Continuous Probability Distributions
Understand probability density functions, cumulative distribution functions, and the uniform distribution for continuous random variables in HSC Advanced Mathematics.
Probability Density Functions (PDF)
A continuous random variable can take any value in an interval. Unlike discrete variables, probabilities are found using areas under a curve called the probability density function (PDF), denoted f(x).
Properties of a PDF f(x):
1. f(x) ≥ 0 for all x (the curve is never negative).
2. The total area under the curve equals 1: ∫−∞∞ f(x) dx = 1.
3. P(a ≤ X ≤ b) = ∫ab f(x) dx (probability = area under the curve between a and b).
Key insight: For continuous random variables, P(X = a) = 0 for any single value a. Probabilities only make sense over intervals.
This means P(X ≤ a) = P(X < a) since the probability at a single point is zero.
Cumulative Distribution Functions (CDF)
The cumulative distribution function (CDF) F(x) gives the probability that X takes a value less than or equal to x:
F(x) = P(X ≤ x) = ∫−∞x f(t) dt
The CDF is the integral of the PDF from −∞ to x.
The CDF is a non-decreasing function with F(−∞) = 0 and F(∞) = 1. To find the PDF from the CDF, differentiate: f(x) = F'(x).
Useful property: P(a ≤ X ≤ b) = F(b) − F(a).
This avoids re-integrating the PDF every time — just use the CDF values.
The Uniform Distribution
The continuous uniform distribution on [a, b] assigns equal probability to all values in the interval. It is the simplest continuous distribution.
PDF: f(x) = 1/b − a for a ≤ x ≤ b, and f(x) = 0 otherwise.
Mean: μ = a + b/2
Variance: σ2 = (b − a)212
Example: A bus arrives uniformly at random between 8:00 and 8:20. What is the probability you wait less than 5 minutes (arriving at 8:00)?
X ~ Uniform(0, 20). P(X ≤ 5) = (5 − 0)/(20 − 0) = 5/20 = 0.25.
Mean wait = (0 + 20)/2 = 10 minutes.
Key Vocabulary
PDF (f(x))
The probability density function whose area under the curve over an interval gives the probability.
CDF (F(x))
The cumulative distribution function giving P(X ≤ x), obtained by integrating the PDF.
Uniform Distribution
A distribution where every value in [a, b] is equally likely, with constant PDF f(x) = 1/(b−a).
Continuous Random Variable
A variable that can take any value in a continuous interval, measured rather than counted.
Worked Examples
A continuous random variable has PDF f(x) = 3x2 for 0 ≤ x ≤ 1. Verify this is a valid PDF and find P(X > 0.5).
Step 1: Check: f(x) = 3x2 ≥ 0 for 0 ≤ x ≤ 1. ∫01 3x2 dx = [x3]01 = 1. Valid PDF.
Step 2: P(X > 0.5) = ∫0.51 3x2 dx = [x3]0.51 = 1 − 0.125.
Answer: P(X > 0.5) = 0.875 or 7/8.
Find the CDF for Example 1 and use it to find P(0.2 ≤ X ≤ 0.8).
Step 1: F(x) = ∫0x 3t2 dt = x3 for 0 ≤ x ≤ 1.
Step 2: P(0.2 ≤ X ≤ 0.8) = F(0.8) − F(0.2) = 0.83 − 0.23 = 0.512 − 0.008.
Answer: P(0.2 ≤ X ≤ 0.8) = 0.504.
X ~ Uniform(2, 8). Find the mean, variance, and P(3 ≤ X ≤ 6).
Step 1: Mean = (2 + 8)/2 = 5. Variance = (8 − 2)2/12 = 36/12 = 3.
Step 2: P(3 ≤ X ≤ 6) = (6 − 3)/(8 − 2) = 3/6 = 0.5.
Answer: Mean = 5, Variance = 3, P(3 ≤ X ≤ 6) = 0.5.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
For a continuous random variable, what is P(X = 5)?
Question 2
X ~ Uniform(0, 10). What is P(X > 7)?
Question 3
Which property must a valid PDF satisfy?
Question 4
If F(x) = x2 for 0 ≤ x ≤ 1 is a CDF, what is the PDF f(x)?
Question 5
X ~ Uniform(5, 15). What is the variance?
Key Concepts Summary
- ● A PDF f(x) defines probabilities as areas: P(a ≤ X ≤ b) = ∫ f(x) dx from a to b.
- ● A valid PDF must satisfy f(x) ≥ 0 and integrate to 1 over its domain.
- ● The CDF F(x) = P(X ≤ x) and f(x) = F'(x). Use F(b) − F(a) for interval probabilities.
- ● For continuous variables, P(X = a) = 0 — probabilities only exist over intervals.
- ● The uniform distribution on [a, b] has f(x) = 1/(b−a), mean (a+b)/2, variance (b−a)2/12.