Nahuatl names and natality in aztec empire
Budelberger, Richard
budelberger.richard at TISCALI.FR
Wed Jan 21 02:41:00 UTC 2004
30 nivôse an CCXII (le 21 janvier 2004 d. c.-d. c. g.), 03h34.
----- Message d'origine -----
De : Davius Sanctex <davius_sanctex at hotmail.com>
À : <nahuat-l at server2.umt.edu>
Envoyé : samedi 30 décembre 2000 01:09
Objet : Nahuatl names and natality in aztec empire
> A nice mathematical argument relating name "Teyacapan"
> to average number for children in Aztec Empire.
> ____________________________
> Bob McCaa points that in a census of 1540 in a group
> of 1205 women, 313 were named "Teyacapan" (= first
> born), i.e. 25,04%:
>
>
>
> I will show that these data implies that the average number per family
> was at most 5,135
> and that 6,88%
0,588 %
> of couples had not any child!
>
> _________________________________________
> This number can be related to the average number of
> children in a family. We assume:
>
> 1) Nearly all first born female babies was named
> "Teyacapan"
> 2) The probability of borns by unit of time remain
> uniform for a community and population is stationary.
>
> First step:
> Second hypothesis implies borns can be well modelized by
> a Poisson distribution, thus the probability of a couple
> to have k kids is P(k):
>
> P(k)= exp(-m)*(m^k)/k!
>
> [Where m is the average number of children]
>
> Second step:
> Thus if the probability of a child to belong to a familiy
> with exactly k kids is p(k):
>
> p(k) = P(k)/(1-P(0))
>
> [p(0) is the % of families that have no kid].
>
> Thirst step:
> If we take a woman at random the probability of being the
> first kid in a family of k kids is just 1/k [= q(k)].
> And thus the probability that a woman to be the first baby
> of a family is r:
>
> r = p(1)*q(1) + p(2)*q(2) + ...+ p(k)*q(k) + ... =
> = P(1)/1 + p(2)/2 + ...+ p(k)/k + ... = 25,04%
>
> This last equation enable us to evaluate m. For m = 5,135
>
> k P(k) p(k) p(k)/k
>
> 0 0,00588 __
> 1 0,03022 0,03039 0,03039
> 2 0,07759 0,07805 0,03902
> 3 0,13282 0,13361 0,04453
> 4 0,17053 0,17154 0,04288
> 5 0,17515 0,17618 0,03523
> 6 0,14991 0,15079 0,02513
> 7 0,10997 0,11063 0,01580
> 8 0,07059 0,07101 0,00887
> 9 0,04028 0,04052 0,00450
>
> sum 0,2498 = 24,98%
>
> This shows that the average number must be of order 5,135.
> Moreover, of this table we deduce that 5,88 % = P(0)
P(0) = 0,588 %
> of couples have not babies and the majority (17,51%) have 5 babies.
> The number of families with 9 is 4,02% ...
> _________________________________________________________________________
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