IQR(x, na.rm = FALSE)
x values.quantile function rather than following Tukey's recommendations,
i.e., IQR(x) = quantile(x,3/4) - quantile(x,1/4).
For normally N(m,1) distributed X, the expected value of
IQR(X) is 2*qnorm(3/4) = 1.3490, i.e., for a normal-consistent
estimate of the standard deviation, use IQR(x) / 1.349.
fivenum, mad which is more robust,
range, quantile.data(rivers) IQR(rivers)