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Given that 1% of a population has atleast $x$ amount of wealth, can you work out the average wealth of the entire population, or would you need more information.

I tried working it myself but don't know where to start, so I'm thinking more information would be required, such as what percentage $x$ wealth is, out of the total wealth that the population holds.

user217339
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  • You need some information on the distribution. Is it normal, say? Uniform? As it stands...well (assuming you had $100$ people) you could have $1$ fellow with $x$ dollars and $99$ with $0$, or $99$ with $x-1$, or, really, anything you like (so long as the $99$ stays below $x$). – lulu Jul 28 '16 at 13:54
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    You have a lower bound assuming every "poor" people have $0$ wealth (and cannot have negative wealth). The average is $\frac{1}{n} \sum_{i=1}^n w(i) \geq \frac{1}{n} (\sum_{i=1}^{n/100} x + \sum_{i=n/100+1}^n 0) = \frac{1}{n} \frac{n}{100} x = x/100$. Where $w(i)$ represents the wealth of person $i$ and we suppose that the richest people appear first in the enumeration. – Zubzub Jul 28 '16 at 14:01
  • Just to emphasize: even if you assume something like "the distribution of wealth is normal", that is not enough. For example, take the two distributions $N(100,10)$ and $N(110,5.701)$. You can check that in each case approximately $1%$ of the population has wealth at least $123.26$. – lulu Jul 28 '16 at 14:06

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No, you can't. One distribution that fits your data is $99\%$ having nothing and $1\%$ having $x$, for an average of $\frac x{100}$. Another is the same except one of the $1\%$ has $10^{12}x$ for a much higher average. There are many other distributions that will fit your data. The average could be anything above $\frac x{100}$

Ross Millikan
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