Median and Percentiles of Continuous Random Variables

The median of a continuous random variable is the value that splits the area under its density function exactly in half. Quartiles and percentiles work the same way with other fractions of the area. On Paper 6 the median is usually the last part of a continuous random variable question: having found $k$, a probability and the mean, you are asked to find the median, or to show that it satisfies an equation and check where it lies. This note shows how to set up those equations, how to solve them, and how to compare the median with the mean.

The median splits the area in half

For a set of data, the median is the middle value: half the data lie below it and half above. For a continuous random variable there is no list of values, only a density curve. "Half the values lie below it" becomes "half the probability lies below it", and probability is area. So the median is the point on the $x$-axis where a vertical line cuts the area under the curve into two equal parts, each of area $\tfrac{1}{2}$.

Here "lower end" means the left-hand end of the interval on which $f$ is non-zero. Below that point the density is zero and contributes nothing to the area.

The syllabus says that explicit knowledge of the cumulative distribution function is not required, but that you may need to locate the median "by direct consideration of an area". That is exactly the definition above: write down the area from the left-hand end up to $m$, set it equal to $\tfrac{1}{2}$, and solve for $m$.

The picture below shows $f(x) = \tfrac{3}{8}x^2$ on $0 \le x \le 2$. The shaded region has area $\tfrac{1}{2}$, so its right-hand edge is the median.

Because the density rises to the right, most of the area is near $x = 2$, so the median ($1.587$) is well to the right of the midpoint of the interval.

Quartiles and percentiles

The same idea works for any fraction of the area.

Questions do not always use these names. "Find the value of $a$ such that $P(X < a) = 0.9$" is asking for the 90th percentile. "Find the time exceeded by $5\%$ of customers" means $P(T > t) = 0.05$, so $P(T < t) = 0.95$: the 95th percentile.

Integrating from the other end

Sometimes the upper tail is easier to integrate, especially on an infinite interval. Since $P(X \ge m) = \tfrac{1}{2}$ too, you may write

$$\int_{m}^{\text{upper end}} f(x)\,dx = \frac{1}{2}$$

and, for the upper quartile, $\displaystyle\int_{q_3}^{\text{upper end}} f(x)\,dx = \frac{1}{4}$. Both routes give the same answer. Choose the one with the simpler integral and limits.

Infinite intervals and exponential densities

When the interval is infinite, for example $t \ge 0$, the area up to $m$ is a definite integral with ordinary limits $0$ and $m$, so nothing special is needed. Exponential densities lead to equations of the form $e^{-cm} = \tfrac{1}{2}$, which you solve with natural logarithms.

Symmetry gives the median for free

If the density is symmetrical about a line $x = c$, then the area to the left of $c$ equals the area to the right, so the median is $c$. The mean is also $c$. A symmetrical density therefore has mean = median, and you can quote both without integrating, provided you state the symmetry as the reason.

Symmetry also links the quartiles: if $f$ is symmetrical about $x = c$ and the lower quartile is $q_1$, then the upper quartile is the same distance above $c$, so $q_3 = 2c - q_1$.

Equations you cannot solve exactly

For a quadratic density such as $kx(4 - x)$ or $k(4 - x^2)$, the area up to $m$ is a cubic in $m$. Cubic equations usually have no neat solution, so examiners phrase the question as "show that $m$ satisfies the equation ..." followed by "verify that $m$ lies between ... and ..." or "hence find $m$ correct to 2 decimal places".

To verify that a root lies between two values $a$ and $b$, substitute both into the expression and show that the values have opposite signs. Because a polynomial is continuous, a change of sign means it passes through zero somewhere between $a$ and $b$. This is the same sign-change argument as in Pure Mathematics 2 and 3.

A cubic has up to three real roots. Only one will lie in the interval where $f$ is defined, and that is the median; the others are meaningless here.

Comparing the median and the mean

Examiners like to ask whether the median is greater or less than the mean, and to make you justify it. Three facts cover nearly every case.

• Symmetrical density: mean $=$ median.
• Density with a long tail to the right (for example a decreasing density, or one with its peak near the left end): the mean is pulled towards the tail, so mean $>$ median.
• Density with a long tail to the left (for example an increasing density such as $\tfrac{3}{8}x^2$): mean $<$ median.

You can always check numerically. A neat test: if $P(X < \text{mean}) > \tfrac{1}{2}$, then more than half the area lies below the mean, so the median is below the mean.

Unknown constants from a median

A given median or quartile is one more equation, so it can be used to find an unknown constant, in the same way as a given probability or a given mean.

Practice questions

Answers

1. $\displaystyle\int_0^m \frac{x}{2}\,dx = \frac{m^2}{4} = \frac{1}{2}$, so $m^2 = 2$ and $m = \sqrt{2} = 1.41$ (the negative root is outside $[0, 2]$). Lower quartile: $\dfrac{q^2}{4} = \dfrac{1}{4}$, so $q = 1$.

2. $\displaystyle\int_0^m 4x^3\,dx = m^4 = \frac{1}{2}$, so $m = 0.5^{1/4} = 0.841$. 80th percentile: $a^4 = 0.8$, so $a = 0.8^{1/4} = 0.946$.

3. $\displaystyle\int_1^m \frac{2}{x^2}\,dx = 2\left(1 - \frac{1}{m}\right) = \frac{1}{2}$, so $\dfrac{1}{m} = \dfrac{3}{4}$ and $m = \dfrac{4}{3} = 1.33$. $q_1$: $2\left(1 - \dfrac{1}{q_1}\right) = \dfrac{1}{4}$, so $\dfrac{1}{q_1} = \dfrac{7}{8}$, $q_1 = \dfrac{8}{7}$. $q_3$: $2\left(1 - \dfrac{1}{q_3}\right) = \dfrac{3}{4}$, so $\dfrac{1}{q_3} = \dfrac{5}{8}$, $q_3 = \dfrac{8}{5}$. IQR $= \dfrac{8}{5} - \dfrac{8}{7} = \dfrac{16}{35} = 0.457$.

4. $\displaystyle\int_0^m 0.5e^{-0.5t}\,dt = 1 - e^{-0.5m} = \frac{1}{2}$, so $m = 2\ln 2 = 1.39$ years. $P(T > 2m) = e^{-0.5 \times 4\ln 2} = e^{-2\ln 2} = \dfrac{1}{4}$.

5. $\displaystyle\int_0^\pi k\sin x\,dx = k[-\cos x]_0^\pi = 2k = 1$, so $k = \tfrac{1}{2}$. The graph is symmetrical about $x = \frac{\pi}{2}$, so the median is $\frac{\pi}{2}$. Upper quartile: $\tfrac{1}{2}[-\cos x]_0^{q} = \tfrac{1}{2}(1 - \cos q) = \tfrac{3}{4}$, so $\cos q = -\tfrac{1}{2}$ and $q = \dfrac{2\pi}{3} = 2.09$.

6. $\displaystyle\int_0^q \tfrac{3}{4}(2x - x^2)\,dx = \tfrac{3}{4}\left(q^2 - \frac{q^3}{3}\right) = \frac{1}{4}$, so $q^2 - \dfrac{q^3}{3} = \dfrac{1}{3}$, $3q^2 - q^3 = 1$, and $q^3 - 3q^2 + 1 = 0$. With $g(q) = q^3 - 3q^2 + 1$: $g(0.65) = 0.274625 - 1.2675 + 1 = 0.00713 > 0$ and $g(0.66) = 0.287496 - 1.3068 + 1 = -0.0193 < 0$. The sign changes, so $0.65 < q < 0.66$.

7. $\displaystyle\int_0^2 \frac{3x^2}{a^3}\,dx = \frac{8}{a^3} = \frac{1}{2}$, so $a^3 = 16$ and $a = 16^{1/3} = 2.52$. $P(X < 1) = \dfrac{1}{a^3} = \dfrac{1}{16}$.

8. (a) $f(x) = \frac{1}{x} > 0$ on $[1, e]$, and $\displaystyle\int_1^e \frac{1}{x}\,dx = [\ln x]_1^e = 1 - 0 = 1$. (b) $[\ln x]_1^m = \ln m = \tfrac{1}{2}$, so $m = e^{1/2} = \sqrt{e}$ $(= 1.649)$. (c) $E(X) = \displaystyle\int_1^e x \cdot \frac{1}{x}\,dx = e - 1 = 1.718$. This is greater than $\sqrt{e} = 1.649$, so the mean is greater than the median. (The density decreases, so the tail is to the right.) (d) $P(\text{all three below the median}) = \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$. $P(X < e - 1) = \ln(e - 1) = 0.5413$. With $N \sim B(3, 0.5413)$, $P(N = 1) = 3(0.5413)(0.4587)^2 = 0.342$.