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Continuous Random Variables

Understand continuous random variables, probability density functions, and uniform distributions.

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📈 Continuous Random Variables

What is a Continuous Random Variable?

A continuous random variable takes any value in an interval. Unlike discrete random variables (which take specific values like 1, 2, 3), continuous variables can be any real number within a range.

Examples:

  • Height of students (150 cm to 210 cm)
  • Time to complete a test (0 to 180 minutes)
  • Temperature (could be 20.5°C, 20.57°C, 20.571°C, ...)

Probability Density Functions (PDF)

Instead of listing probabilities at specific values, continuous distributions use a probability density function f(x).

Key Properties:

  1. f(x)≥0f(x) \geq 0 for all x (function is non-negative)
  2. Total area under the curve = 1: ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x) \, dx = 1
  3. Probability is area: P(a≤X≤b)=∫abf(x) dxP(a \leq X \leq b) = \int_a^b f(x) \, dx
  4. Point probability is zero: P(X=c)=0P(X = c) = 0 for any single value

Why Area Under Curve = Probability

For continuous variables, probability is determined by area, not by the height of the function at a point.

Graphically:

  • The total area under the curve represents 100% probability (or 1.0)
  • The area between two x-values represents the probability the variable falls in that range
  • A curve that never touches the x-axis means those x-values have zero probability

Uniform Distribution

The uniform distribution on interval [a, b] has constant density.

f(x)=1b−a for a≤x≤bf(x) = \frac{1}{b - a} \text{ for } a \leq x \leq b

Mean: μ=a+b2\mu = \frac{a + b}{2} (midpoint)

Standard Deviation: σ=b−a12\sigma = \frac{b - a}{\sqrt{12}}

Probability: P(c≤X≤d)=d−cb−aP(c \leq X \leq d) = \frac{d - c}{b - a} (rectangular area)

Uniform Distribution Example

Buses arrive every 15 minutes. Your arrival time is uniformly distributed between buses. What is the probability you wait less than 5 minutes?

Given: X ~ Uniform(0, 15) where X = wait time in minutes

P(0≤X≤5)=5−015−0=515=13≈0.333P(0 \leq X \leq 5) = \frac{5 - 0}{15 - 0} = \frac{5}{15} = \frac{1}{3} \approx 0.333

Normal Distribution as Continuous RV

The normal distribution (bell curve) is the most important continuous distribution in statistics.

Properties:

  • Symmetric about the mean μ
  • Determined by two parameters: μ (mean) and σ (standard deviation)
  • Approximately 68% of data within 1σ, 95% within 2σ, 99.7% within 3σ
  • Used to model many natural phenomena (heights, test scores, errors)

Notation: X∼N(μ,σ2)X \sim N(\mu, \sigma^2)

Common Mistakes

  1. Confusing PDF height with probability: The y-axis value is not probability; only area is
  2. Thinking P(X=c)≠0P(X = c) \neq 0: For continuous distributions, the probability of any exact value is always 0
  3. Not recognizing when to use continuous vs discrete: Discrete has specific values; continuous has intervals

AP Exam Tip

When dealing with continuous variables on the AP exam, always think "area under the curve." If asked for a probability, you need to find an area (using tables, calculators, or normal approximation). Never calculate probability for a single point; always use an interval.

📚 Practice Problems

1Problem 1easy

❓ Question:

A uniform distribution on [0, 4] has probability density f(x) = 0.25. Find P(X ≤ 1).

💡 Show Solution

For a continuous distribution, probability is area under the density curve. P(X ≤ 1) = ∫₀¹ 0.25 dx = 0.25 × 1 = 0.25. This means 25% of the probability mass lies between 0 and 1 in a uniform distribution on [0, 4].

2Problem 2medium

❓ Question:

Heights of adult males follow approximately N(μ=70, σ=3). What is P(X = 70)?

💡 Show Solution

For any continuous random variable, P(X = c) = 0 for any single point c. P(X = 70) = 0, even though 70 is the mean. Probability is area, and a single point has no width. Instead, we ask P(69.5 ≤ X ≤ 70.5) ≈ 0.1326, which is the probability of a small interval.

3Problem 3hard

❓ Question:

Daily rainfall X is modeled by a continuous distribution with density f(x) = 0.4e^{-0.4x} for x ≥ 0. Find P(X ≤ 2).

💡 Show Solution

P(X ≤ 2) = ∫₀² 0.4e^{-0.4x} dx. Let u = -0.4x, then du = -0.4dx. ∫ 0.4e^{-0.4x} dx = -e^{-0.4x} + C. P(X ≤ 2) = [-e^{-0.4x}]₀² = -e^{-0.8} + e⁰ = 1 - e^{-0.8} ≈ 1 - 0.449 = 0.551. About 55% probability that rainfall is at most 2 units.

Explain using:

⚠️ Common Mistakes: Continuous Random Variables

Avoid these 3 frequent errors

📌 Related Topics in Unit 4: Probability, Random Variables, and Probability Distributions

❓ Frequently Asked Questions

What is Continuous Random Variables?▾
Understand continuous random variables, probability density functions, and uniform distributions.
How can I study Continuous Random Variables effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Continuous Random Variables study guide free?▾
Yes — all study notes, flashcards, and practice problems for Continuous Random Variables on Study Mondo are free to access. No account is needed.
What course covers Continuous Random Variables?▾
Continuous Random Variables is part of the AP Statistics course on Study Mondo, specifically in the Unit 4: Probability, Random Variables, and Probability Distributions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Continuous Random Variables?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.