Well the sample mean $\bar{X}$ follows a normal distribution with mean $\mu$ and variance $\frac{\sigma^2}{n}$, i.e. $\bar{X}\sim N(\mu,\frac{\sigma^2}{n}) $ however if you standardize the sample mean (by writing it in the form in the question), you find that it is Normal with mean $0$ and variance $1$, i.e. $Z \sim N(0,1)$.

Well the sample mean $\bar{X}$ approaches a normal distribution with mean $\mu$ and variance $\frac{\sigma^2}{n}$, i.e. $\bar{X}\sim N(\mu,\frac{\sigma^2}{n}) $ however if you standardize the sample mean (by writing it in the form in the question), you find that approaches a Normal distribution with mean $0$ and variance $1$, i.e. $Z \sim N(0,1)$.