finding $Pr(X>\frac{16-Z^4}{X+Z^2})\leq\frac{3}{8}$

if $X$,$Z$ be two independent random variable that $X\sim PO(1)$ and $Z\sim \mathcal{N}(0,1)$ how can show $\displaystyle\Pr(X>\frac{16-Z^4}{X+Z^2})\leq\frac{3}{8}$

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(0% accept rate) + (long history of not even answering queries about the fact) = (low level of motivation to provide answers). – Did Apr 1 '12 at 9:37
Hint: Look separately at the cases $X=0,1,2,3,4$ and $X\ge 5$. A simple approach uses the fact that $\Pr[X\ge 5] \lt 1/4!$ (from Taylor's Theorem) and $\Pr[|Z| \gt \sqrt{\sqrt{37}/2 - 3/2}] \gt \Pr[|Z| \gt 1] \lt 1/3$ (an extremely well known approximation); notice that $1/3+1/4!=3/8$. The actual probability is close to 0.0923709. – whuber Apr 1 '12 at 18:46
(Erratum: please reverse one inequality within the last set of inequalities to read $\Pr[|Z| \gt \sqrt{\sqrt{37}/2 - 3/2}] \color{red}\lt \Pr[|Z| \gt 1] \lt 1/3$.) – whuber Apr 1 '12 at 20:09