MATH U113 · Probability & Statistics · Module 3

Continuous Random Variables & Probability Distributions

Devore (9th ed., Metric) §4.1–4.4 — pdfs, cdfs and expected values, the normal distribution, exponential, gamma and chi-squared. The compressed revision map. (The handout's extra topics — continuous MGFs, Chebyshev's inequality R1 §4.5 and transformation methods R1 §6.7 — are in Part 2.)

Start here

First time with this material? The Module 3 lesson teaches it slowly, with the interactive z-table curve. Grounding facts: this chapter is brand new to everyone — no board or entrance exam covers it; and the calculus load is real but bounded (polynomial and e^−λx integrals only — the MATH U101 drill covers the toolkit). The chapter is Module 2 with Σ swapped for ∫; the dictionary below makes that literal. Handout scope check: §4.1–4.4 are in (with the normal's mean/variance proofs officially omitted — a small mercy); §4.5–4.6 are out (kept below, clearly marked, for curiosity only); and the syllabus adds MGFs, Chebyshev's inequality (R1 §4.5, stated without proof) and transformation methods (R1 §6.7) — those live in Part 2 notes (lesson).

4.1–4.2 · The framework, continuous edition §4.1–4.2

Concept	Discrete (Module 2)	Continuous (this module)
Distribution	pmf p(x): bars	pdf f(x): curve; probability = area
Legitimacy	p ≥ 0, Σp = 1	f ≥ 0, ∫_−∞^∞f = 1
Interval probability	F(b) − F(a−1)	F(b) − F(a) — no off-by-one; P(X = c) = 0, so ≤ and < agree
cdf ↔ distribution	running total; jumps	F(x) = ∫_−∞^xf; smooth; F′ = f (FTC)
Mean, variance	Σ x p(x); E(X²) − μ²	∫ x f(x) dx; same shortcut, same rescaling rules
New tool	—	Percentiles: solve F(η) = p; median at p = 0.5

Special case worth naming: uniform on [A, B] — f = 1/(B−A), probabilities are lengths ÷ total length. And remember f(x) is a density, not a probability — it may exceed 1; only areas are probabilities (same rule as the R-lab page's density histograms).

Probability = area under the density: the shaded strip is P(1 ≤ X ≤ 1.5) = F(1.5) − F(1); a single point is a line with no area, so P(X = c) = 0.

Worked example 1The full §4.1–4.2 machine on one pdf

Machining error X (mm) has pdf f(x) = kx² on [0, 2], 0 elsewhere. Find k, F, P(X ≤ 1), P(1 ≤ X ≤ 1.5), the median, E(X), and σ.

k from total area 1: ∫₀² kx² dx = k·x³3]₀² = 8k3 = 1 ⟹ k = 3/8.
cdf: F(x) = ∫₀^x (3/8)y² dy = x³/8 for 0 ≤ x ≤ 2 (0 below, 1 above).
Probabilities by two lookups: P(X ≤ 1) = 1/8 = 0.125; P(1 ≤ X ≤ 1.5) = F(1.5) − F(1) = 3.375/8 − 1/8 ≈ 0.297.
Median: solve η³/8 = 0.5 ⟹ η³ = 4 ⟹ η = ∛4 ≈ 1.587 mm.
Mean: E(X) = ∫₀² x · (3/8)x² dx = (3/8)·x⁴4]₀² = (3/8)(4) = 1.5 mm. (Mean < median 1.587: this density's thin tail points left — the descriptive-statistics rule, still working.)
Variance by shortcut: E(X²) = (3/8)∫₀² x⁴ dx = (3/8)(32/5) = 2.4, so V = 2.4 − 1.5² = 0.15, σ ≈ 0.387 mm.

4.3 · The normal distribution §4.3

X ~ N(μ, σ): bell centred at μ, width set by σ. Its cdf has no formula, so every computation routes through the standard normal Z ~ N(0, 1) and its tabulated cdf Φ (the lesson's slider is the table, animated). The recipe never changes:

standardise z = x − μσ → look up P(X ≤ x) = Φ(z) → assemble (right tail: 1 − Φ; between: two lookups)

Φ(z) is the whole area left of z — one lookup; the right tail is 1 − Φ, and symmetry gives Φ(−z) = 1 − Φ(z).

Symmetry: Φ(−z) = 1 − Φ(z) — tables often print only z ≥ 0.
Percentiles backwards: find z with Φ(z) = p in the table's body, then x = μ + zσ. Notation: z_α has area α to the right (z_0.05 = 1.645, z_0.025 = 1.96).
68–95–99.7: within 1σ / 2σ / 3σ of μ — the universal sanity check.
Binomial approximation: if np ≥ 10 and nq ≥ 10, Bin(n,p) ≈ N(np, √(npq)) with continuity correction: P(X ≤ x) ≈ Φ((x + 0.5 − np)/√(npq)).

Worked example 2Normal computations, both directions

A batch of "1 kΩ" resistors is N(μ = 1000 Ω, σ = 20 Ω). (a) P(X ≤ 1025)? (b) P(960 ≤ X ≤ 1040)? (c) The 99th percentile of resistance?

(a) z = (1025 − 1000)/20 = 1.25 → Φ(1.25) = 0.8944.
(b) Two lookups: z = ±2 → Φ(2) − Φ(−2) = 0.9772 − 0.0228 = 0.9544. (The 95-rule promised ≈ 95% within 2σ ✓.)
Unpack Φ(−2)

Symmetry: Φ(−2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228 — the left tail mirrors the right.
(c) Backwards: Φ(z) = 0.99 → table body gives z = 2.33 → x = 1000 + 2.33(20) = 1046.6 ≈ 1047 Ω. Only 1% of resistors exceed it.
Exam habit: write the standardising step explicitly every time — it's a method mark, and it catches σ/σ² slips before they propagate.

4.4 · Exponential and gamma §4.4

Exponential(λ) — waiting time to the next event of a Poisson process with rate λ (Module 2's promised handshake):

f(x) = λe^−λx (x ≥ 0) · F(x) = 1 − e^−λx · E(X) = σ = 1/λ · P(X > t) = e^−λt

Closed-form cdf — no table needed. Memoryless: P(X > s+t | X > s) = e^−λt = P(X > t) — survival so far never changes the outlook. Unique to the exponential (among continuous distributions), loved by examiners, and the reason it can't model wear-out.

Gamma(α, β) generalises it (α = 1 ⟹ exponential with λ = 1/β): shape α, scale β, E(X) = αβ, V(X) = αβ², built on Γ(α) where Γ(n) = (n−1)! for whole n and Γ(α) = (α−1)Γ(α−1). Interpretation: waiting time to the α-th Poisson event. Compute probabilities via the incomplete-gamma table (Devore Appendix) when asked; know E, V, and the α = 1 reduction cold.

Chi-squared(ν) — named in the handout, and just a gamma in uniform: it is the gamma with α = ν/2, β = 2, where the whole number ν is called the degrees of freedom. So its facts come free from the gamma formulas: E(X) = αβ = ν and V(X) = αβ² = 2ν. Nothing more is needed now — it exists in this module so that it's an old friend when it returns as the backbone of variance inference in the statistics half of the course.

Worked example 3Exponential lifetimes + the memoryless move

A sensor's lifetime (in thousands of hours) is exponential with λ = 0.5. (a) Mean and SD? (b) P(it lasts beyond 3)? (c) P(it fails within its first 1)? (d) Given it has already run 2, P(it lasts at least 3 more)?

(a) E(X) = 1/λ = 2 (i.e. 2000 h), and for the exponential σ = 1/λ = 2 as well — mean and SD coincide, its own signature.
(b) P(X > 3) = e^{−0.5 × 3} = e^−1.5 ≈ 0.223.
(c) P(X ≤ 1) = 1 − e^−0.5 ≈ 0.393 — nearly 40% die before the "half-mean" mark; the exponential front-loads its failures.
(d) Memoryless: P(X > 5 | X > 2) = P(X > 3) = e^−1.5 ≈ 0.223 — identical to (b). Two thousand hours of faithful service changed nothing. State the property by name for the mark, then apply it in one line.

4.5–4.6 · The other families & probability plots §4.5–4.6 off-syllabus — handout confirmed

Skip for exams. The handout leaves §4.5–4.6 out of the lecture plan. The table stays for one honest reason: these names (Weibull, lognormal) appear constantly in engineering practice, and knowing they exist costs two minutes. Spend zero practice problems here.

Family	One-line identity	Use it for
Weibull	Exponential with an age-dependent failure rate (shape α tunes wear-out vs. early failure)	Lifetimes, reliability
Lognormal	ln(X) is normal → take logs, then it's §4.3	Multiplicative growth: incomes, particle sizes, repair times
Beta	Lives on a bounded [A, B]	Proportions, completion fractions
Probability plot §4.6	Sorted data vs. theoretical percentiles: straight line ⟹ the model fits	"Is my data plausibly normal?"

Hold this section at reading depth until the handout confirms scope — then, if included, the skills to drill are Weibull/lognormal probability computations (both reduce to plugging into F or standardising after a log) and reading straightness in a plot.

Classic traps in Chapter 4

1) Treating f(x) as a probability — it's a density; it may exceed 1; only areas are probabilities. 2) Standardising with σ² instead of σ — if a "z" comes out enormous, this is usually why. 3) Forgetting the table gives the left area — right tails are 1 − Φ, and Φ(−z) = 1 − Φ(z). 4) λ↔mean confusion in the exponential: the mean is 1/λ, and λ's units must match x's (per hour with hours). 5) Dropping the continuity correction when approximating a binomial — half a unit that moves the third decimal, and marked accordingly.

Your minimal prerequisite kit for this module

Power-rule integrals: ∫xⁿ dx = xⁿ⁺¹/(n+1), evaluated between limits — the engine of Worked Example 1. Drill: MATH U101 daily drill.
Exponential integrals: ∫λe^−λx dx = −e^−λx + C — one antiderivative covers all of §4.4.
Improper integrals as limits: ∫₀^∞ means "integrate to b, let b → ∞" — conceptual comfort is enough here.
FTC: differentiate the cdf to get the pdf, integrate the pdf to get the cdf.
Roots & logs on a calculator: cube roots for percentiles, ln for exponential inversions ("solve e^−λt = 0.1").

What to practise in Devore

Skill	Where	How many
pdf validity, find-k, uniform probabilities	§4.1 exercises	3–4
cdf both directions, percentiles/medians, E & V by integration	§4.2 exercises	5–6
Normal: probabilities, percentiles backwards, binomial approximation	§4.3 exercises	6–8
Exponential (incl. memoryless & Poisson-process link); gamma basics	§4.4 exercises	4–5
Weibull/lognormal/beta & probability plots — off-syllabus (handout confirmed)	skip §4.5–4.6 entirely	0

Prefer odd-numbered exercises (answers to selected odd ones in the back). Metric Version numbering may differ from the US edition — choose by section and skill, not by numbers copied from elsewhere.