Compute expected standard deviation from proportion data This example is from the first chapter of Gelman & Hill 2007:


The following are the proportions of girl births in Vienna for each month in     1908 and 1909 (out of an average of 3900 births per month):
.4777 .4875 .4859 .4754 .4874 .4864 .4813 .4787 .4895 .4797 .4876 .4859 .4857 .4907 .5010 .4903 .4860 .4911 .4871 .4725 .4822 .4870 .4823 .4973
The data are in the folder girls. von Mises (1957) used these proportions to claim that the sex ratios were less variable than would be expected by chance.
Compute the standard deviation of these proportions and compare to the standard deviation that would be expected if the sexes of babies were inde- pendently decided with a constant probability over the 24-month period.



The standard deviation is: 0.064
I'm wondering how to calculate the expected standard deviation? Should I use a binomial model? but where do the probabilities come from then?
EDIT:
The next step would be to assess whether the observed (0.0064) and chance (0.0080) deviations significantly differ. 
The exercise gives a hint: "under the randomness model, the actual variance should have a distribution with expected value equal to the theoretical variance, and proportional to a chi^2 with 23 degrees of freedom"
I'm wondering how I could assess whether the two deviations are significantly different. The only solution I can come up with is 
(observed-expected^2)/expected = (0.0064-0.0080)^2/0.0080 = 0.00032 and then compare this to a chi^2 distribution with 23 degrees of freedom. Is this the correct way to do it?
 A: I think your calculated standard deviation for the proportions is $0.0064$ (note extra 0).  
Yes,  to calculate the expected standard deviation, you are probably expected to use a binomial model and the formula $\sqrt{\dfrac{p(1-p)}{n}}.$ You can take the mean of your sample data to give you an estimate of $p$, though taking $0.5$ will give much the same result in practice. $n$ comes from "out of an average of 3900 births per month".
A: To assess whether difference between the observed (0.0064) and chance (0.0080) deviations is statistically significant, you can also compute the 95% confidence interval of the chance standard deviation. This is where the chi^2 distribution with 23 degrees of freedom comes to the picture; you need the quantiles of this distribution, among others, to form this confidence interval. Once you have the confidence interval, check if the observed deviation is inside the interval; if it is, then the difference between the observed and chance deviations is not statistically significant.
A: Very late response, but regarding the goodness-of-fit test to the binomial distribution, it turns out that this
$$
s^2(n-1)/\sigma^2
$$
has a $\chi^2$ distribution with $n-1$ degrees of freedom, where $s$ is the sample standard deviation and $\sigma$ is the expected standard deviation.
The formula above will produce a test statistic whose p-value you can compute against that distribution. Because the observed variance is “too small”, you want the left-tailed probability (ie, the probability of getting a statistic smaller than the calculated value).
