How can 10th (or any other sufficiently low) percentile be equal to the mean number?

Question

For example, a 3rd percentile equal to n means that values of no more than 3% of observations in a group of observations are strictly less than n and that values of no less than 3% of observations are less than or equal to n.

However, if n is also the mean value, than the sum of all observations divided by their number has to equal n - and this does not appear possible since the value of any observation in the top 93% has to be more than n. I might be missing something very obvious, but it looks like having a certain (low) percentile function as a mean is basically impossible.

Answer 1

You're very used to thinking of nice normal distributions when it comes to statistics - in the case of a normal distribution, the median (the 50th percentile) is equal to the mean, and obviously, the 50th percentile is greater than the 3rd percentile for a normal distribution.

But all you need is a something deviating from a normal distribution to show an example of what you have described.

As an example, suppose we have a data set consisting of one $1$, ninety-eight $2$s, and one $3$. The mean is $2$, and the third percentile is $2$.

Answer 2

The mean isn't about the ordering, it's about the values. The values don't have to be similar to the ordinal position. For example, the ten values might be

$$ 1, 2, 3, 4, 5, 5, 6, 7, 8, 9999999999$$

While the median is 5, the mean is nowhere near that value.