Triangular Distribution — Definition, Formula & Examples

The triangular distribution is a continuous probability distribution shaped like a triangle, defined entirely by three parameters: a minimum value, a peak (mode), and a maximum value. It is often used when you have limited data but can estimate the lowest, most likely, and highest possible outcomes.

A continuous random variable $X$ follows a triangular distribution with parameters $a$ (minimum), $b$ (maximum), and $c$ (mode), where $a \le c \le b$ , if its probability density function is piecewise linear, rising from zero at $x = a$ to a maximum of $\frac{2}{b - a}$ at $x = c$ , then falling linearly back to zero at $x = b$ .

Key Formula

f(x) = \begin{cases} \dfrac{2(x - a)}{(b - a)(c - a)} & a \le x \le c \\[6pt] \dfrac{2(b - x)}{(b - a)(b - c)} & c < x \le b \\[6pt] 0 & \text{otherwise} \end{cases}

Where:

$a$ = Minimum value (left endpoint of the distribution)
$b$ = Maximum value (right endpoint of the distribution)
$c$ = Mode (most likely value, where the peak occurs)
$x$ = A value of the random variable

How It Works

You use the triangular distribution when you can reasonably estimate three values: the worst case (

a

), the best case (

b

), and the most likely case (

c

). The PDF forms a triangle over the interval

[a, b]

with its apex at

c

. Because the total area under any PDF must equal 1, the height at the peak is always

\frac{2}{b-a}

, regardless of where

c

falls. The mean of the distribution is simply the average of the three parameters:

\frac{a + b + c}{3}

. This makes it a quick tool for rough estimates in project management, risk analysis, and simulation when precise data is unavailable.

Worked Example

Problem: A task takes at least 2 hours, at most 10 hours, and most likely 4 hours. Find the mean and the value of the PDF at x = 4.

Identify parameters: Set the minimum, mode, and maximum.

a = 2,\quad c = 4,\quad b = 10

Compute the mean: The mean of a triangular distribution is the average of its three parameters.

\mu = \frac{a + b + c}{3} = \frac{2 + 10 + 4}{3} = \frac{16}{3} \approx 5.33 \text{ hours}

Evaluate the PDF at the mode: The peak height of the triangle depends only on the base length

b - a

f(4) = \frac{2}{b - a} = \frac{2}{10 - 2} = 0.25

Answer: The expected task duration is approximately 5.33 hours, and the PDF reaches its maximum value of 0.25 at x = 4.

Why It Matters

Project managers use the triangular distribution in PERT (Program Evaluation and Review Technique) to estimate task durations when historical data is scarce. It also appears in Monte Carlo simulations for risk analysis, where analysts need a simple distribution that captures asymmetry without requiring detailed data fitting.

Common Mistakes

Mistake: Assuming the mode must be at the midpoint of the interval.

Correction: The mode

c

can be anywhere from

a

b

. When

c

is not centered, the triangle is asymmetric (skewed), which is one of the distribution's advantages over the uniform distribution.