How to calculate Z score without standard deviation

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I have been scratching my head about this for ages. I am studying intro stats for a Psychology Masters, and I can't get my head round this example problem:

Example S.6.1 from https://newonlinecourses.science.psu.edu/statprogram/reviews/statistical-concepts/proportions

I get Z scores and all that, but they seem to have calculated the Z score here without a standard deviation, which has me flummoxed and I can't find a good explanation of how this works anywhere online. Grateful if someone could explain it to me!

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If the proportion of people with a given characteristic is $p_0$, then when you take a sample of size $n$ then (under the right assumptions) the number of people in your sample with the characteristic is a Binomial random variable, i.e. $n_0 \sim Bin(n, p_0)$.

Under that model, we know that $E(n_0) = n \times p_0$ and $Var(n_0) = np_0(1 - p_0)$. If we want to look at the proportion of the sample rather than the straight counts, then we divide $n_0$ and $E(n_0)$ by $n$ and $Var(n_0)$ by $n^2$, which gives the required values for the text.