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Q: Life Expectancy Statistics ( Answered,   0 Comments )
Question  
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?
Answer  
Subject: Re: Life Expectancy Statistics
Answered By: hedgie-ga on 11 Mar 2006 22:08 PST
 
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  = 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
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