Continuous random variables
Objective. Use density and distribution functions.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
A continuous random variable is described by a probability density function f(x), where probabilities correspond to areas under the curve rather than individual point values, so P(X=x)=0 for any specific x. The total area under f(x) over its entire domain must equal 1, and probabilities of intervals are found by integrating f(x) between the relevant bounds: P(a<X<b) = ∫ₐ^b f(x)dx. The cumulative distribution function F(x) = P(X≤x) is the integral of f(x) from the lower limit of the domain up to x, and f(x) is recovered by differentiating F(x).
Worked example
A continuous random variable X has probability density function f(x) = kx for 0 ≤ x ≤ 4, and 0 otherwise. Find the value of k and then P(1 ≤ X ≤ 2).
- 1Total area must be 1: ∫₀^4 kx dx = k[x²/2]₀^4 = 8k = 1, so k = 1/8
- 2Density function is f(x) = x/8 for 0≤x≤4
- 3P(1≤X≤2) = ∫₁^2 (x/8)dx = [x²/16]₁^2
- 4Evaluate: 4/16 − 1/16 = 3/16
Answer: k = 1/8; P(1≤X≤2) = 3/16
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Start practisingCommon mistake. Students treat f(x) itself as a probability, e.g. stating P(X=2) = f(2), when in fact probability requires integrating f(x) over an interval, and any single point has probability zero. Always integrate the density function to find probabilities.
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