Obtaining different CDF of non-standard normal distribution using two methods

Question

I was reading about the derivation of the CDF of a non-standard normal distribution.

One particular version goes like this:

Let $X \sim \mathcal{N}(\mu, \sigma^2), Z \sim \mathcal{N}(0,1).$

$\therefore X = \sigma Z + \mu$

$$ P(X\le x) = P\left(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}\right) = \Phi\left(\frac{x-\mu}{\sigma}\right) $$

Source: https://www.youtube.com/watch?v=9vp1Ll2NpRw&t=14m54s (Harvard stats 110)

Now, I thought of another variation, this time arriving at a different conclusion:

$$ P(X\le x) = P\left(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}\right) = P(Z\le z) $$

My version looks grossly incorrect. How could the CDF of $X$ and $Z$ be the same? Could someone please advise me on what I am doing wrong?

Answer 1

They're not the same because $z$ isn't $x$, but rather is $(x-\mu)/\sigma$.

Answer 2

In their working $\Phi$ is CDF of $Z$.

In your working, you did not define what is $z$. I believe you want to define $z$ such that $z=\frac{x-\mu}{\sigma}.$

We have $$P(X \le x) =P(Z \le z)= P(Z \le \frac{x-\mu}{\sigma}).$$

Nothing wrong about that.