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