Ch 3 · 6.3 Continuous Random Variables · A Level Mathematics 9709
Topic 3 of 5 in Probability & Statistics 2 — A Level Mathematics 9709 — 48 lessons.
Introduction
The time a customer waits, the lifetime of a light bulb, the distance a seed travels: these can take any value in a range, not just whole numbers. For such a continuous random variable, probabilities are areas under a curve called the probability density function. Integration from Paper 3 does the work. (Syllabus section 6.3, Paper 6.)
Key ideas — The probability density function
A continuous random variable X is described by a pdf f(x) with
🧑🏫 Teacher's explanation
Imagine you are throwing a dart at a long, thin ruler. Unlike a discrete variable (like rolling a die) where you land exactly on '3', with a continuous variable, you land on a specific point like 3.14159.... Because there are infinite points, the probability of hitting one exact, infinitely precise point is actually zero. Instead, we talk about the probability of landing within a certain range, like between 3 and 4.
📖 Definitions
🔍 Worked example — Easy
A continuous random variable X has a pdf given by f(x) = kx for 0 ≤ x ≤ 2, and f(x) = 0 elsewhere. Find the value of k.
🔍 Worked example — Standard
Given f(x) = 3/x^4 for x ≥ 1, find P(2 < X < 3).
🔍 Worked example — Exam-style
A continuous random variable X has pdf f(x) = c(4x - x^2) for 0 ≤ x ≤ 4. Find the median of X.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Show that ∫_0^∞ f(x) dx = 1. [2 marks] (b) Find the value of λ such that P(X > 1) = 0.5. [3 marks] [5 marks]
Key ideas — Finding an unknown constant
If f(x) = kx^2 for 0 ≤ x ≤ 3, then ∫_0^3 kx^2 dx = 9k = 1, so k = tfrac19. Questions also ask you to sketch a pdf: draw it only over its domain (it is zero elsewhere).
🧑🏫 Teacher's explanation
Imagine you have a lump of clay shaped like a hill. The total amount of clay you have is exactly 1 unit. If I tell you the shape of the hill follows a specific mathematical rule, like f(x) = kx^2, the constant k is simply the 'scaling factor' that ensures the total volume under that hill equals exactly 1. If k were too large, you'd have too much clay; if k were too small, you'd have too little.
📖 Definitions
🔍 Worked example — Easy
A continuous random variable X has a pdf given by f(x) = kx for 0 ≤ x ≤ 4, and f(x) = 0 elsewhere. Find the value of the constant k.
🔍 Worked example — Standard
A continuous random variable Y has a pdf f(y) = c(4 - y) for 0 ≤ y ≤ 4. Find the value of c.
🔍 Worked example — Exam-style
The pdf of a continuous random variable Z is f(z) = k(z^2 + 1) for 0 ≤ z ≤ 2, and f(z) = 0 elsewhere. Find k and sketch the shape of the function.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Mean and variance
E(X) = ∫ x f(x) dx Var(X) = ∫ x^2 f(x) dx - big[E(X)big]^2
🧑🏫 Teacher's explanation
In discrete probability, you find the average by multiplying each value by its probability and summing them up. For a continuous random variable, the variable can take any value in an interval (like 1.234...), so we can't sum individual points. Instead, we use integration to 'sum up' the values weighted by the probability density function f(x). Think of it like finding the center of mass of a thin metal rod where the density varies along its length.
📖 Definitions
🔍 Worked example — Easy
A continuous random variable X has the probability density function f(x) = 3x^2 for 0 ≤ x ≤ 1, and f(x) = 0 elsewhere. Find E(X).
🔍 Worked example — Standard
For the same variable X where f(x) = 3x^2 for 0 ≤ x ≤ 1, find Var(X).
🔍 Worked example — Exam-style
A continuous random variable X has pdf f(x) = kx e^-x for x ≥ 0. Find the value of k such that E(X) = 2.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Median and percentiles
The median m splits the area in half:
🧑🏫 Teacher's explanation
Imagine you have a long, unevenly shaped piece of dough. If you want to find the 'median' point, you aren't looking for the middle of the length, but the point where the volume of dough on the left side exactly equals the volume on the right side. In probability, the 'volume' is the area under the curve of our probability density function (pdf), f(x).
📖 Definitions
🔍 Worked example — Easy
A continuous random variable X has a probability density function f(x) = 2x for 0 ≤ x ≤ 1, and f(x) = 0 elsewhere. Find the median of X.
🔍 Worked example — Standard
A continuous random variable X has a pdf f(x) = 3/8x^2 for 0 ≤ x ≤ 2. Find the upper quartile.
🔍 Worked example — Exam-style
A continuous random variable X has a pdf f(x) = kx(4-x) for 0 ≤ x ≤ 4. Find the value of k and the 90th percentile.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples
f(x) = kx^2 for 0 ≤ x ≤ 3, zero otherwise.
🔍 Worked example — Easy
A continuous random variable X has a probability density function f(x) = kx^2 for 0 ≤ x ≤ 3, and f(x) = 0 elsewhere. Find the value of k, the mean E(X), and the variance Var(X).
🔍 Worked example — Standard
A continuous random variable X has a pdf f(x) = k/x^4 for x ≥ 1, and f(x) = 0 otherwise. Find the value of k and the median of the distribution.
🔍 Worked example — Exam-style
For the distribution f(x) = kx^2 for 0 ≤ x ≤ 3 (where k = 1/9), find the upper quartile q.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) k (8 - tfrac83 ) = 1 ⇒ k = 3/16; P(X < 1) = 3/16 (4 - tfrac13 ) = 11/16; E(X) = 3/16(8 - 4) = tfrac34. (2) E(X^2) = 3/16 (32/3 - 32/5 ) = tfrac45; Var(X) = tfrac45 - 9/16 = 19/80. (3) ∫_1^∞ 2x^-3 dx = 1 and f ≥ 0 ✓; 1 - m^-2 = tfrac12 ⇒ m = sqrt2; E(X) = ∫_1^∞ 2x^-2 dx = 2. (4) The pdf is symmetric about x = π/2, so the median and the mean are both π/2.