A statistical question
1. Let's say that our midwife has 4 clients a month (i.e. 4 births a month)
2. Let's say that the due dates are uniformly distributed over the month (i.e roughly one due date a week)
3. Let's say that the distribution of actual births around the formal due date is .... I do not really know, but you can click here or here
4. Let's say that the length of "a birth" (from the moment the midwife shows up) is 8 hours
THEN
what is the probability that our midwife will have two active births at the same time exactly?
hmmmmm
posted by Yori's @ 10:43 PM
3 Comments:
The average number of clients per day will be 4/30. The distribution of actual births around the formal due date is unimportant here, because if the due dates are uniformly distributed and the actual births are independent of each other, then the distribution of actual births will also be uniform.
Given an average number of clients, the actual distribution will be Poisson (as long as everything is independent of everything else and uniformly distributed and so on). So the probability of exactly two births at the same time will be exp(-4/30)*(4/30)^2/2, which is 0.0078. This is however not what you want - what you want is the probability of having two or more active births simultaneously, which is 1-p(0)-p(1), where p(0)=exp(-4/30) and p(1)=exp(-4/30)*(4/30). The answer is 0.0081. So you have about 1% chance of having more than one birth at the same time.
This is still not exactly what you want to know, because is states the probability from the point of view of the midwife. Instead, what you want to know is what is the chances that while you are having birth, another birth will take place. For this, the question is what is the probability that on a given day (specifically, the day you are giving birth), another birth will happen. Assuming that the midwife has 3 clients in addition to you, this means that the average number of 'other clients per day' is 3/30. The probability of having at least one of them on the day you chose for your birth is 1-exp(-3/30), which is 0.095 - almost 10%.
From the point of view of the midwife, the rate of occurrence of cases in which she has two or more births simultaneously will be about 1%. However, the chances of having to deal with a second birth or more while she is already occupied is 10%. roughly speaking, the chances of having a birth on a certain day is 10%, so the chances of having two is 1%. You are interested in the first number (10%), not in the second one (1%).
Our midwife said 2 birhs at once did not happen to her in 25 years. This is not consistent with 10%, or even with 1%
I do not understand why the distribution of births around the due date is unimportant... If, let's say, 100% of the births would occur exactly on the due date, then the probability for colision would be 0% - if it is really uniformly distributed, then of course the probability is higher.
The standard deviation of the due date is of at least one week, possibly two weeks (from the papers you referred in the blog).
Therefore, even If the due dates are every sunday, still the distribution of actual birth days would be rather uniform over the month. So the answer is that your nurse should have encountered about one double birth per year. This is true for long births - it may well be that most births are indeed shorter. Her claim that in 25 years and 3000 births she never had a double birth is however unlikely (but not impossible, of course!). It may well be that in cases where double births were possible, she moved patients to colleagues and things like that. Such manipulations are of course outside the validity of the statistical calculations.
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