I am a nurse working in a hospital, and from my observation, out of $p$ number of babies born in year 2016, 70% of them are girls.

Given this, what are the chances that the ratio of boy:girl remain $50:50$ for new born baby in year 2016? Assuming that the total number of born baby in the world in year 2016 is $X$

PS: I am asking this because I suspect that there might be a deeper reason why more girls are born than boys this year.

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    $\begingroup$ We would first need to know how many births occurred at your facility. $\endgroup$ – Michael McGovern Dec 18 '16 at 4:00
  • $\begingroup$ @MichaelMcGovern, as mentioned in the question, my observation sample is $p$, total number of newborn baby is $X$ $\endgroup$ – Graviton Dec 18 '16 at 4:00
  • $\begingroup$ You gave the total number of births in the world, not the total number of births at your facility. $\endgroup$ – Michael McGovern Dec 18 '16 at 4:02
  • $\begingroup$ The short answer is: given how you have phrased to question, there doesn't seem to be enough information to answer it. $\endgroup$ – Michael McGovern Dec 18 '16 at 4:08
  • $\begingroup$ @graviton, you have mentioned about p babies born in your facility but then you have not mentioned the total no of babies born at your facility at that year rather you have mentioned the total no of babies born in the world so this makes your problem unsolvable $\endgroup$ – Navin Dec 18 '16 at 4:43

Very, very, likely there is no deeper reason.

You are probably experiencing what Daniel Kahneman calls "the law of small numbers" (in his "Thinking, Fast and Slow"). We expect small samples to have the same behaviour as the underlying, large, universe they are drawn from. So we may expect to have also roughly 50:50 birth distribution on a small sample, because we know that is so on the large scale. Yet that is not the case, "strange" things this happen, more so on the small scale.

A couple of quick reasonings.

There are some 135 million births per year (2016 data, from wikipedia). IF you split those in batches of 100 (I do not know how many you have observed), that means we have about 1.35 million batches. Now note that something as (un)likely as 1 in 1.35 million (0.00007%) is quite likely (63%) to happen in at least one of those batches.

Another approach, a quick Bayesian estimate of the birth distribution updated by your observation. Using a prior knowledge of binomial distribution with 67.5 million newborns of each gender, that is a probability of 50% of having a girl on a birth. A updated best guess for the newborn girl probability, using 0.7*p girls out of p births, would be:


For $p=100, P(girl,p) \approx 50.000015%$.

Finaly, last assume in the previous 4 years you had a birth distribution of 45:55, (girls:boys). I bet you would not take any special note of it. Now if you join those more or less normal years and add them to this strange year, you get $(4*45+1*70):(4*55+1*30) \rightarrow 250:250 \rightarrow 50:50$, that is, a completely normal 5 year period.

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I'm not sure this question can be answered as asked, but...

If we assume male/female births are equally likely (not true [males are born more frequently], but close), we would expect p/2 female births with a variance of p/4 and a standard deviation of Sqrt[p]/2.

Percentagewise, this is 50% female births with a 1/2/Sqrt[p] standard deviation.

So, 20% would be (20/100)/(1/2/Sqrt[p]) or (2/5)*Sqrt[p] deviations from the mean.

Thus, if more than 25 babies were born total, having 17.5 (ie, 18) or more females is only 5% likely.

If more than 56.25 (ie, 56) babies were born total, having more than 39.375 (ie, 40) born female is only 0.5% likely.

So, depending on the value of p, this circumstance may or may not be remarkable (though I suspect many other factors apply, so this is only a theoretical answer).

Of course, unless p is very large compared to X (unlikely in a single hospital), the world population, it won't have a significant effect on the world ratio of male/female births.

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  • $\begingroup$ Given the equation $$z_\alpha=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}\left(1-p_{0}\right)}{n}}}$$ where $\hat{p}$ is the observed sample proportion, p_{0} is the expected population proportion, $z_\alpha$ is the critical $z$ at the desired confidence level $\alpha$, and $n$, which value for $p$ are we missing? $\endgroup$ – Michael McGovern Dec 18 '16 at 17:30
  • $\begingroup$ For the purpose of this exercise, we assume that $p=500$. BUt the main point is, I want to derive a formula connecting the probability of this happening with the $p$ and $X$ value. Can this be done? $\endgroup$ – Graviton Dec 19 '16 at 5:13
  • $\begingroup$ @Graviton Yes. For p=500, we'd expect about 250 females with a variance of 125 and a standard deviation of about 11.18. 70% females would be 350 females, which is (350-250)/11.18 or about 8.95 standard deviations above the norm. The likelihood of this happening purely by chance is about 2*10^-19 or virtually zero. However, keep in mind the caveats other people have given before using this result. $\endgroup$ – user2469 Dec 19 '16 at 5:52

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