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Understand continuous random variables, probability density functions, and uniform distributions.
Learn step-by-step with practice exercises built right in.
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:
Instead of listing probabilities at specific values, continuous distributions use a probability density function f(x).
Key Properties:
A uniform distribution on [0, 4] has probability density f(x) = 0.25. Find P(X ≤ 1).
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].
Heights of adult males follow approximately N(μ=70, σ=3). What is P(X = 70)?
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Explore more AP Statistics topics
For continuous variables, probability is determined by area, not by the height of the function at a point.
Graphically:
The uniform distribution on interval [a, b] has constant density.
Mean: (midpoint)
Standard Deviation:
Probability: (rectangular area)
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
The normal distribution (bell curve) is the most important continuous distribution in statistics.
Properties:
Notation:
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.
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.
Daily rainfall X is modeled by a continuous distribution with density f(x) = 0.4e^{-0.4x} for x ≥ 0. Find P(X ≤ 2).
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.