# Do 50% of people die on either side of an average life expectancy?

I am no statistician so this may come across as very naive, but I've been trying to get my head around how to interpret life expectancy data.

Say you have a population of $$n$$ and a range of possible ages from 0–$$m$$.

Am I right in thinking that given an average life expectancy of $$x$$ years, roughly 50% of the population must fall on either side of that line (ignoring those who fall directly on it)?

As such, given an average life expectancy of 75, a maximum age of 120 and a population of 10,000, 5,000 (ish) would be expected to die between the ages of 0 and 74 and 5,000 (ish) between 76 and 110? And that that second half will be more densely packed than the first half (as it covers 34 years rather than 74 with the same number of people)?

Finally, if the above is broadly correct (and if we assume for simplicity's sake that each age has an equal likelihood of fatality), does that mean you are more likely to die on the left side of the average until you make it halfway between 0 and $$x$$, whereupon you are equally as likely to die on either side of the average, and each subsequent year survived increases your likelihood of outliving the average?

My thinking being that if you have a $$\frac{1}{120}$$ chance of dying each year (0.83%), each year of the left side is $$\frac{1}{74}$$ (1.35%) whereas each year on the right side is $$\frac{1}{34}$$, or ~$$\frac{2}{74}$$ (2.94%).

• How do you define "average"? Do you know what a median is, or any other averages?
– J.G.
Commented Apr 16, 2019 at 15:54

The mean and median of a data set are NOT equal in general. In fact, the data show that life expectancy is skewed left with the median (82) roughly three years greater than the mean (79). So, interestingly most people can expect to live longer than their actual life expectancy.

• Yep I would expect accidents in youth to be outliers which drag down arithmetic average quite a bit compared to median. Commented Apr 16, 2019 at 16:01

Not for arithmetic averages, but for median averages.

Assume we have a probability density function $$p(x) s.t. \int_{-\infty}^\infty p(x)dx = 1$$

Arithmetic mean is then defined as $$\int_{-\infty}^{\infty} x\cdot p(x)dx$$

Median is defined as $$x_0 \text{ s.t. } \int_{-\infty}^{x_0} p(x)dx = \frac{1}{2}$$

So median is defined to be the value for the random variable for which half of the cases are below and the other half is above.