How many grandmothers will die in two months?
December 8, 2009 10:55 PM Subscribe
Actuarial / statistics geeks: a puzzle for you.
You are on a two month training program with a group of Korean teachers of English.
There are 20 teachers, ranging in age between late 20s to early 50s.
How many of their grandmothers are likely to die during the course?
Consider the current months (October-November) as the timeframe. Assume the grandmothers are all residents of Daegu, Korea.
I probably cannot provide any more specifics than those, but if you want clarification I will try.
You are on a two month training program with a group of Korean teachers of English.
There are 20 teachers, ranging in age between late 20s to early 50s.
How many of their grandmothers are likely to die during the course?
Consider the current months (October-November) as the timeframe. Assume the grandmothers are all residents of Daegu, Korea.
I probably cannot provide any more specifics than those, but if you want clarification I will try.
Response by poster: Sorry, but this is entirely outside of my area of expertise. I've never heard of the term "mortality table", so I guess my answer is "whichever one seems right to you given the question".
posted by Meatbomb at 11:37 PM on December 8, 2009
posted by Meatbomb at 11:37 PM on December 8, 2009
Well I'm just about to head to bed, so I'll lay out the actuarial principles and then anyone who's taken a freshman stats course can take over and flesh out the details.
A quick google yielded this WHO link -- select Republic of Korea to get their 2006 Life Table.
From left to right, the first 3 columns in these tables are:
- Age Range: self-explanatory
- nMx: portion of population that will die within this age range. E.g. 0.12622 (i.e. ~13%) of Korea females will die after reaching age 85 and before reaching age 89.
- nqx: portion of population that will be dead having reached the upper bound of this age range. E.g. 0.47971 (i.e. ~48%) of Korean females die before reaching age 89.
The rest aren't really of much interest to you.
So let's say that your age range is 25-54. To calculate the probability that a Korean female will die after age 25 and before age 55:
1) Add up nMx for 25-29, 30-34, etc, 50-54
OR
2) Subtract nqx for 20-24 from nqx for 50-54.
Either way should yield the same number.
After that, it becomes a binomial probability calculation adjusted for your 2-month window.
Wikipedia says that Daegu has 2.5 million people -- more than enough to justify it collecting data for local life tables. Especially in union-dominated Korea where defined-benefit pensions are still common. You're on your own for sourcing that info.
posted by randomstriker at 12:04 AM on December 9, 2009
A quick google yielded this WHO link -- select Republic of Korea to get their 2006 Life Table.
From left to right, the first 3 columns in these tables are:
- Age Range: self-explanatory
- nMx: portion of population that will die within this age range. E.g. 0.12622 (i.e. ~13%) of Korea females will die after reaching age 85 and before reaching age 89.
- nqx: portion of population that will be dead having reached the upper bound of this age range. E.g. 0.47971 (i.e. ~48%) of Korean females die before reaching age 89.
The rest aren't really of much interest to you.
So let's say that your age range is 25-54. To calculate the probability that a Korean female will die after age 25 and before age 55:
1) Add up nMx for 25-29, 30-34, etc, 50-54
OR
2) Subtract nqx for 20-24 from nqx for 50-54.
Either way should yield the same number.
After that, it becomes a binomial probability calculation adjusted for your 2-month window.
Wikipedia says that Daegu has 2.5 million people -- more than enough to justify it collecting data for local life tables. Especially in union-dominated Korea where defined-benefit pensions are still common. You're on your own for sourcing that info.
posted by randomstriker at 12:04 AM on December 9, 2009
Very impressive, by the way, that more than 40% of Korean males and more than 50% of Koreean females will live past age 90.
posted by randomstriker at 12:10 AM on December 9, 2009
posted by randomstriker at 12:10 AM on December 9, 2009
Oh, I just misread the question...it seems you're asking about the teacher's grandmothers, not the teachers themselves!!!
That introduces a whole bunch of other assumptions and calculations based on the distribution of age differences between teachers and their grandmothers.
If you need a real answer, I think this data would be very hard to source.
posted by randomstriker at 12:14 AM on December 9, 2009
That introduces a whole bunch of other assumptions and calculations based on the distribution of age differences between teachers and their grandmothers.
If you need a real answer, I think this data would be very hard to source.
posted by randomstriker at 12:14 AM on December 9, 2009
If the ages of the grandmothers are known and do not have to be calculated based on trends, then it becomes an easy problem again, solved by the same method that I provided.
posted by randomstriker at 12:20 AM on December 9, 2009
posted by randomstriker at 12:20 AM on December 9, 2009
It should also be pointed out that if the answer is x and your actual experience is 2x (or 10x or whatever), it doesn't mean something nefarious has occurred. You've just experienced an above-the-curve time span.
posted by yerfatma at 4:07 AM on December 9, 2009 [1 favorite]
posted by yerfatma at 4:07 AM on December 9, 2009 [1 favorite]
We'll also need to know the standard deviation to get a meaningful answer.
posted by Obscure Reference at 5:54 AM on December 9, 2009
posted by Obscure Reference at 5:54 AM on December 9, 2009
If the sample statistic is a proportion P, then its standard deviation is sqrt(P(1-P)) and its SE is sqrt(P(1-P)/N).
posted by ROU_Xenophobe at 6:00 AM on December 9, 2009
posted by ROU_Xenophobe at 6:00 AM on December 9, 2009
Best answer: It should also be pointed out that if the answer is x and your actual experience is 2x (or 10x or whatever), it doesn't mean something nefarious has occurred.
Indeed. If you're managing a bunch of English teachers and you're suspicious because more than a couple of them are simultaneously absent due to "a death in the family", then a statistical projection is not the right way to confirm the veracity of their claims.
The Law of Large Numbers yields the probability figures that are used in the suggested calculations. But on an individual level, events and non-events are far from evenly distributed. You might roll a whole bunch of sixes in a row, but that does not mean you have fixed dice unless this happens thousands or millions of times.
Sorry but if you really want to know if something is up with your teachers, you're gonna have to figure it out the old-fashioned way...i.e. ask for documentation.
posted by randomstriker at 10:13 AM on December 9, 2009
Indeed. If you're managing a bunch of English teachers and you're suspicious because more than a couple of them are simultaneously absent due to "a death in the family", then a statistical projection is not the right way to confirm the veracity of their claims.
The Law of Large Numbers yields the probability figures that are used in the suggested calculations. But on an individual level, events and non-events are far from evenly distributed. You might roll a whole bunch of sixes in a row, but that does not mean you have fixed dice unless this happens thousands or millions of times.
Sorry but if you really want to know if something is up with your teachers, you're gonna have to figure it out the old-fashioned way...i.e. ask for documentation.
posted by randomstriker at 10:13 AM on December 9, 2009
Response by poster: yerfatma and randomstriker: yes, I agree and understand. This question was initially generated by a slight WTF moment with the absences of a couple teachers during the course.
And now I realize that any number will need a lot of guessing / assuming, and will be so ballpark as to be kind of useless, it seems.
But my main motivation now, going forward, is to have a more realistic expectation about this sort of thing... Given a group of 20 people, over a month (2 months, 3 months...) how likely is it that there will be (1,2,10...) family deaths? If I knew that statistically one or two is likely it will be something to accept and plan for more easily...
posted by Meatbomb at 1:04 PM on December 9, 2009
And now I realize that any number will need a lot of guessing / assuming, and will be so ballpark as to be kind of useless, it seems.
But my main motivation now, going forward, is to have a more realistic expectation about this sort of thing... Given a group of 20 people, over a month (2 months, 3 months...) how likely is it that there will be (1,2,10...) family deaths? If I knew that statistically one or two is likely it will be something to accept and plan for more easily...
posted by Meatbomb at 1:04 PM on December 9, 2009
Best answer: Statistics are a useful tool for some scenarios. This is clearly not one of them.
A sample size of 20 is too small and an event window of 2 months is too short to make a meaningful comparison with population trends.
posted by randomstriker at 4:25 PM on December 9, 2009
A sample size of 20 is too small and an event window of 2 months is too short to make a meaningful comparison with population trends.
posted by randomstriker at 4:25 PM on December 9, 2009
Consider, for example, the birthday paradox:
In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will have the same birthday.
Such statistical realities are counter-intuitive "WTF moments" for the average layman. So you should tread carefully before making judgements about your English teachers.
posted by randomstriker at 4:30 PM on December 9, 2009
In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will have the same birthday.
Such statistical realities are counter-intuitive "WTF moments" for the average layman. So you should tread carefully before making judgements about your English teachers.
posted by randomstriker at 4:30 PM on December 9, 2009
Response by poster: That is why I talk to random Internet strangers about it rather than the people themselves, randomstriker. Thanks for your input. :)
posted by Meatbomb at 6:05 PM on December 9, 2009
posted by Meatbomb at 6:05 PM on December 9, 2009
Best answer: Taking the figures from the link posted by randomstriker, we can do some handwaving:
If the teachers are in their late 20s, then their grandmothers are probably no younger than 70. Probability of living to age 70: 0.88. Probability of dying over a two months period (on average) between 70 and 74: 0.0027. The chance of a 70 year old grandmother being alive initially and dying in the two months is the product of the two.
For age 70: 0.88 * 0.0027 = 0.0024
For age 75: 0.81 * 0.0052 = 0.0042
For age 80: 0.68 * 0.0097 = 0.0066
For age 85: 0.49 * 0.016 = 0.0078
For age 90: 0.25 * 0.022 = 0.0055
For age 95: 0.09 * 0.026 = 0.0023
So the highest chances are if all their grandmothers are (or would have been) 85 years old, with each grandmother death occurring with p = 0.0078. Taking 20 of those gives us about a 13% chance of having one grandmother death in the two months, 1% of two deaths, and a 0.05% chance of three. Assuming the grandmother deaths are independent.
posted by parudox at 11:05 PM on December 9, 2009 [1 favorite]
If the teachers are in their late 20s, then their grandmothers are probably no younger than 70. Probability of living to age 70: 0.88. Probability of dying over a two months period (on average) between 70 and 74: 0.0027. The chance of a 70 year old grandmother being alive initially and dying in the two months is the product of the two.
For age 70: 0.88 * 0.0027 = 0.0024
For age 75: 0.81 * 0.0052 = 0.0042
For age 80: 0.68 * 0.0097 = 0.0066
For age 85: 0.49 * 0.016 = 0.0078
For age 90: 0.25 * 0.022 = 0.0055
For age 95: 0.09 * 0.026 = 0.0023
So the highest chances are if all their grandmothers are (or would have been) 85 years old, with each grandmother death occurring with p = 0.0078. Taking 20 of those gives us about a 13% chance of having one grandmother death in the two months, 1% of two deaths, and a 0.05% chance of three. Assuming the grandmother deaths are independent.
posted by parudox at 11:05 PM on December 9, 2009 [1 favorite]
Best answer: Scratch that. Since each person has two grandmothers, the probability that at least one of them will die is basically doubled to p = 0.0155. That brings it to a 23% chance of one grandmother death, 3% chance of two, 0.3% chance of three, and 0.02% chance of four deaths.
posted by parudox at 11:41 PM on December 9, 2009
posted by parudox at 11:41 PM on December 9, 2009
Response by poster: Very thank you parudox, those ballpark figures are just the thing I was needful of. So 3 unrelated dead grandmothers is certainly an unlikely outlier case.
posted by Meatbomb at 1:00 AM on December 10, 2009
posted by Meatbomb at 1:00 AM on December 10, 2009
Very thank you parudox, those ballpark figures are just the thing I was needful of. So 3 unrelated dead grandmothers is certainly an unlikely outlier case.
Not so hasty now, remember the all important caveat:
Assuming the grandmother deaths are independent.
*Cue ominous granny slaughtering music*
posted by atrazine at 2:29 AM on December 10, 2009
Not so hasty now, remember the all important caveat:
Assuming the grandmother deaths are independent.
*Cue ominous granny slaughtering music*
posted by atrazine at 2:29 AM on December 10, 2009
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posted by randomstriker at 11:17 PM on December 8, 2009