Bivariate Data: Correlation and Lines of Best Fit
A senior study guide to analyzing two numerical variables: describing scatterplots, interpreting Pearson's r and the coefficient of…
- Difficulty
- Level 3 · Challenge
- Time
- 25 min
A senior math worksheet on continuous random variables: checking and completing probability density functions, finding probabilities by integration, calculating the mean, variance and median, moving between a density function and its cumulative distribution function, and applying the uniform and normal distributions. Finishes with an extended problem on the lifetime of a component and its warranty. Full working is given for every answer.
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Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.
Completed: Continuous Random Variables and Probability Density Functions
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12th Grade students (typically ages 17–18) working on statistics. It is pitched at level 3 · challenge.
Yes. The PDF includes a full answer sheet with the working for every question. You can also view the answers online.
About 50 minutes for 17 questions. Short, regular sessions work best: two or three a week beats one long one.
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A continuous random variable can take any value in an interval (a time, a mass, a length). It is described by a probability density function (pdf) f(x), with f(x) ≥ 0 everywhere and total area 1: ∫ f(x) dx = 1 over all x.
Probabilities are areas: P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx. Because a single point has no area, P(X = a) = 0, and ≤ and < give the same probability.
Worked example. f(x) = kx for 0 ≤ x ≤ 2. ∫ from 0 to 2 of kx dx = 2k = 1, so k = . Then P(X < 1) = ∫ from 0 to 1 of dx = [frac 1/4].
Mean: E(X) = μ = ∫ x f(x) dx. Variance: Var(X) = E(X2) − μ2, where E(X2) = ∫ x2 f(x) dx. Standard deviation: σ = √Var(X).
The median m splits the area in half: ∫ from the lowest value to m of f(x) dx = .
Worked example. f(x) = 2x on 0 ≤ x ≤ 1. E(X) = ∫ 2x2 dx = ; E(X2) = ∫ 2x3 dx = ; Var(X) = − = . Median: m2 = , so m = √0.5 ≈ 0.707.
The cumulative distribution function (cdf) is F(x) = P(X ≤ x), the area under f to the left of x. It rises from 0 to 1, and f(x) = F′(x).
With a cdf, probabilities need no integration: P(a < X ≤ b) = F(b) − F(a), and the median solves F(m) = .
Worked example. f(x) = 2x on 0 ≤ x ≤ 1 has F(x) = x2 for 0 ≤ x ≤ 1. So P(0.5 < X < 0.8) = 0.64 − 0.25 = 0.39.
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