How old is the SECOND oldest person in the world compared to the oldest? Same for the united states?

I bogged down long before I got the answer. Below is the gibberish I generated towards bogging down.

So OK, I don't even know offhand how old is the oldest, but I would bet it is in the 114 years old (yo) to 120 yo range.

Then figure in some hand-wavey way that people die at ages normally distributed with a mean of 75 yo. We can estimate how many sigma (standard deviations) away from that is the oldest person.

Figure there are 6 billion people now, but I know this number has grown a lot in my lifetime, it was less than 4 billion when I was born 55.95 years ago. So say the 75 yo's come from a population consistent with 3 billion people. 1/2 die younger than 75, 1/2 die older, so the oldest person in the world is 1 in 1.5 billion on the distribution.

OK what do I know about normal distributions? Normal distribution goes as exp ( -((mean-x)/(2sigma))^2 ). So at what x is exp( -(x/2sigma)^2 ) = 1e-9? (x / 2sigma) ^ 2 = -ln ( 1e-9). How to estimate natural log of a billionth? e = 2.7 is close enough for government work to the sqrt(10). So ln(z) = 2log10(z). Then -ln(1e-9) = -2*log10(1e-9) = 2*9 = 18. So (x/2sigma)^2 = 18, sqrt(18) = 4 so

So I got 1 in a billion is 4 sigma. I didn't trust that so I looked that up, Maybe I should have trusted it, in fact 1 in a billion is (slightly more than ) 6 sigma.

mean of 75 yo, x=115 yo, x-mean = 40 years. 6 sigma is 40 years. 1 sigma=6 years.

So do I have ANYTHING yet? I am looking for dx where exp(-((x+dx)/(2sigma))^2) - exp( -(x/2sigma)^2)