Continuous Random Variables

Main idea

When a quantity can land anywhere in an interval, no single exact value has any real chance of happening. Probability lives in ranges instead, as the area under a density curve, and the whole area adds up to 1.

{\htmlData{sym=1}{P(a \le X \le b)}} = \int_a^b {\htmlData{sym=0}{f(x)}}\,dx, \qquad \int_{-\infty}^{\infty} {\htmlData{sym=0}{f(x)}}\,dx = 1

What the symbols mean

$f(x)$
Probability density
The height of the curve. Areas under it, not heights, are probabilities.
$P(a \le X \le b)$
Probability of a range
The area under the density between a and b.

Explanation

A continuous random variable is simply a random quantity that can land anywhere in a range, like a reaction time or a height, rather than on a short list of values like a die roll. "Continuous" means there are no gaps between the possible values. All we're really saying in this lesson is that the chance of landing in a range is the area under a curve over that range. The formula $P(a \le X \le b) = \int_a^b f(x)\,dx$ looks like calculus homework, but it's just three pieces: $f(x)$ (a curve that's high where values are common and low where they're rare), $a$ and $b$ (the ends of the range you care about) and $\int$ (the area between them). The second formula, total area $= 1$ , just says that something has to happen. It's the same "area under a curve" you met in calculus, wearing a new outfit.

From bars to curves. With a discrete random variable, like a die roll, each value got its own bar and the bars' heights added to 1. Now picture a measurement like a reaction time, which could be 0.31 s or 0.3107 s or anything in between. There are too many possible values to give each one a bar. So instead we draw a smooth curve and let area play the role the bar heights used to play.

Single values have probability 0. The area above a single point is zero, since it's a slice with no width. So $P(X = 2) = 0$ and $P(X \le 2) = P(X < 2)$ . That isn't saying 2 can't happen; it's saying any one exact value is infinitely unlikely compared with the whole range.

The mean of a continuous random variable is $\mu = \int x\,f(x)\,dx$ , the balance point of the area. It's the same idea as $\sum x\,P(x)$ for discrete variables, with the sum turned into an integral.

For the common distributions, areas come from formulas or tables rather than fresh integrals.

A common mistake. It's tempting to read the height $f(x)$ as a probability, because bar heights were probabilities in the discrete case. It isn't one. $f(x)$ can even exceed 1, for a narrow, tall density. Only areas are probabilities.

See it

The density here is f(x) = x/8 on [0, 4], a straight ramp. As the shaded area sweeps right, its size P(X ≤ x) = x²/16 traces out the cumulative distribution function F. Notice that F only reaches 0.25 by x = 2, halfway across, because the ramp is low on the left. It finishes at exactly 1 at x = 4, since the total area is 1.

Worked examples

Example 1

$X$ has density $f(x) = \frac{x}{8}$ on $[0, 4]$ . Find $P(X \le 2)$ .

1
The probability is the area under the density from 0 to 2: $\int_0^2 \frac{x}{8}\,dx = \left[\frac{x^2}{16}\right]_0^2 = \frac{4}{16}$ .

$\frac14$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

For $f(x) = \frac x8$ on $[0, 4]$ , find $P(X \le 2)$ .

Problem 2+5 XP

For the same density, find $P(1 \le X \le 3)$ .

Problem 3+5 XP

For the same density, find the mean $\mu = \int_0^4 x\,f(x)\,dx$ .

Problem 4+5 XP

Find $c$ so that $f(x) = cx^2$ on $[0, 3]$ is a density.

Problem 5+5 XP

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

Problem 6+5 XP

A machine part's lifetime, in years, has density $f(x) = \frac{3x^2}{1000}$ on $[0, 10]$ . What's the probability it lasts more than 8 years?

Problem 7+5 XP

Find the mean lifetime for the same density.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the median lifetime for the same density.

Where it's used

Heights, reaction times, rainfall amounts and the lifetimes of components are continuous; engineers and actuaries work with their densities.

Waiting for the Bus: Schedule vs RandomStats · Economics & Finance

The Poisson Distribution The Uniform Distribution