Revise Continuous random variables and probability density functions for IB Mathematics: Analysis and Approaches (4.8)

A continuous random variable $X$ can take any value in an interval of the real line. Because the possible values cannot be listed, we describe its behaviour with a probability density function (pdf), written $f(x)$. The single most important idea in the whole topic is this: for a continuous variable, probability is area. The height $f(x)$ of the curve at a point is a density, not a probability — it tells you how concentrated the probability is near $x$, but on its own it is not a chance of anything. You only obtain a probability once you integrate the density across an interval, sweeping out an area.

This immediately explains a fact that surprises students: the probability of any single exact value is zero, $P(X=c)=\int_c^c f(x)\,dx=0$, because a point has no width and so bounds no area. As a result, for a continuous variable it never matters whether endpoints are included: $P(a\le X\le b)=P(a<X<b)=P(a\le X<b)=P(a<X\le b)$.

The probability that $X$ falls in an interval is then the area under the pdf over that interval, obtained by a definite integral.

P(a \le X \le b) = \int_a^b f(x),dx

The summary numbers you met for discrete variables all carry over — but the sum $\sum$ becomes an integral $\int$. The expected value (mean) weights each value $x$ by its density $f(x)$ and integrates; the variance measures spread using $E(X^2)$ and the mean.

$E(X) = \mu = \int_{-\infty}^{\infty} x\,f(x)\,dx, \qquad \mathrm{Var}(X) = E(X^2)-[E(X)]^2, \qquad E(X^2)=\int_{-\infty}^{\infty} x^2 f(x)\,dx$

The pattern is always the same: to average any quantity of $X$, integrate that quantity multiplied by the density $f(x)$ across the interval. Multiply by $x$ for the mean, by $x^2$ for $E(X^2)$, and the variance follows by subtracting the square of the mean. As with discrete variables, the mean need not be a value $X$ is especially likely to take — it is a balance point of the area.

The median $m$ is the value that divides the area under the pdf into two equal halves, so it satisfies $\int_{-\infty}^{m} f(x)\,dx=0.5$. The mode is the value of $x$ where the density $f(x)$ is greatest — the peak of the curve. If $f$ has a turning point inside the interval you find the mode with $f'(x)=0$; if $f$ is increasing or decreasing throughout, the mode is at an endpoint. The next example finds the mode, median and variance of one distribution in a single pass.

It is often convenient to accumulate probability from the left. The cumulative distribution function (cdf) is $F(x)=P(X\le x)=\int_{-\infty}^{x} f(t)\,dt$. It starts at 0, rises to 1, and never decreases. Two relationships make it useful: the pdf is the derivative of the cdf, $f(x)=F'(x)$, and any interval probability can be read off directly as $P(a\le X\le b)=F(b)-F(a)$. The median condition can then be stated compactly as $F(m)=0.5$.

$F(x)=P(X\le x)=\int_{-\infty}^{x} f(t)\,dt, \qquad f(x)=F'(x), \qquad P(a\le X\le b)=F(b)-F(a)$

For the density $f(x)=\tfrac38 x^2$ on $0\le x\le 2$, integrating gives $F(x)=\tfrac{x^3}{8}$ for $0\le x\le 2$ (with $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>2$). Notice $F(2)=1$ as required, and setting $F(m)=\tfrac{m^3}{8}=0.5$ recovers the median $m=\sqrt[3]{4}$ found above — the same result reached through the cdf instead of a fresh integral.

Every technique in this lesson is a single idea worn many ways: integrate the density against whatever you want to average. That is exactly the machinery behind the continuous distribution you will meet most often — the normal distribution, whose bell-shaped pdf you already handle on the GDC and whose probabilities are these same areas under a curve. The cdf you met briefly here is precisely what a GDC evaluates when it returns a normal probability, and the mean, variance, median and mode transfer unchanged. Master the routine now — check the pdf, enforce total area 1, then integrate to find probabilities and summary numbers — and continuous probability becomes one method you apply again and again.