(I hope the question in in scope, please see my question on Meta about that)

My 15 yo son (2nde in France, this is the first year of the equivalent of a High School) is going through basic statistics. One of the exercises in his book made me wonder about why some simplifications are made in an exercise.

While I would love to be corrected and learn something, the core of my question is rather whether this exercise is

  • simplified because it will help to understand something later (and one can live with the simplification)
  • simplified because this is the right thing to do (this is the moment where I would learn something about statistics)
  • a bad exercise (and the problem is over)

The problem(1) is :

In 2016 women were asked about their age at the time they got married. The table below contains their answers:

$$ \begin{array} {|r|r|}\hline Age & 20 \le A \lt 30 & 30 \le A \lt 40 & 40 \le A \lt 50 & 50 \le A \lt 60 & 60 \le A \lt 70 \\ \hline Amount & 10 & 35 & 15 & 2 & 1 \\ \hline \end{array} $$

Calculate the mean age of marriage for these women. Calculate the standard deviation of this series.

The answer then proceeds with

In order to determine the mean, we assume that a person who is between 20 and 30 years old is 25 years old.

Then the problems is solved as expected for a series.

The assumption is, I believe, wrong, it depends on the distribution within the range. There are no reasons for the distribution to be symmetric.

A very similar problem was given later, where the range were apartment surfaces. For the range 0 to 20 m2, I expect that I should have stated that the average is 10 sqm, which makes even less sense as there obvious limits on the lower size of an apartment and the distribution will be heavily skewed to the right.

My problem with these exercises is that they make an artificial issue (= having to decide on the value to use in the ranges) while it is easy to have exercises such as "satisfaction level from 1 to 10 over a month, calculate the average & deviation" or "high jump - which athletes are more consistent in their results? (some would have low and high results, some more around the average for instance)).

My core questions are:

  • Was there in the exercise I quoted a reason to do the assumption?
  • Is this a good assumption?
  • In what will it help later in their curriculum?

(1) Déclic 2nde Math, Programme 2019, Hachette. Ex. 3 p. 311

  • 3
    $\begingroup$ Maybe they mean: "we assume that a person who is between 20 and 30 years old is 25 years old, on average". Such an assumption would make perfect sense. $\endgroup$ Commented Apr 25, 2020 at 12:09
  • 1
    $\begingroup$ I wonder if it's a particular problem for ages 30 -40. I would guess that more 30-35 year-olds get married than 35-40 year-olds. $\endgroup$
    – Amy B
    Commented Apr 26, 2020 at 8:41

8 Answers 8


This is a real-life situation: Sometimes you receive data in groups (bins) like this; that's a pretty common result from using automatable multiple-choice survey forms. There is no way to retrieve the original discrete data points, or the distribution of the data in the bins. So what are you going to do?

Formulas for generating statistics from grouped data like this are very standard. Here's the presentation in Weiss, Introductory Statistics, Sec. 3.2:

Weiss: Grouped-Data Formulas

Note the text that says, "these formulas yield only approximations to the actual sample mean and sample standard deviation", which is accurate and addresses the OP's point. Each term $x_i$ "denotes class midpoint", matching the OP's given exercise. In some sense, this represents the average of all possible distributions over the unknown individual bin data, so it's the best approximation we can make in the general case (or at least a reasonable default model, such that the burden of proof is on one suggesting some different improved model).

Is this some kind of error or mistake? No; the whole essence of statistics is to take limited data about the world, and make some kind of estimate or approximation to the larger truth that we can't see. Deductive reasoning is relatively easy; inferential reasoning is much harder, and that's exactly what the field of statistics attempts to formalize. Recall the famous words of Bishop Joseph Butler (1736):

Probable Evidence, in its very nature, affords but an imperfect kind of Information; and is to be considered as relative only to Beings of limited Capacities. For nothing which is the possible object of Knowledge, whether past, present, or future, can be probable to an infinite Intelligence; since it cannot but be discerned absolutely as it is in itself, certainly true, or certainly false: but to us, Probability is the very Guide of Life.


The assumption is, I believe, wrong, it depends on the distribution within the range. There are no reasons for the distribution to be symmetric.

It's not wrong, it's an approximation. It's equivalent to the rectangle rule for approximating an integral: https://en.wikipedia.org/wiki/Numerical_integration

It's good in my opinion that your kid's text does an example like this where you have to come up with an approximation. Students should be exposed to the realities of life, not sheltered from those realities by only having them work on sanitized examples that fit perfectly with some theoretical framework.

It would be good if the book correctly described it as an approximation.

  • 2
    $\begingroup$ +1; I was about to explain it in such calculus-based terms, but you've saved me the need to do it. The only thing I'd add is that this approximation is reasonable for sufficiently narrow interviews. $\endgroup$
    – J.G.
    Commented Apr 25, 2020 at 19:39

It is worth considering that, if the ages would have been recorded as integers, rather than intervals, the assumption would have still been wrong in a similar but less obvious way. That is, a 25 year old and a 25.25 year old would have been grouped together and the 'binning' will affect the mean to some extent.

Any finite presentation of numerical data, such as ages, will bin them into imprecise groups and will produce the same issue. The important thing to recognize is that the end result has error bars, and we can calculate them! You can simply replace the midpoint with the lower and upper ends of the intervals to get the minimum and maximum possible means.

while it is easy to have exercises such as "satisfaction level from 1 to 10 over a month, calculate the average & deviation"

Though these seem numerical, one could argue that these are actually categorical data and that means and deviations are not meaningful. For example, imagine the 3 state satisfaction scale happy :) neutral :| sad :(

What is the mean of :) and :| ?

  • $\begingroup$ Since you could argue that :( < :| < :) and that the delta between sad and neutral is about the same as between happy and lucky (by definition so, as neutral is supposed to be in the middle), I'd rather say the smiles are pictographic representations of the selectable integer values, and that other locations on the numerical scale simply lack such representations. $\endgroup$ Commented Apr 27, 2020 at 20:11
  • $\begingroup$ @ZsoltSzilagyi I agree that there is some ordering and that you could argue that the differences are the same, but you could also argue that the differences are different. Different personalities or cultures might be biased to :) or :( or even :| Saying that :| is in the middle doesn't mean that people respond that way. $\endgroup$
    – Adam
    Commented Apr 29, 2020 at 18:40
  • $\begingroup$ Good observation, though you have cultural biases even in numeric scales. "How much do you like that (mediocre) vendor?" might give you a harsh 3 in Eastern Europe and a polite 8 in Japan. $\endgroup$ Commented May 4, 2020 at 15:30
  • $\begingroup$ @ZsoltSzilagyi This is actually the point I was trying to make. $\endgroup$
    – Adam
    Commented May 4, 2020 at 16:31

The situation you give is:

A woman got married when she was between 20 and 30 years old. At what age did she get married?

But let me give a different situation:

A woman is between 20 and 30 years old. What is her age?

Given no information, assume the value is in the middle.

For the situation I gave, a good estimate is 25 (midway). Why not, say, 22? Because we have no idea if the person is relatively young (nearer 20). Why not, say, 27? Because we have no idea if the person is relatively old (nearer 30). We do not know "the distribution within the range." Is the distribution asymmetric? If so, should you assume it is positively skewed? Why not assume it is negatively skewed? There is no information. So we assume the skewness is zero (in the middle).

If there is information, then the middle might not be a good assumption.

If I say the height of a certain person is between 5 feet and 15 feet, "common sense" tells us that 10 feet is not a good assumption. This is because in our experience, there are no people who are ten feet tall. There is information in this case, even though it is not explicitly stated. Experience tells us that the majority of people who are 5 or more feet tall are not 7 or more feet tall, so probably a good guess for the height is 6 feet.

In the situation that you gave, what is the information?

The woman got married when she was between 20 and 30 years old. Other women got married at an older age. And, it seems (from your data), that no other women got married when they were less than 20 years old. So a person with "common sense" might assume that "the middle" in this case is more than 25; there are fewer women who got married at a younger age and more women who got married at an older age. Okay, so it's more than 25. Is it 26? 27? How is a student expected to know?

You assume that the student doesn't know. You assume that there is no information. You assume that the value is in the middle.

  1. I don't think it's a good example, given the students ability. Better to give them simpler examples and just work with crunching numbers. Since they're just encountering the topic for the first time, no reason to draw in concepts like good/bad/middling assumptions in data analysis.

  2. Nevertheless, I don't think it's the end of the world that your boy had a suboptimal question given to him. Many commenters here are logicians by training and mindset, therefore expecting perfection, often in areas not of the most importance. But life goes on. C'est la vie. ;-)

  3. I deal with datasets like this, for practical use, at times. For instance, here is another similar data set. EIA data on L48 oil production by EIA gravity bin:


    You can see that 40-45 is the largest bin, but that the data is non-normal (skew). And it's even fuglier if you try looking at an individual state. (The data is only collected in 5 degree bins so that is as good as it gets, absent doing your own, very expensive survey. And then the stat data is obscured for reasons of commercial confidentiality given some states are dominated by a small number of producers, buyers.)

    In addition, you have the confusion of a greater than or less than bin at the end of the distributions that is not an exact 5 degree bin, but extends to 0 or infinity (in theory, but practically to 8 or 70 or so in terms of known crude strains). I think I used 2.5 degrees below the LT or above the GT as an approximation for those bins.

    I have actually crunched this data, a couple ones. One is to just assume midpoint. The next, probably better, is to linearly weight versus the surrounding bins. However, the answer I got was within a half degree of the simpler assumption.

    You could even do higher order fits, but I would caution against it given the small degrees of freedom. For that matter, I think even the linear weighting introduces an unfortunate complexity when discussing the answer and perhaps not worth the "chaff" given that a half degree is not important functionally in terms of insight. For that matter basic trend analysis over time is also insensitive to the choice of within bin average.

    [Note that API gravity is actually sort of a bastard's reciprocal of density, so you can't/shouldn't average it per barrel anyways. That said, you can convert to densities and back again. And it doesn't "change the story" in terms of useful insights either.]

    But all this sort of minutia/thinking is a distraction to kids that should just be learning to add the stuff and divide, for now.


The approximation isn't bad — in fact, it's quite reasonable, if the binned data is all we have:

Follow-up exercise for advanced students (and curious parents): show that the answer you got using the approximation must be within ±5 years of the true average, regardless of how the actual ages of marriage are distributed within the ranges. Can you provide any intuitive arguments for why the actual error is likely to be considerably less than that? (Try to come up with at least two reasons.)

As for the pedagogical merit of using such an approximation in this exercise, I would say it depends on how it's presented in the broader context of the course.

It's good that students are exposed to such approximations, because in the real world we have to make them all the time. If we couldn't do that, and be reasonably confident that the answer is still close to correct, statistics would be all but useless.

On the other hand, I would say that it's not good if the students are simply told to use that particular approximation without being given any intuition on why it's a reasonable or safe thing to do.

In particular, if the student is left with the same impression as you seem to have been, i.e. that the approximation is being made just because that's what you're instructed to do in this math class, even though it intuitively feels wrong to them, it can contribute to the all too common feeling that the mathematics one learns in school is somehow divorced from reality, just a collection of meaningless rote formulae only useful for passing the exam.*

Ideally, either the textbook or the teacher of the class (or, preferably, both) would've taken a moment to mention that taking the midpoint of the range is indeed an approximation, and that the average obtained using it won't be exactly right, but that it also won't be too far from the true average, either. They could also use something like the follow-up exercise I suggested above to illustrate that, perhaps as a shared in-class discussion activity.

In the context of such a discussion, it could also be good to discuss why, in practice, data from surveys like this is often binned into such wide ranges, and why someone might feel uncomfortable providing e.g. their exact birthdate or the date of their marriage in a survey like this. I would also recommend emphasizing the fact that, in practice, all data is approximate, and that even if we knew the ages of marriage in the exercise down to, say, one year, that would still be an approximation — just a more fine-grained one than grouping them in 10-year ranges.

(As a slight tangent, it might also be worth noting that, if we followed the common everyday practice of rounding ages down to a whole year before taking the average, that would introduce a systematic bias to the average. For statistical purposes, a person known to be at least 22 but less than 23 years old should really have their approximate age counted as 22.5 years.)

Of course, in case your son's math textbook and/or their teacher haven't explicitly brought this up, this is also a good opportunity to do it yourself while going through the exercises with your son. You could even bring it up in a discussion with the teacher — not confrontationally, but just mentioning that you found the suggested assumptions in this particular exercise a bit confusing, and that it might be something worth discussing in class, if they haven't already done so.

*) A topic that I've previously commented on here in the context of unreasonable word problems.

  • Was there in the exercise I quoted a reason to do the assumption?

Yes, as pointed out in the other answers, it gives a simple way to answer the question and is clearly the intended solution.

  • Is this a good assumption?

No. Because you don't have the actual data, the mean and standard deviation are actually random variables, so it is not even clear what the question is asking. In order to calculate the expected value of the mean and standard deviation, you need to make some assumptions about how the data are generated.

For example, you could use the maximum entropy assumption, and assume that each sample of ages is uniformly distributed in the given range. (There are strong arguments that this is what you should do if you have zero knowledge.) This will give the same formula for the expected mean as in the question, but the expected standard deviation will be smaller than the standard deviation you get from assuming that all the ages are equal to the midpoint of the range.

Alternatively, you can assume that the ages are randomly sampled from a particular underlying distribution, say a beta distribution in this example, and then use inference to fit this distribution, from which you can get the expected mean and standard deviation.

As you point out in your example of apartment area, your choice of distribution will depend on your knowledge of the subject matter.

Having said this, I admit that in a lot of cases, you will see that the naive formulas do actually give results that are pretty close to a more statistically-valid approach. There are some examples on Crossvalidated.

  • In what will it help later in their curriculum?

In real life, it is very common to have anonymised data. Statistical agencies often bin data like this in order to reduce the danger of people being identifiable (e.g. if I have your birth year, it is easier for me to steal your identity than if I only know your age to within 10 years).

So, it is very likely that the students will have to deal with data like this in real life. Therefore, it is good to give them a quick and dirty strategy for dealing with it, rather than just giving up or doing something excessively complicated.

However, the standard deviation part of the question could be improved by mentioning that in practice, statisticians would probably use something like Sheppard's Correction to get a more accurate estimate of the standard deviation.


As some other answers have mentioned, this kind of assumptions may not be statistically valid but are nevertheless useful in their simplicity and convenience for calculation. A more accurate term for them is "heuristic". Note that if nothing at all is known about the distribution, then it is an unbiased heuristic. However, if we expect the underlying distribution to be continuous, then it is clearly incorrect, but there is no easy way to get the 'correct' answer. For example, if the distribution is known to be normal, we can estimate the true mean by finding the best parameter $m$ such that the probability of observing the given data has maximum likelihood. But if the distribution is not known, we still need a heuristic since there are many continuous distributions that would yield the same data on average.


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