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Continuous random variables
Continuous random variables
CatalogueDPMathematicsMathematics: Analysis and Approaches HLStatistics and probability

Continuous random variables

Interpret a probability density as area and use integration to find constants, probabilities and summary values.

  • AA HL
  • Continuous random variables
  • Papers 1 and 2
  • GDC optional
  • About 25 min

By the end of this lesson

  • normalize a probability density function
  • calculate interval probabilities and expectation
  • distinguish mean, median and mode for a density

Why this matters. For a continuous variable, height is density rather than probability. Probability is accumulated area over an interval.

Probability is area

A probability density function $f$ satisfies $f(x)\ge0$ and has total area $1$ over its support.

$$P(a\le X\le b)=\int_a^b f(x)\,dx,\qquad \int_{-\infty}^{\infty}f(x)\,dx=1.$$
Probability as area under a densityA nonnegative density curve is drawn above the x axis. The area between a and b is shaded to represent the probability that X lies in that interval.abP(a ≤ X ≤ b)
The shaded area is an interval probability. The height $f(x)$ at one point is not itself a probability.

Checkpoint

For a continuous random variable, what is $P(X=2)$?
Show explanation

A single point has zero width and therefore zero area.

Example 1

Find the normalizing constant

Make the total area one

Let $f(x)=kx$ for $0\le x\le2$, and $f(x)=0$ otherwise.

$$\int_0^2kx\,dx=1\Rightarrow 2k=1\Rightarrow k=\frac12.$$

So $f(x)=x/2$ on the support.

Find interval probabilities

For the density above,

$$P(1\lt X\lt 2)=\int_1^2\frac{x}{2}\,dx=\left[\frac{x^2}{4}\right]_1^2=\frac34.$$

Endpoint choices do not change a continuous probability.

Checkpoint

If $f(x)=c$ on $0\le x\le5$, what is $c$?
Show explanation

The rectangle has area $5c=1$.

Example 2

Calculate expectation

Weight values by density

For $f(x)=x/2$ on $0\le x\le2$:

$$E(X)=\int_0^2x f(x)\,dx=\int_0^2\frac{x^2}{2}\,dx=\frac43.$$

The mean lies to the right of the midpoint because the density increases with $x$.

Why multiply by $x$?

Expectation is a weighted average. The density supplies the weight assigned near each value $x$.

Separate mean, median and mode

  • The mean is $E(X)$.
  • A median $m$ satisfies $P(X\le m)=1/2$.
  • A mode occurs where the density is greatest.

Checkpoint

For $f(x)=x/2$ on $0\le x\le2$, which equation finds the median $m$?
Show explanation

The median divides total probability area in half.

Example 3

Find a median and mode

An increasing density

For $f(x)=x/2$ on $[0,2]$, solve

$$\int_0^m\frac{x}{2}\,dx=\frac12\Rightarrow \frac{m^2}{4}=\frac12\Rightarrow m=\sqrt2.$$

The negative root is outside the support. Since $f$ increases throughout $[0,2]$, the mode is $2$.

Here mean $4/3$, median $\sqrt2$, and mode $2$ are different.

In the exam

Quick check

Five questions to check the main decisions from this lesson.

Your score is not saved.

  1. What must the total area under a density equal?
    Show explanation

    A density distributes one unit of total probability.

  2. $P(a<X<b)$ is found by…
    Show explanation

    Interval probability is accumulated density.

  3. For continuous $X$, $P(X<c)$ and $P(X\le c)$ are…
    Show explanation

    The single endpoint has probability zero.

  4. Which integral gives $E(X)$?
    Show explanation

    Expectation weights each value by its density.

  5. A median $m$ divides what into two halves?
    Show explanation

    $P(X\le m)=1/2$.

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