Very old man

5830 /www/nrich/html/content/id/5830/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/7220">Warm-up</a></li> <li><a href="http://nrich.maths.org/6680">Try this next</a></li> <li><a href="http://nrich.maths.org/5937">Think higher</a></li> <li><a href="http://understandinguncertainty.org/node/1286">Read: mathematics</a></li> <li><a href="http://understandinguncertainty.org/node/76">Read: science</a></li> <li><a href="http://en.wikipedia.org/wiki/Li_Ching-Yuen">Explore further</a></li> </ul> <div> </div> Li Ching-Yuen was a Chinese herbalist and longevity expert who was known to have died in 1928. He claimed to have been born in 1734, giving him a lifespan of 196 years. Investigations into birth records indicated that he was actually born in 1678, giving an even longer lifespan of 250 years! Whilst this may seem unbelievable, is it? In this question we use statistics to look into the lifespan of very old people. Whilst there is no conclusive historical evidence to support the birth date of Li Ching-Yuen, the following data concerning lifespans are known [at the time of writing this question (October 2008); sources given below] <ul> <li>There were about 450000 people in the world aged over 100.</li> <li>There were 82 living people who were known to be over the age of 110</li> <li>There were 2 people known to be over the age of 115 (ages 115 and 116)</li> <li>There are 31 unverified claims of people over the age of 110, two of whom claimed to be aged 115 and 116.</li> <li>In the past 50 years, 25 people are known for certain to have lived beyond the age of 115.</li> <li>In the past 50 years, 2 people are known for certain to have lived beyond the age of 120 (dying at ages 120 and 122).</li> </ul> <div>A hypothesis $H$ is made saying: Once you make it to your 100th birthday there is a fixed probability $p$ of surviving to your next birthday on any given subsequent birthday. For example, if $p$ were $0.05$ then the hypothesis says that on my 100th birthday there is a $5$% chance of surviving until I am $101$; on my $101$st birthday there would be a $5$% chance of surviving until I am $102$ and so on.</div> <div>Does the data approximately fit this hypothesis? What values of $p$ would seem most appropriate?</div> <div>Assume that the hypothesis is true with a generous value of $p=0.5$. With this hypothesis, how many 100 year olds would need to be in a room before we might feel confident that one would live to the age of 196 suggested by Li Ching-Yuen himself? How does this number compare with the number of people on earth today (6.7 billion)?</div> Extension: There are many statistical complications involved in predicting death rates. How many can you think of? How might these effect these statistics in future? <div class="framework">The data in this problem was collected from the websites of <a href="http://www.grg.org/">The Gerontologists</a> and <a href="http://www.thecentenarian.co.uk/">The Centenarians</a> . Living is a risky business. To see more about the statistics concerning living and for an estimate of your life expectance, see the <a href="http://understandinguncertainty.org/node/76">Understanding Uncertainty</a> pages.</div> </mdoxml> <mdoxml version="1.0"> This problem is can be tackled in a number of different ways - both algebraically and graphically. Because of the different ways of grouping the given data for use graphically, this solution will be solely algebraic. Using the first two given pieces of data, it can be seen that the proportion of people over the age of 100 who are also over the age of 110 is $\frac{82}{450000}$. Thus, it can be approximated that the probability of a person people over the age of 110, given that they are over the age of 100, is $\frac{82}{450000}$. $p(Age > 110) = 1 - p(100< Age< 110)$ $ = 1-(p + p^2 +p^3 +p^4 +...+p^{10})$ The term shows in brackets is a geometric series, and can be calculated as follows: Let $X = p +p^2 +p^3 +p^4 +...+p^{10}$ Therefore, $pX = p^2 +p^3 +...+p^{11}$ Subtracting the first of these from the second: $pX - X = p^{11} -p$ $X(p-1) = p(p^{10} -1)$ $X = \frac{p(p^{10} -1)}{p-1}$ Therefore, using this expression for the geometric series gives: $p(Age > 110) = 1 - \frac{p(p^{10} -1)}{p-1}$ Since $p(Age > 110) = \frac{82}{450000}$ the equation can be rearranged to give: $p^{11} -\frac{899918p}{450000} +\frac{449918}{450000} =0$ This polynomial equation is clearly VERY difficult to solve, and so it is best solved numerically. This can be done any number of ways, but is best done using either the Newton-Raphson technique by hand, or by using the 'Solve' function on some calculators, or by using an internet polynomial solver program. Using these gives a valid root of: $$p = 0.500...$$ Additional data is given that there are another 31 unverified claims of people over the age of 100. Using this data gives a total 113 people over the age 110, which gives: $$p = 0.556...$$ A similar calculation can be carried out for people over the age of 115. Using the given data of 2 people over this age gives a polynomial equation: $p^{16} - \frac{899998p}{450000} + \frac{499998}{450000} = 0$ $$\therefore p = 0.556...$$ Also, using the additional unverified data reveals 4 people over the age of 115, and thus: $$p = 0.556...$$ From the data so far the values of p are more consistent when including the potentially spurious data, and so it seems justifiable that it should be included. Doing so gives a consistent value of $\mathbf{p =0.556}$. There are also two final pieces of data given about ages over the last 50 years. This data is somewhat more difficult to use as no data is given for the total number of people that have been aged over 100 years old in the 50 year period. Therefore, the first aim is to estimate the number of total people who have been aged over 100 years: Firstly, it could be assumed that at any one time there are a constant 450000 people who are aged over 100. Each year, 450000p of these die, and are replaced by the same number. Therefore, over the 50 year period , there is a turnover of $450000 \times 50p$. Thus, the overall number of people over 100 in this time period is give by $450000 + (450000 \times 50p) = 450000(1 +50p)$. Therefore, for those aged over 115, an equation can be formed: $\frac{25}{450000(1 +50p)} = 1 - \frac{p(p^{15}-1)}{p-1}$ $50p^{17} + p^{16} - 100p^2 + \frac{864001p}{18000} + \frac{499975}{450000} = 0$ $$p =0.500...$$ For those aged over 120: $\frac{2}{450000(1+50p)} = 1 - \frac{p(p^{20}-1)}{p-1}$ $50p^{22} + p^{21} -100p^2 + \frac{21600002p}{450000} + \frac{449998}{450000} = 0$ $$p = 0.500...$$ Using this additional data seems to support the exclusion of the unverified data from previously, since both values of p here are the same as that for the original calculation for ages greater than 110. It is difficult to comment on whether it is justifiable to include or exclude the unverified data: its exclusion relies heavily on the calculation for the number of people aged of 100 in the 50 year period as being flawed. The final part of the question asks what $p(Age > 196)$ is. An equation can be set up as follows: $p(Age> 196) = 1 - \frac{p(p^{96} -1)}{p-1}$ With a value of $p=0.5$, this yields: $p(Age > 196) = 2^{-96}$ To calculate the number of people that would need to be in the room before we feel confident that the age of 196 is possible, is given by: $\frac{1}{2^{-96}} = 7.92 \times 10^{28}$ people in the room. The total number of people in the world is 6.7 billion, which is $6.7 \times 10^9$, and so we would need $1.2 \times 10^{19}$ times more people than in the whole world to be confident that Li Ching-Yueng's claim is true. </mdoxml> <mdoxml version="1.0"> <h3>Why do this problem?</h3> This <a href="http://nrich.maths.org/5830&part=">problem</a> provides an excellent scenario for making and testing statistical hypothesis. It can be attempted at a variety of levels of statistical sophistication ranging from making sense of the data to a full statistical analysis. <h3>Possible approach</h3> <div>There is quite a lot of information in this problem to digest. Students might need to spend some time reading through the problem and making sense of the data. It would work best as a homework or task where students are given time to think about the problem and then to come up with their own analysis. Students could then compare their answers at a later time. From these, the concensus for a 'best' analysis might emerge.</div> <h3>Key questions</h3> <div>Before starting this problem, how might we organise or represent the data? Are you clear as to how we might test the hypothesis?</div> <h3>Possible extension</h3> <div>This question naturally raises its own extension: students might try to improve their answers by searching for more data on in the internet. pose their own questions, make their own hypothesis or pursue similar ideas through the materials on the Understanding Uncertainty site.</div> <h3>Possible support</h3> <div>Suggest students simply make plots of the life expectancy data and extrapolate these graphs.</div> </mdoxml> <mdoxml version="1.0"> Li Ching-Yuen himself gave a hint as to the secret of long life. It was * Keep a quiet heart * Sit like a tortoise * Walk sprightly like a pigeon * Sleep like a dog If this hint is not enough to help, try plotting the data on a chart. If there is a 50% chance, say, of death each year, by what factor might the expected numbers of living very old people reduce over 5 years? Can you find a percentage which might fit the data? </mdoxml> 2 3 0 0 0 0 1 Very old man Is the age of this very old man statistically believable? Handling, Processing and Representing Data Handling data Advanced Probability and Statistics Sampling and hypothesis tests Probability Experimental probability Advanced Probability and Statistics Statistics generally Applications biology Applications biology Using, Applying and Reasoning about Mathematics Investigations