Median and IQR of a CRV

The median (or \(50\)th percentile), usually denoted by \(m\), is the value of \(X\) such that the line \(x = m\) splits the area under the curve \(y = \mathrm{f}(x)\) into two halves.

\[\text{i.e.}\hspace{0.5em} \int_{a}^{m} \mathrm{f}(x) \, \mathrm{d}x = 0.5 \text{ or}\hspace{0.5em} F(m) = 0.5\]

The lower quartile (or \(25\)th percentile), \(r\), is the value of \(X\) such that \(\int_{a}^{r} \mathrm{f}(x) \, \mathrm{d}x = 0.25\) or \(F(r) = 0.25\).

The upper quartile (or \(75\)th percentile), \(s\), is the value of \(X\) such that \(\int_{a}^{s} \mathrm{f}(x) \, \mathrm{d}x = 0.75\) or \(F(s) = 0.75\).

In general, the \(k^{\mathrm{th}}\) percentile, \(t\), is the value of \(X\) such that \(\int_{a}^{t} \mathrm{f}(x) \, \mathrm{d}x = \frac{k}{100}\) or \(F(t) = \frac{k}{100}\).

The interquartile range = upper quartile – lower quartile.

For a piecewise density, locate the median before choosing an equation. Add the area of each complete earlier piece; if their cumulative area is still below \(0.5\), continue into the next piece. Solve for the remaining area there and check that the answer lies in that piece of the support.

The median divides probability into equal halves. It need not be halfway between the endpoints of the support and need not equal the mean or the mode. The same area reasoning can find another stated percentile.