Couldn't you solve this by using binomial distribution?
* In hospital A, the number of boys follows a binomial distribution with n = 230 and p = 0.51
* In hospital B, the number of boys follows a binomial distribution with n = 560 and p = 0.51
If you know the formula for binomial distribution, you can just plug in the values and compute:
* P(55% boys in hospital A) = ... ?
* P(55% boys in hospital B) = ... ?
I haven't done the computations myself, but intuitively I would say that your final paragraph makes sense. The larger the sample size is, the closer we expect the real mean and the sample mean to be. To have a sample mean of 55%, we need more variation, since this is different from the 51% that we expect the sample mean to converge to.