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!

  • $\begingroup$ How do you compute standard deviation on a probability? You don't. There's an alternate definition of z-score for probabilities $\endgroup$ Jul 16, 2019 at 22:40
  • $\begingroup$ OK so I think I get this now.. Basically the variance of any distribution is sum of squares / sample size, but if you have a predictable distribution function (e.g. binomial, Poisson, whatever) you can calculate the variance from the distribution function essentially, so you don't need to know the std dev. In the example given the sample size is massive so you can assume it is binomial distribution. Is that right? $\endgroup$
    – Tzen
    Jul 17, 2019 at 8:14
  • $\begingroup$ Essentially, that's the right premise, but I'd suggest reading into the subject a bit more. $\endgroup$ Jul 17, 2019 at 16:01

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.