# Relationship between cdf of normal distribution and uniform distribution defined on $[0,1]$

So the problem is, given $$X\sim N(1,2^2)$$, $$Y=e^X$$, if the cdf of standard normal distribution is $$\Phi$$, to show that

$$\Phi\left(\frac{\ln Y-1}{2}\right)\sim U[0,1],$$

where $$U[0,1]$$ is the uniform distribution defined on $$[0,1]$$.

What I can tell now is that

$$\frac{\ln Y-1}{2}=\frac{X-1}{2}=Z\sim N(0,1^2),$$

is a standard normal distribution. So basically the problem is reduced to showing that $$\Phi(Z)\sim U[0,1].$$

But as you can see in here, the cdf of snd (the red curve) is not "uniform" on $$[0,1]$$. Rather $$N(-2,0.5)$$ (the green curve) is. So I'm certainly misunderstanding here something. Please help me to understand.

$$\Phi (Z)$$ is uniformly distributed on $$(0,1)$$: $$P(\Phi (Z) \leq u)=P(Z \leq \Phi^{-1}(u))=\Phi (\Phi^{-1}(u))=u$$ for all $$u \in (0,1)$$. I have used the fact that $$\Phi$$ is a continuous strictly increasing function.