How to Find Average and Median Age From an Aggregated Frequency Table

General comments: If the union of all bins was a finite interval, you could compute a mean only with certain assumptions. A common (though often untenable) assumption is uniformity within a bin.With an open upper-end bin (75), you can't compute an average without some strong assumptions. It would be useful to explore sensitivity of the mean estimate to those assumptions.Usually you can compute a median category, and it's straightforward, so let's begin there.Median: The median age is the age of the "middle person" (or any value between the middle two people if there's an even number - with binned data you want those two to be in the same bin; fortunately it's rare for a bin boundary to be between them, in which case either bin could be regarded as the median bin; you might choose the boundary itself as the median in that case). With 107769 people, the age of the (107769 1)/2 53885-th oldest person is the median age.there are 46259 people aged 34 or younger and 63608 aged 44 or younger, so the median age group is 35-44. You could go further by making some assumptions to try to make an estimate of the year within that - e.

g. if you assume uniform age distribution within bins, the median age would be (53885-46259)/17349 43.96% of the way through the range of ages in that age group, which suggests a median age of about 39.4. However, you would need to assess the reasonableness of that assumption. Being close to the mode with what looks like (and probably is) a fairly smooth distribution, it may not be so bad an assumption for a rough approximationSome books give formulas by which to calculate an estimate of the median which amount to doing pretty much what I just did, such as a formula like this: median $L wfrac(fracn2 c)f$ (where $L$ is the lower limit of the bin containing the median, $w$ is the width of that bin, $n$ is the total population, $c$ is the cumulative count (cumulative frequency) up to $L$ (the end of the previous bin), and $f$ is the count (frequency) in the median bin does pretty much the same thing (aside from the (n1)/2 vs n/2, it is the same).Mean: The mean is most often calculated by treating the data as if it occurred at the bin-centers. For the mean, this is equivalent to assuming the data are uniformly spread within each bin. Clearly this presents a problem with the last category which has no upper bound. Even if you imposed one ("well, let's say nobody lives past 120"), the midpoint is still a terrible estimate of the mean within the group. You can do things like assume the distribution is similar to some population and get estimates from life tables (many countries make life tables available, which allow calculation of the proportion of people alive at each age, say).

You could also simply assume some average (say 80, or 85), and then see how much difference it made. Nine year old (or so) figures from one Western country (one with longer average lifespan than the US) suggests that the average age of males 75 is 82.2 - If you can't get suitable figures, I'd think assuming 82 and trying 80 and 85 to get some idea of sensitivity to the assumption would be reasonable.

(More complicated assumptions than the ones described here are possible but not as often used)

• Related Questions

Help:to prove increasing function (formula and curves plot)

I will take a slightly different approach with this function by relating it to the result established in Help : How to prove the following simple function is decreasing function? .We will call the function in this post $$ f(x) frac(x-1)cdot c^x - 2xcdot c^x-1 x 12^x - x - 1 $$ and the function in the linked post $$ g(x) fracc^x - ccdot x x - 12^x - x - 1 , $$which, since they have the same denominator, we will add to produce $$ h(x) f(x) g(x) fracleft( 1 - frac2c right) x c^x (2 - c) x2^x - x - 1 . $$To save a bit of writing, we will let $ K left( 1 - frac2c right) , $ for which we find that $ 1 2 . $ The second term contains the quadratic factor $ 1 (ln c) x (ln c) x^2 $ , which has the discriminant $ (ln c)^2 - 4 (ln c)^2 1 $ , the term $ - K 1 (ln c) x (ln c) x^2 c^x $ is also always positive.It is the first and third terms here that must be examined more carefully. The factor $ K 1 x ln(fracc2) $ is negative for $ x 2 e^-1/2 approx 1.

213 . $ In the third term, the factor $ 1 - (ln 2) x $ is negative for $ x 2 $ , we conclude that $ h(x) $ is monotonically increasing. From this, we have $ h'(x) f '(x) g '(x) > 0 $ , and since we demonstrated in the linked post that $ g ' (x) - g '(x) > 0 . $$ Hence $ f (x) $ is monotonically increasing.

Similarly to what we observed in the other post, this function has the interesting "parametric" behavior that for $$ f(c) frac(x-1)cdot c^x - 2xcdot c^x-1 x 12^x - x - 1 , $$we have $ f(1) equiv 0 $ and $ f(2) equiv -1 $ . It took some little trouble to find a way to work with the derivative for this function. I had also looked at $ j( x ) f(x) - g(x) $ , but while more terms cancel in this difference function, and it has the pleasant parametric behavior $ j(c1) j(c2) equiv 0 $ , it is not monotonic, but rather has an absolute minimum that shifts with the value of $ c $ , making it very difficult to disentangle information about $ f'(x) $ . .


As a team lead how can I bring changes into the team without resistance from team?

Here's my best process for getting people in charge of subcomponents to agree with me. I had to do the same sort of thing in my first management job:As the leader, this is really your home turf. As the top of this particular section of the org chart, it falls to you to decide how overall resources should be spent and what needs improvement most urgently. Do take the position that the team leads have to convince you that a different issue is more urgent, but be open to being convinced, as this is a two way conversation. They see things you don't... you see things they don't... so you need to collaborate to figure out the real highest priorities.But in the end, the call on what you do and how it impacts your team's obligations is your call. So you need to know:The sketchiness is as important as the plan... here's why:Chances are good that you will get the plan wrong. You'll be overly specific in places where you thought you had a clue and not specific enough in an area that someone on the team will call "absolutely vital". OK - well then the job of your leads is to help you refine the plan. If they won't even try - ask why? Why is this so crazy? Why is it impossible? Why is there no value here? Keep going back to the problem you see and ask for better ideas. Maybe coding standards AREN'T the way to solve this problem or they are too expensive to be feasible. So challenge them to help you fix the problem. I have had very few cases of people refusing to admit that there WAS a problem. However, I've been corrected (many many times) on the fact that the problem didn't have the root cause that I believed. Another reason for that plan to be sketchy --- it gives you the opportunity to throw it out the window without much time investment.The challenge is, they have to convince you. Don't accept a proposition just to avoid confict. Ask questions and get alernate solutions vetted. Be convinced that your team is right before you agree.Once you've got a solution, you can hammer out the actual plan - maybe it's fixing the plan you walked into the meeting with. Maybe it's creating a whole new plan together...The goal here is to be just flexible enough to get the job done. Know what you care about and what you don't care about. Know what you're willing to spend money and time on and what is unreasonable.There are times when people will argue for no obvious reason. They may be overly emotional, illogical or simply non-sensical. This is when one on one meetings are a good backup. People will voice fears and concerns differently when talking to you privately. And sometimes smart people with good ideas won't voice them in meetings.

When you're getting truly inexplicable resistance, eliminate the chaos of group communications and take it to a one on one. It also lets you force feedback. There is nothing like asking a provocative question and then waiting in awkward silence until your person steps up and says something - particularly since your "awkward silence" may be your persons "moment of blesssed relief" while they collect their thoughts and come up with something truly insighful. It's hard to pull this off in a team setting, since all team members add to the mix.

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