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 Subject: Life Expectancy Statistics Category: Science Asked by: bill0917-ga List Price: \$8.00 Posted: 11 Mar 2006 20:05 PST Expires: 10 Apr 2006 21:05 PDT Question ID: 706270
 ```I am looking for the mean and standard deviation of the life expectancy of an American male and female in the United States. Where did you find it?``` ```Overview with references: http://www.tonysteuer.com/HowMuch_LifeExpect_Mortalty.html SEARCH TERM: Mortality tables``` Request for Answer Clarification by bill0917-ga on 12 Mar 2006 05:50 PST ```I appreciate your current efforts, but I'm not looking for a chart on life expectancy. I'm looking specifically for the mean (average expected age of death) and standard deviation from that mean. I would imagine that if that data is not available some place, then it would need to be calculated from the CDC chart. What price do I need to give for this question?``` Clarification of Answer by hedgie-ga on 12 Mar 2006 21:07 PST ```Life Expectancy News (AP) U.S. life expectancy has hit another all-time high ? 77.6 years ? Also, Americans' life expectancy increased again in 2003, up from 77.3 the year before. By comparison, it was 75.4 in 1990. http://www.cbsnews.com/stories/2005/12/08/health/main1109413.shtml Us is first in the health spending Health > Spending > Per person 4,271 [1st of 137] but is not first in life expectancy, Its rank is #48 United States 77.14 year http://www.nationmaster.com/graph-T/hea_lif_exp_at_bir_tot_pop That's the expectancy at birth - a property of the whole population. Here the change of this parameter with time is shown graphically http://www.mchb.hrsa.gov/whusa02/Page_26.htm More specificaly we ask about life expectancy of given person of age x: Life expectancy is the average number of years remaining for a living being (or the average for a class of living beings) of a given age to live. Life expectancy is also called average life span or mean life span. http://en.wikipedia.org/wiki/Gompertz-Makeham_law_of_mortality So, life expectancy already is a 'mean' of a probabilty distribution Do we want the 'standard deviation' if this distribution? This is how the mean is calculated: Calculating life expectancy The starting point for calculating life expectancy is to calculate the crude death rates of people in the population at each age. For example, if one observed a group of people who were alive at their 90th birthday, and 10% of them were dead by their 91st birthday, then the crude death rate at age 90 would be 10%. These crude death rates can be used to calculate a life table, from which one can calculate the probability of surviving to each age. In actuarial notation the probability of surviving from age x to age n+x is denoted n.P(x) The "curtate" life expectancy at age x, denoted e(x) is then calculated by adding up these probabilities at every age. This is the expected number of complete years lived (one may think of it as the number of birthdays they celebrate). e(x) = Sum(over t) t.P(x) ... Average life expectancy is almost always calculated as an arithmetic mean as above. The median life expectancy is very occasionally used instead. Note that no allowance has been made in this calculation for expected changes in life expectancy in the future. Usually when life expectancy figures are quoted, they have been calculated like this with no allowance for expected future changes. This means that quoted life expectancy figures are not generally appropriate for calculating how long any given individual of a particular age is expected to live, as they effectively assume that current death rates will be "frozen" and not change in the future. Instead, life expectancy figures can be thought of as a useful statistic to summarise the current health status of a population. http://en.wikipedia.org/wiki/Life_expectancy So, the standard deviation of this distribution would be sigma(x) = Sum(over t) ( t - e(x) ) ^ 2 * t.P(x) Is that would you mean by standard deviation? The reason that this measure is not often used has to do with shape of this distribution n.P(x) : For a given x the distribution is exponetial. Different distributions and there parameters are described here http://astronomy.swin.edu.au/~pbourke/other/distributions/ The higher momenta of the distribution are well defined and can be epressed in terms of the rate parameter lambda: lambda= " ? > 0 is a parameter of the distribution and can be thought of as the multiplicative inverse of the rate parameter ... In this specification, ? is a survival parameter in the sense that if a random variable X is the duration of time that a given biological or mechanical system M manages to survive and X ~ Exponential(?) then . That is to say, the expected duration of survival of M is ? units of time. http://en.wikipedia.org/wiki/Exponential_distribution mean = 1/ lambda is expected value or men and Variance = E (t - mean) ^2 =1/ lambda ^2 then sigma = standard deviation is square root of variance would be same as the mean. This curious result is specific to exponential distribution and applies to your question to the degree to which t.P(x) can be described by an exponential. Gompertz law of mortality states exponential increase in death rates with age. http://mathworld.wolfram.com/GompertzCurve.html Further reading Leonid A. Gavrilov & Natalia S. Gavrilova (1991), The Biology of Life Span: A Quantitative Approach. New York: Harwood Academic Publisher, ISBN 3718649837``` Clarification of Answer by hedgie-ga on 13 Mar 2006 01:26 PST ```On second thought, based on fairly old data, mean is µ = 70.778 and the standard deviation is = 16.5119. The probability distribution od lifespan is not exponential, nor normal. It's shape, and aproximations are shown here in this article http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/cprob/cprob4.html The distributions for more specific groups ( white males, now aged 56 ..) can be derived from the mortality tables (which you have) using the method described in the above article. It would reguire to transfer data to a spreadsheet and/or writing a simple program to do the calculation. I guess someone here may be able to do that for \$100 - \$150 You can safely skip the previous clarification. It ws not too clear I am afraid. Hedgie```  