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This is a problem I had in school that I think I remember how to solve except my son does not agree with me.

Given a population with a life expectancy that fits a normal distribution, mean of 75 and a standard deviation of 5. What percentage of the people that live to age 80 will live to age 85?

As soon as I'll figure out how to produce suitable graphics to answer the question but I'll post it now in case anybody can come up with a good explanation.

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    $\begingroup$ You want $P(X>2\sigma | X > \sigma)$. Where are you stuck? $\endgroup$ Mar 14 at 17:35
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    $\begingroup$ Bayes' theorem is needed. Hint : The probabilities to become (at least) $85$ and that to become (at least) $80$ must be divided. $\endgroup$
    – Peter
    Mar 14 at 17:42
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    $\begingroup$ @preferred_anon I looked up standard deviation tables on the web and my answer is about 14% will survive to 85. If I am reading your notation correctly, I want the probability that they will reach 2σ given they have reached 1σ, which is the solution I remember. The idea that areas under the curve are what is of interest is the big thing I learned. Lots of people have magic formulas and lookup numbers in the tables, very few understand what they are looking for areas under the curve. $\endgroup$ Mar 14 at 17:48
  • $\begingroup$ @peter I had to look that up, I remembered the name but not what it was for. Yes, Bayes' theorem is what is needed. Now I am going to have to double check my 14%. $\endgroup$ Mar 14 at 17:53
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    $\begingroup$ Within this accuracy, $14$% is correct. $\endgroup$
    – Peter
    Mar 14 at 18:09

2 Answers 2

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If the random variable $X$ represents the age at which a randomly selected person died, you want to compute $P(X>\mu+2\sigma | X > \mu+\sigma)$ where $X$ follows a normal distribution of parameters $\mu=75$ and $\sigma=5$.

By definition of conditional probability, we have

$$ p=P(X>\mu+2\sigma | X> \mu+\sigma) = \frac{P(X> \mu+2\sigma \text{ and } X>\mu+\sigma)}{P(X> \mu+\sigma)} = \frac{P(X> \mu+2\sigma)}{P(X> \mu+\sigma)} $$

Now, if $f$ be the pdf of the normal distribution $\mathcal N(\mu,\sigma^2)$, that is

$$ f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} = \frac{1}{5\sqrt{2\pi}}e^{-\frac{(x-75)^2}{50}} $$

we have

$$ p = \frac{\int_{\mu+2\sigma}^\infty f(x)\; dx}{\int_{\mu+\sigma}^\infty f(x)\; dx} $$

The function $f$ has no elementary antiderivative, so you will have to rely on computers or tables to find approximations. Using Maple, I got $\int_{\mu+2\sigma}^\infty f(x)\; dx\simeq 0.02275$, $\int_{\mu+\sigma}^\infty f(x)\; dx\simeq 0.15866$ so

$$ P(X>\mu+2\sigma | X> \mu+\sigma) \simeq 0.14339 \simeq 14.34\% $$

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The way I remember this as being presented by my teacher in Probability and Statistics class is:

The explanation part, First, The area under the curve shows the probability of an event. If we say that everybody dies with an age greater than Negative Infinity and less than Positive Infinity, the area under the whole normal distribution curve will be 1. (Gee, We have a probability of 100% that people will die eventually, that's profound. (Doubly so since the area under the curve is 100% by definition.))

Next, 50% of people die before they reach the mean value and 50% die after. (Since that is the definition of the median (which matches the mean for a normal distribution), again, not very profound.)

Where this gets interesting is when we want to see for specific values.

We all remember that 68% or so of the population dies within one standard deviation of the mean. We can see this, if I go into Excel and type the magic formula:

=NORM.DIST(80,75,5,TRUE)-NORM.DIST(70,75,5,TRUE)

or 68%. When arg4 is true, we get the cumulative value from Negative Infinity.

So the formula says, given a normal distribution curve, centered on 75 with a S.D. of 5, tell me the probability somebody will die before age 80 minus the probability that somebody will die before age 70. So Excel can tell us what we already know. I used excel for all the numbers given here and the graphic.

The solution part, The problem presentation is how many people that reach age 80, one standard deviation past the mean, will reach age 85, two standard deviations past the mean. We have, say 100 people of age 80, how many will reach age 85? ALL of our sample has reached age 80. Now we want to know how many will last another five years. We will want the area under the curve from 80 to 85 or the fill in portion of this image, enter image description here

  • The area under the curve before age 80 is 0.8413 so we are left with 1-0.8413 or 0.1587 of the original population.
  • The area under the curve before age 85 is 0.9772 so we are left with 1-0.9772 or 0.0228 of the original population.

If we want to know what percentage of people that reach age 80 will reach age 85, we can take as 100% the area under the curve from 80 to Positive Infinity. The population that lived to age 85 is represented by the nonshaded area under the curve from 85 to Positive Infinity. The shaded area is what percentage will die before age 85 so we need to divide the non-shaded after 85 by the sum of the shaded and un-shaded after age 80 or:

$\dfrac{0.02275}{0.1587} \times 100 = 14.34\% $

Comments? Clarifications? Corrections?

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    $\begingroup$ "Next, 50% of people die before they reach the mean value and 50% die after. (Since that is the definition of the mean, again, not very profound.)". This is the definition of the median. For the normal distribution, the median and the mean are equal (by symmetry of the distribution), but not in general (e.g. exponential distribution) $\endgroup$
    – Taladris
    Mar 15 at 2:17
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    $\begingroup$ @Taladris, Quite so. I stand corrected (and have corrected). $\endgroup$ Mar 15 at 3:06
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    $\begingroup$ It's a lot cleaner to just calculate $0.02275/0.1587$. Why write a double complement? $\endgroup$
    – obscurans
    Mar 15 at 6:13
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    $\begingroup$ "The area under the probability curve is 100%. I want to sign up for that Statistics and Probability course! This is how one must start a statistical analysis. Please tell this to folks falling for the "Monty Hall" bait and switch. $\endgroup$ Mar 16 at 10:54
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    $\begingroup$ @user1683793 only meant to be complementary. +1 answer. I was referring to the "bait" of information irrelevant to the final analysis in the "Monty Hall" win the car game. Certainly not in your answer. $\endgroup$ Mar 16 at 18:55

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