[tex]\mathbb P(-z<Z<z)=0.08\implies \mathbb P(Z<-z)+\mathbb P(Z>z)=0.92[/tex]
Because the distribution (normal) is symmetric, you know that [tex]\mathbb P(Z<-z)=\mathbb P(Z>z)[/tex], so
[tex]\mathbb P(Z<-z)+\mathbb P(Z>z)=2\mathbb P(Z<-z)=0.92\implies\mathbb P(Z<-z)=\mathbb P(Z>z)=0.46[/tex]
Now,
[tex]\mathbb P(-z<Z<z)=\mathbb P(Z<z)-\mathbb P(Z<-z)=0.08[/tex]
[tex]\implies\mathbb P(Z<z)=0.08+0.46=0.54[/tex]
If [tex]F_Z(z)[/tex] is the CDF of the normal distribution, so that [tex]F_Z(z)=\mathbb P(Z<z)[/tex], then the z-score satisfies
[tex]\begin{cases}z={F_Z}^{-1}(0.54)\\z=-{F_Z}^{-1}(0.46)\end{cases}[/tex]
so that [tex]z\approx0.1004[/tex].
