MATH U113 · First-time lesson · Module 3
Continuous random variables, taught from zero
Devore (9th ed., Metric) §4.1–4.4 — density curves, cdfs and percentiles, the normal distribution, exponential, gamma and chi-squared. Budget 2–3 hours; section 3 (the normal) deserves the biggest slice.
First time through → this page, in order, attempting each green check before opening. Revising → the notes page. Status check: this chapter is brand new to every person in the hall — no school board or entrance exam teaches continuous distributions. And the calculus it needs is honest but bounded: definite integrals of polynomials and of e−λx — nothing more. That's exactly what the MATH U101 daily drill builds; if integrals feel shaky, do a week of that drill alongside this module and the two courses will reinforce each other.
1 · From bars to curves §4.1
Module 2's variables counted things — the values sat at 0, 1, 2, … with a probability bar on each. But measure something — a resistor, a wait, a lifetime — and the possible values fill a whole interval. Between 1.2 and 1.3 there's 1.25; between 1.25 and 1.26 there's 1.255. No listing, so no bars. What replaces the pmf?
You've already met the answer, in the descriptive-statistics lesson: the density histogram, where area = proportion. Let the sample grow and the classes shrink, and the staircase of bars smooths into a curve — the probability density function (pdf) f(x). Probability is now the area under the curve, and area is what integrals compute:
A legitimate pdf needs f(x) ≥ 0 and total area ∫−∞∞ f(x) dx = 1 — the same two laws every pmf obeyed, with Σ traded for ∫. That trade is the whole chapter: every Module 2 formula survives with the sum replaced by an integral.
Two consequences take a moment to accept:
First: P(X = c) = 0 for every single value c. A single point is an interval of width zero — zero area. Yet some value always occurs! The resolution: probability zero doesn't mean impossible for continuous rvs; it means "a target infinitely too thin to bet on". Ask about intervals, and everything is sensible again. A pleasant side-effect: ≤ versus < no longer matters — P(X ≤ b) = P(X < b) — so Module 2's off-by-one anxiety is retired in this chapter (a genuinely relaxing fact).
Second: f(x) is not a probability. It's a probability density — probability per unit length, exactly like the descriptive-statistics lesson's histogram densities. It can exceed 1 (a uniform rv on [0, ½] has f = 2); only areas are probabilities.
The simplest pdf, the uniform distribution on [A, B], spreads probability evenly: f(x) = 1/(B−A) on the interval, 0 outside. Probabilities are just lengths ÷ total length — no integral tables needed.
Check yourself: X is uniform on [0, 10]. P(2 ≤ X ≤ 5)? And P(X = 7)?
f = 1/10 on [0, 10], so P(2 ≤ X ≤ 5) = 3 × 1/10 = 0.3 — a rectangle, width 3, height 0.1. And P(X = 7) = 0: zero width, zero area.
2 · cdf, percentiles, mean and variance §4.2
The cdf keeps its Module 2 meaning — F(x) = P(X ≤ x) — but the running total is now a running integral: the area accumulated up to x. Instead of a staircase, a smooth S-shaped climb from 0 to 1. And the two directions of travel between f and F are the two halves of the Fundamental Theorem of Calculus (MATH U101's centrepiece, moonlighting here):
Note the last formula: F(a), not F(a−1) — single points carry no probability now. Once you have F, most questions are two lookups, exactly as in Module 2 but cleaner.
Run the machine once on a concrete pdf, f(x) = 2x on [0, 1] (a ramp — larger values more likely):
- Valid? ∫01 2x dx = x²]01 = 1 ✓.
- cdf: F(x) = ∫0x 2y dy = x² for 0 ≤ x ≤ 1 (0 before, 1 after). So e.g. P(X ≤ ½) = ¼ — only a quarter of the probability sits in the left half; the ramp leans right.
- Percentiles come from solving F(η) = p. The median is the 50th: η² = 0.5 → η = √0.5 ≈ 0.707.
- Mean: E(X) = ∫01 x · 2x dx = 2/3 ≈ 0.667 — the balance point of the ramp, and less than… wait, more? No: 2/3 < 0.707. Mean below median — the ramp's long thin tail is on the left, and the tail pulls the mean, exactly as the descriptive-statistics lesson taught.
- Variance by the shortcut: E(X²) = ∫01 x² · 2x dx = ½, so V(X) = ½ − (2/3)² = 1/18 ≈ 0.056, σ ≈ 0.236.
That five-step run — validate, integrate to F, solve for percentiles, integrate for μ, shortcut for σ² — is the §4.1–4.2 exam question, in every variation. The notes page's Worked Example 1 runs it again on a fresh pdf with every antiderivative shown.
Check yourself: f(x) = 3x² on [0, 1]. Find F(x), P(X ≤ 0.5), and E(X).
F(x) = ∫0x 3y² dy = x³. P(X ≤ 0.5) = 0.125 — even more right-leaning than the ramp. E(X) = ∫01 3x³ dx = 3/4.
3 · The normal distribution §4.3
Now the most important distribution in statistics — you watched it being born in Module 2's widget, where binomial bars smoothed into a bell as n grew. Measurement errors, dimensions from a stable process, heights, exam totals — anything that is the sum of many small independent effects comes out bell-shaped (Chapter 5 proves this; for now it's why the bell is everywhere). The pdf is
with two parameters you already own: μ centres the bell, σ sets its width. Here's the practical problem: that pdf has no elementary antiderivative — no formula for its cdf exists to write down. The way out is one brilliant observation: every normal is the same curve, just shifted and stretched. So tabulate one of them — the standard normal Z (μ = 0, σ = 1, cdf written Φ(z)) — and convert every other normal to it:
That number (the z-score) is the universal currency of this chapter. P(X ≤ x) = Φ((x−μ)/σ), and Φ lives in the book's Appendix table. Feel what the table contains — slide z and watch the shaded area:
The Appendix z-table is this slider, printed: one row per z, the shaded area as the entry. Three habits make it error-proof: the table gives the area to the left (right tails need 1 − Φ(z)); negative z works by symmetry (Φ(−z) = 1 − Φ(z)); and a between-two-values probability is two lookups subtracted. Anchor values worth memorising — the 68–95–99.7 rule: about 68% of any normal lies within 1σ of μ, 95% within 2σ, 99.7% within 3σ. They make superb sanity checks on every answer.
Two more standard moves, then you own §4.3:
Percentiles run the table backwards. "Find the 99th percentile" → find the z whose table entry is 0.99 (that's z = 2.33) → un-standardise: x = μ + 2.33σ. (Notation you'll meet: zα is the z with area α to its right — so z0.05 = 1.645.)
The normal can stand in for the binomial. When both np ≥ 10 and n(1−p) ≥ 10, Bin(n, p) ≈ N(μ = np, σ = √(npq)) — with a continuity correction: the discrete value 55 owns the strip [54.5, 55.5], so P(X ≤ 55) ≈ Φ((55.5 − μ)/σ). Try it on Bin(100, 0.5): Φ(5.5/5) = Φ(1.1) = 0.8643 — the exact answer is 0.8644. Half a unit of care, four correct decimals.
Check yourself: bolt lengths are N(μ = 70 mm, σ = 3 mm). P(a bolt is at most 76 mm)?
z = (76 − 70)/3 = 2 → Φ(2) = 0.9772. (Sanity: 76 is 2σ above the mean, and the 95-rule says ~2.5% sticks out above +2σ — consistent. ✓)
StatQuest: The Normal Distribution, Clearly Explained (~5 min). Watch it once, before or after this section — it's all on-syllabus and short enough to leave no room for rabbit holes.
4 · Waiting times: exponential (and gamma) §4.4
Module 2 ended with the Poisson process: events striking at rate λ per unit time, counts Poisson-distributed. Natural follow-up: how long do you wait for the next event? Wait more than t ⟺ zero events in a window of length t, and Module 2 already knows that probability: e−λt. So the waiting time X has cdf F(t) = 1 − e−λt, and differentiating (FTC again) gives the exponential distribution:
A closed-form cdf — no tables! — which is why exponential questions are the friendliest in the chapter. The mean is instinct again: events every 1/λ time units on average. Keep the units of λ and of x consistent (λ per hour with x in hours) and half the section's traps vanish.
The exponential's famous personality trait is memorylessness: given the component has already survived time s, the chance it survives t more is exactly P(X > t) — as good as new, always. (One line shows it: P(X > s+t)/P(X > s) = e−λ(s+t)/e−λs = e−λt.) It's a favourite conceptual exam question, and it's also the model's honest limitation: real machines do wear out, and that's exactly why the section introduces the gamma distribution — a two-parameter family (shape α, scale β; mean αβ, variance αβ²) built on the gamma function Γ(α), which generalises factorials (Γ(n) = (n−1)! for whole numbers). Gamma waits for the α-th Poisson event instead of the first (α = 1 recovers the exponential exactly) — Module 2's negative binomial idea, gone continuous. At this level: know the shape of the family, the mean and variance, and how to use a given Γ value; heavy gamma integrals are not first-course territory.
Check yourself: buses pass a stop as a Poisson process, on average one per 5 minutes (λ = 0.2/min). P(you wait more than 5 minutes)?
P(X > 5) = e−0.2 × 5 = e−1 ≈ 0.368. More than a third of waits exceed the average wait — the exponential's long right tail at work (its median ln 2/λ ≈ 3.5 min sits well below its mean of 5).
5 · The rest of the chapter — now officially off-syllabus §4.5–4.6
The handout confirms §4.5–4.6 are not in the lecture plan — no exam question will come from them. The four bullets below stay only because these names show up all over engineering practice; reading them once costs two minutes. Do zero exercises here.
- Weibull §4.5: the engineer's lifetime workhorse — an exponential whose failure rate is allowed to change with age (a shape parameter tunes wear-out vs. infant-mortality). Reliability courses live on it.
- Lognormal §4.5: X is lognormal when ln(X) is normal — the model for quantities built by multiplying many small effects (incomes, particle sizes, repair times). Computations reduce to §4.3 with a log first.
- Beta §4.5: lives on a bounded interval [A, B] — the tool for proportions and task-completion fractions.
- Probability plots §4.6: the practical question "is my data plausibly normal?" answered by plotting sorted data against theoretical percentiles — a straight-ish line says yes. Concept in one sentence; the skill is reading the plot, and it pairs beautifully with the R-lab page's boxplots.
6 · You're ready — then on to Part 2
The chapter in one sentence: replace Module 2's sums with integrals, and the same framework — pdf, cdf, percentiles, E and V — runs on measured quantities, with the normal (standardise, then one table) and the exponential (closed-form cdf, memoryless) as the two stars. The handout adds smaller topics to this module: chi-squared is covered above, and MGFs for continuous rvs, Chebyshev's inequality (stated, not proved) and transformation methods now have their own page, the Module 3 Part 2 lesson (with notes). The lectured core above is the bulk of the module. Next:
- Work the examples: the notes page has the discrete↔continuous dictionary, three fully computed examples (pdf workout, normal resistors, exponential lifetimes), and the practice table.
- Revise from the notes page, not this one. The course guide now carries the handout's confirmed scope and the evaluation calendar.
Pin it to the syllabus and make it interactive:
"I'm studying Devore 9th ed. §4.3. Quiz me on normal-distribution calculations: alternate between 'find the probability' and 'find the percentile' questions (give me μ and σ each time), one at a time; I'll show my standardising step and table lookup, and you check both before continuing."
"Walk me through why the exponential distribution is memoryless (Devore §4.4), first with the algebra, then with an everyday analogy, then give me one exam-style question that tests whether I've understood it conceptually."
One caution: AI answers can contain confident arithmetic errors — recompute any final number yourself, and cross-check table values against your book's Appendix.