Continuous random variables and probability density functions

Discrete and continuous

A discrete variable takes separate values (0, 1, 2, ...). A continuous random variable takes any value in an interval. Because there are infinitely many values, the probability of any single exact value is zero: P(X = a) = 0, so P(X < a) = P(X ≤ a).

The probability density function

A continuous random variable X is described by a probability density function (pdf) f(x) with:

• f(x) ≥ 0 for all x;
• the total area is 1: ∫ f(x) dx = 1 over all x;
• P(a < X < b) = ∫ from a to b of f(x) dx.

In S2, f(x) is a simple polynomial, possibly defined piecewise, and is 0 outside a stated range. f(x) itself is not a probability, so it can be bigger than 1.

Finding a constant

If f(x) = kx² for 0 ≤ x ≤ 3 (and 0 otherwise), then ∫ from 0 to 3 of kx² dx = 9k = 1, so k = 1/9. For a piecewise pdf, add the integrals over each piece and set the total to 1. Simple shapes (triangles, rectangles) can be done by area, which examiners accept; for anything else, integrate, and show the substitution of the limits.

Worked example

f(x) = kx² for 0 ≤ x ≤ 3 and 0 otherwise. Find k and P(1 < X < 2).

1. ∫ from 0 to 3 of kx² dx = k[x³/3] = 9k = 1, so k = 1/9.
2. P(1 < X < 2) = ∫ from 1 to 2 of x²/9 dx = [x³/27] from 1 to 2
3. = (8 - 1)/27

Answer: k = 1/9 and P(1 < X < 2) = 7/27

Check yourself

1. f(x) = kx² for 0 ≤ x ≤ 3 and 0 otherwise is a pdf. Find k. Give your answer as a fraction in its simplest form.
   Show answer

   1/9. ∫ from 0 to 3 of kx² dx = 9k = 1.

2. f(x) = x²/9 for 0 ≤ x ≤ 3 and 0 otherwise. Find P(1 < X < 2). Give your answer as a fraction in its simplest form.
   Show answer

   7/27. [x³/27] from 1 to 2 = 8/27 - 1/27 = 7/27.

3. f(x) = x²/9 for 0 ≤ x ≤ 3 and 0 otherwise. Find P(X > 2.5). Give your answer to 3 significant figures.
   Show answer

   0.421. 1 - [x³/27] from 0 to 2.5 = 1 - 15.625/27 = 0.4213.

4. A pdf is f(x) = kx for 0 ≤ x ≤ 2, f(x) = 2k for 2 < x ≤ 3, and 0 otherwise. Find k. Give your answer as a fraction in its simplest form.
   Show answer

   1/4. Area = ∫ from 0 to 2 of kx dx + 2k × 1 = 2k + 2k = 4k = 1.

5. f(x) = k(4 - x) for 1 ≤ x ≤ 3 and 0 otherwise is a pdf. Find k. Give your answer as a fraction in its simplest form.
   Show answer

   1/4. ∫ from 1 to 3 of (4 - x) dx = [4x - x²/2] = 7.5 - 3.5 = 4, so 4k = 1.

6. Which of these could be a probability density function on 0 ≤ x ≤ 2 (and 0 otherwise)?
   • f(x) = x
   • f(x) = x - 1
   • f(x) = 1
   • f(x) = x/2
   Show answer

   f(x) = x/2. x/2 is never negative on 0 to 2 and its area is [x²/4] = 1.

7. X is a continuous random variable with pdf f(x) = x/2 for 0 ≤ x ≤ 2. What is P(X = 1)?
   • 0.5
   • 0
   • 0.25
   • 1
   Show answer

   0. For a continuous variable, a single value has zero probability (no area).

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit S2

Last updated 8 November 2026. Spotted a mistake? Tell us.