Comment by fjkdlsjflkds

I stopped reading at this point:

> Roughly half of your measurements will be above average, and the other half below it.

This is simply not true for most definitions of "average" (i.e., for all definitions of "average" except the median).

Example:

> (data <- c(1:10,100))

[1] 1 2 3 4 5 6 7 8 9 10 100

> sum(as.numeric(data > mean(data)))/length(data)

[1] 0.09090909

A 90/10 split is hardly "roughly half of your measurements".