What is the geometrical difference between $Z_{\alpha/2}$ and $E_{a}$?
Let's say we have a generator of toys and the weight is distributed with a standard deviation of $4kg$ and a mean of $5kg$ for example.
I know that the interval of confidence is defined as $[\mu+E_a,\mu-E_a]$ being $E_a=Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ the random error and $\sigma$ the standard deviation. Which then the extreme values of that interval, are $\mu+E_a$ and $\mu-E_a$, which then I suppose $E_a$ is a value with units of $kg$, and then since those are the extreme values, I'm assuming that $E_a$ are these values that would be showed in the graph that limit the area $1-\alpha$:
But then I think, then what is $Z_{\alpha/2}$? Since I thought that it was literally the limit that would separate the area $1-\alpha$. Is $Z_{\alpha/2}$ just the equivalent to the above in a distribution that represents $f(X)$ and not $f(\overline{X})$? Or is it just the equivalent to the above in a normal distribution $N(0,1)$?
Also I notice that the graph it is delimited by $X_{\alpha/2}$, so is $X_{\alpha/2}=E_a$ and then $Z_{\alpha/2}$ is just $Z_{\alpha/2}=\frac{X_{\alpha/2}-\mu}{\sigma}$. In general this question is for me to join spots and make it have sense.