| If you believe that age is discrete (you are 0 years old from the day you are
born until your first birthday, then 1 year old until your second birthday,
etc) then if you denote the probability of dying at age A as p(A), then the
specified conditions together with sum(p(A))=1 give you a solvable set of
simultaneous equations whose solution is:
p(0) = 1/20
1 - sum(I=0 to A-1)(P(I))
p(A) = --------------------------, A > 0
20
i.e. 5% of the people die every year, independent of age. I believe that if you
treat age as a continuum, you will get a similiar answer in continuous terms,
such as p(A) = exp(-A/20)/20, but I haven't worked it out.
BUT, it isn't possible to figure out what percentage of the population are
teenagers without some other condition, e.g. that the population is constant,
since the problem conditions and the above derivations would hold true even if
there were no births in Teeny-Bop in the last 20 years!
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I agree with Richie's approach in .1. and let's assume that the
population distribution is static.
Let p(x) be the probability density function for a person's age.
We are told that int[x=y,%]x.p(x).dx = (y+20).int[x=y,%].p(x).dx.
Differentiate by y to get:
-y.p(y) = -(y+20).p(y) + int[x=y,%].p(x).dx
20.p(y) = 1-int[x=0,y].p(x).dx
Let
q(y) = int[x=0,y].p(x).dx
20dq/dy = 1-q
20ln(1-q) = -y+C
When y=0,q=0, so C =0.
Therefore:
q = 1-exp(-y/20),
p = exp(-y/20)/20.
and in particular:
q(20)= 1-1/e
Multiply this by 100 to get the percentage of the population which are
teenagers.
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