送交者: mangolasi 于 2006-2-02, 16:05:34:
Basically you don't need to translate them back to how many people/hour/year etc. You just treat that (.226*10^(-6))*(1/3) as the instanous hazard rate (and constant for that matter if you don't believe old workers are easier to die) for a particular person. (1/3) means every 3 hour you work for one hour.
Basically wasguru did that.
But with mistake:
P(alive at 60|alive at starting working in the mine)=exp( \int_{start}^{60} (original rate+.226*10^(-6))dt)=exp(-\int_{start}^{60}original rate dt)*exp(-.226*10^(-6)*(1/3)*60000)
Suppose the first exponential is 85% (normal people), the 2nd exponetial is the 1/3 power wasguru gave out so it is (.837)^(1/3)=.9424.
Hence the prob a person working in the mine has .8010 chance to survive to 60%. Then even the whole society's expected life is 60 (the last date of 60), 85% people survive to 60 (first date), miners has about .05 chance less to reach that age, and the expected life lost for miners are 60*.05=3 year.
Suppose the expected life in the whole society is 61 (last date), with 85% people survive to 60(first date). Then among those survive, 60% people survive to 60 (last date), then the expected life lost for a miner retire on 60 would be 60*.05+61*(.05*.60)=4.83.
The beginning/end of a year is a bit tricky, but the reasonning is the same.
Mr. He is too opmistic.