Before the method: the idea behind it

Two shapes worth naming

Most continuous distributions need their density supplied by the question. Two, though, appear so often that they have names, standard formulas, and their own question types. Both are made of straight lines, which means their probabilities can be found with geometry rather than calculus.

The uniform distribution

Every value in an interval is equally likely, so the density is a horizontal line.

If a bus arrives at some unknown moment between 8:00 and 8:20, with no reason to favour any particular minute, the arrival time is uniformly distributed. The density must be a rectangle of total area 1, so its height is fixed by the width alone.

\[ f(x)=\frac{1}{b-a} \text{ on } [a,b], \qquad F(x)=\frac{x-a}{b-a}, \qquad E(X)=\frac{a+b}{2}, \qquad \text{Var}(X)=\frac{(b-a)^2}{12} \]

Because the density is constant, probabilities are simply lengths: the chance of landing in any sub-interval is its length divided by the total length. No integration is ever needed for a uniform probability, and the cumulative function is a straight line rising from 0 to 1.

A useful consequence: conditioning a uniform distribution on a sub-interval leaves it uniform on that sub-interval. Given the bus has not arrived by 8:10, its arrival is now uniform on the remaining ten minutes.

The triangular distribution

Values near the middle are more likely than those at the extremes, and the density rises to a peak then falls.

This models situations where a central value is most plausible but the extremes remain possible: project durations, measurement errors, or estimates built from a minimum, a maximum and a most likely value.

The whole density is a single triangle of area 1, so its peak height is determined by the width: if the base runs from a to b, then \(h=\frac{2}{b-a}\). Notice the height depends only on the width, not on where the peak sits.

\[ h=\frac{2}{b-a}, \qquad E(X)=\frac{a+b+c}{3} \text{ for vertices } (a,0),\ (c,h),\ (b,0) \]

A symmetric triangular density has its peak at the midpoint, so the mean, median and mode all coincide there and no integration is needed for the centre. An asymmetric one has the peak off-centre, and then all three differ: the mean is pulled towards the longer side, away from the peak.

The great practical advantage of both distributions in this lesson is that every piece is a straight line. That makes probabilities the areas of rectangles, triangles and trapeziums, which is faster and far less error-prone than integrating. Save integration for the mean and variance, or for when a question explicitly asks you to show it.

Where this shows up in real life

Project planning

Project managers estimate task durations with a best case, a worst case and a most likely value, which defines an asymmetric triangular distribution exactly. Software then simulates thousands of schedules from these to estimate the chance of finishing on time.

Rounding and measurement error

When a value is rounded to the nearest unit, the error is uniform on an interval half a unit either side. Add two independent rounding errors and the result is triangular, which is the simplest case of errors accumulating towards a bell shape.

Try it yourself

The working method, twenty practice questions, ten harder ones, and the ClassPad guide.