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Mathematics (AS & A Level) 9709 · Chapter 30

Continuous random variables

Paper 6. Probability density functions and their properties, finding probabilities, the mean, the variance and the median by integration.

Written for the 2026–2027 syllabus · 4 min read

6.3 Continuous random variables

Probability density functions

A continuous random variable can take any value in a range. Its probability density function (pdf), f(x)\mathrm{f}(x), describes it, and an area under the graph gives a probability.

f(x)\mathrm{f}(x) itself is not a probability, and it can be larger than 1. For a single value, P(X=a)=0\mathrm{P}(X = a) = 0, so P(X<a)\mathrm{P}(X < a) and P(X≤a)\mathrm{P}(X \leq a) are the same.

A probability density function on an interval, with a shaded area between two values giving a probability, and the median splitting the total area in half

A probability density function on an interval, with a shaded area between two values giving a probability, and the median splitting the total area in half

Finding a constant

∫03kx2 dx=[kx33]03=9k=1⇒k=19\int_0^3 kx^2\,\mathrm{d}x = \left[\frac{kx^3}{3}\right]_0^3 = 9k = 1 \quad\Rightarrow\quad k = \frac{1}{9}

Finding probabilities

∫01x29 dx=[x327]01=127\int_0^1 \frac{x^2}{9}\,\mathrm{d}x = \left[\frac{x^3}{27}\right]_0^1 = \frac{1}{27}

Mean and variance

E(X)=∫03x39 dx=[x436]03=94\mathrm{E}(X) = \int_0^3 \frac{x^3}{9}\,\mathrm{d}x = \left[\frac{x^4}{36}\right]_0^3 = \frac{9}{4}

∫03x49 dx=[x545]03=275\int_0^3 \frac{x^4}{9}\,\mathrm{d}x = \left[\frac{x^5}{45}\right]_0^3 = \frac{27}{5}

Var(X)=275−(94)2=0.3375\mathrm{Var}(X) = \frac{27}{5} - \left(\frac{9}{4}\right)^2 = 0.3375

The median and percentiles

The median mm splits the area in half.

[x327]0m=12⇒m3=13.5⇒m=2.38\left[\frac{x^3}{27}\right]_0^m = \frac{1}{2} \quad\Rightarrow\quad m^3 = 13.5 \quad\Rightarrow\quad m = 2.38

The same method gives any percentile. The 90th percentile uses an area of 0.9 from the bottom.

Infinite ranges

A pdf can run to infinity. The upper limit is handled by seeing what the term tends to as xx grows.

∫1∞kx−4 dx=[−k3x3]1∞=0+k3=1⇒k=3\int_1^\infty kx^{-4}\,\mathrm{d}x = \left[-\frac{k}{3x^3}\right]_1^\infty = 0 + \frac{k}{3} = 1 \quad\Rightarrow\quad k = 3

E(X)=∫1∞3x−3 dx=[−32x2]1∞=32\mathrm{E}(X) = \int_1^\infty 3x^{-3}\,\mathrm{d}x = \left[-\frac{3}{2x^2}\right]_1^\infty = \frac{3}{2}

∫1∞3x−2 dx=[−3x]1∞=3⇒Var(X)=3−94=34\int_1^\infty 3x^{-2}\,\mathrm{d}x = \left[-\frac{3}{x}\right]_1^\infty = 3 \quad\Rightarrow\quad \mathrm{Var}(X) = 3 - \frac{9}{4} = \frac{3}{4}

Sketching a pdf

  • The graph must be zero outside the stated range, so show it at zero there or stop at the limits.
  • Label the limits on the xx-axis, and any height that the function gives at those ends.
  • The graph may jump to zero at the ends of its range, as f(x)=19x2\mathrm{f}(x) = \dfrac{1}{9}x^2 does at x=3x = 3.

Reading the shape

FactHow to use it
Symmetric pdfThe mean and median are on the line of symmetry
The peak of the pdfThis is the mode
Most of the area is on the left, with a long tail to the rightMean is greater than median
Most of the area is on the right, with a long tail to the leftMean is less than median
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