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I've a general question: Can you take the population of the U.S. and divide it by average life expectancy (both female and male) in the U.S. to get the average number of deaths per year? The number I get is too high, and yet it seems this basic division should give me a closer result.

Thanks!

Addition: This is from the Columbia Common Core class Frontiers of Science http://ccnmtl.columbia.edu/projects/mmt/frontiers/web/index2.html [14:] I frequently use envelope backs to debunk, or place in perspective, sensationalized news stories. For example, every few years the media gets excited about "killer sharks." By the beginning of fall term a few years ago, despite weeks of headline coverage concerning the "shark menace," precisely two people had died in the US from shark bites. What fraction is that, you might ask yourself, of all the people who have died during the year? The answer can be easily determined as follows:

[15:] There are about $300$ million ($3 \times 10^8$) people in the US, and the average life expectancy in this country is about $75$ years (averaging men and women -- note that average life expectancy is just what we want here because it tells us how long the average person lives). This means that:

[16:] $3.0 \times 10^8 \text{ people} / 75 \text{ years} = 4.0 \times 10^6$ people die each year.

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    $\begingroup$ The number of deaths would be something you could estimate by taking the population at each age times their probability of dying at that age, and summing it. I get that the population part of each term sums to the total population, I don't get how life expectancy directly gives you anything like an average of the other part of the term. I also don't get what life expectancy you can use, the full name of the published statistic is the life expectancy at birth. Can you show us why your conjecture seems right to you? $\endgroup$ – hkr Apr 6 '17 at 16:38
  • $\begingroup$ I put an additional comment in the prompt. $\endgroup$ – David Marlowe Apr 6 '17 at 21:21
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    $\begingroup$ Why do you think 4 million is too high? Your approach is good. It gets a little off because the age distribution is not what would be steady state, but that is not far off. Using 300 million says you are thinking in terms of a $\pm 10\%$ number and you should certainly be within that. $\endgroup$ – Ross Millikan Apr 6 '17 at 23:45
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This is good back-of-an-envelope estimation, well known (in some circles) as Little's Law:

The long-term average number of customers in a stable system $L$ is equal to the long-term average effective arrival rate, $λ$, multiplied by the average time a customer spends in the system, $W$; or expressed algebraically: $L = λW$. ( https://en.wikipedia.org/wiki/Little%27s_law)

In your example the stability is the (roughly) constant population size: in the short term the effective arrival (birth) and departure (death) rates are about the same.

You have $L$ and $W$ and use them to find $λ = L/W$.

The cdc says

Number of deaths: 2,626,418
Death rate: 823.7 deaths per 100,000 population
Life expectancy: 78.8 years
Infant Mortality rate: 5.82 deaths per 1,000 live births

(https://www.cdc.gov/nchs/fastats/deaths.htm)

so $4 \times 10^6$ is the right order of magnitude.

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