We have shown how to find a cumulative probability, \(P(X \le x)\). It is sometimes useful to work in the opposite direction — when given a cumulative probability, we may want to find the corresponding value of \(x\).
Definition
The \(p\)'th quantile of a continuous distribution is the value, \(x\), such that
\[ P(X \le x) \;\; = \; \; p \]When \(p\) is expressed as a percentage, the value is called the \(100p\)'th percentile.
Definition
These three values split the probability density function into four equal areas, and the corresponding probabilities are 0.25 for each of these ranges of values.
Example
What are the median and quartiles of the \(\RectDistn(1, 5)\) distribution?
The cumulative distribution function of \(X\) is
\[ F(x) = \begin{cases} 0 & \quad \text{for } x \lt 1\\[0.2em] \dfrac {x-1} 4 & \quad \text{for } 1 \le x \le 5 \\[0.2em] 1 & \quad \text{for } x \gt 5 \end{cases} \]To find the \(p\)'th quantile, we need to solve
\[F(x) \;\; = \; \; \frac {x-1} 4 \;\;=\;\; p\]giving
\[x \;\; = \; \; 4p + 1\]The quartiles and median are found by inserting \(p = \) 0.25, 0.5 and 0.75 into this equation giving the values \(x\) = 2, 3 and 4.
The next example is a little harder.
Example
Find a formula for the \(p\)'th quantile of the exponential distribution with probability density function
\[ f(x) = \begin{cases} 4\;e^{-4x} & \quad \text{for } x \ge 0\\[0.2em] 0 & \quad \text{otherwise} \end{cases} \]The cumulative distribution function of \(X\) is
\[ F(x) = \begin{cases} 0 & \quad \text{for } x \lt 0\\[0.2em] 1 - e^{-4x} & \quad \text{for } x \ge 0 \end{cases} \]To find the \(p\)'th quantile, we need to solve
\[F(x) \;\; = \; \; 1 - e^{-4x} \;\;=\;\; p\]With a bit of algebraic manipulation,
\[x \;\; = \; \; - \frac {\log(1 - p)} 4\]