x)$ and an independent $Z\sim N(0,1)$
I am trying to write the probability
\begin{align}
\Pr(Z\ge p(X))
\end{align}
as a single Q-function $Q(x)\triangleq \Pr(Z\ge x)$. A similar upper bound can also be useful, but it should take into consideration the probability function $p(x)$ (and not the trivial one with minimum over $p(x)$).
I initially thought that $\Pr(Z\ge p(X))\le Q(E[p(X)])$ due to the convexity of $Q(x)$ on $x>0$, but an answer below shows that I used Jensen's in the incorrect direction and the opposite is true, i.e.
\begin{align}
\Pr(Z\ge p(X))\ge Q(E[p(X)]).
\end{align}
 A: Writing $p()$ for the pmf of $X$ and defining $Q(z):= P(Z\ge z),$ we have
$$\begin{align}P(Z\ge p(X))
&=E[P(Z\ge p(X)\mid X)]\\[2mm]
&=E[Q(p(X))]\\[2mm]
&\ge Q\left(E[p(X)]\right)\quad\text{by Jensen's inequality}\\[2mm]
&\ge Q\left(\sum_x p(x)^2\right).
\end{align}$$
Note that Jensen's inequality applies because $p(X)\ge 0$, and $Q()$ is convex on $[0,\infty)$.
Now $Z\ge p(X)\implies Z\ge p_{\inf}$, where $p_{\inf}=\inf\{p(x): p(x)>0, x\in\mathbb R\};$  hence, $P(Z\ge p(X)) \le P(Z\ge p_{\inf})=Q(p_{\inf})\le Q(0)={1\over 2},$ giving the following bounds:
$$\color{blue}{Q\left(\sum_x p(x)^2\right)\quad \le\quad P\left(Z\ge p(X)\right)\quad \le\quad Q(p_{\inf})}.$$
NB: It also follows that  $\sum_x p(x)^2\ge p_\inf$ for any discrete distribution.

Aside from the above "trivial" upper bound, here is another ...
Samford (1953) proves that $\forall t\ge 0,\ Q(t)\le B(t),$ where $B(t)={4\phi(t)\over 3t+\sqrt{8+t^2}}$, and $\phi(t)={1\over\sqrt{2\pi}}\,e^{-t^2/2}.$ Therefore,
$$P(Z\ge p(X))=\sum_x p(x) Q(p(x))\le\sum_x p(x)\, B(p(x)).$$
Example (binary case, with $p(x)\in\{p, 1-p\}$):
$$\begin{align}P(Z\ge p(X))&= p\,Q(p) + (1-p)\,Q(1-p)\\
&\le p\,B(p)+(1-p)\,B(1-p)\end{align}$$
Here's a picture showing this upper bound as the dashed line (red) and $P(Z\ge p(X))$ as the solid line (blue):

