Title: Week 6 — Lecture Outline · Random Variables
Course: Introduction to Statistics (18-Week Generic Edition)

Course: Introduction to Statistics (18-week generic edition) Objectives covered: Objective 4 — Use probability rules, conditional probability, and random variables to quantify uncertainty (this week: the random-variables half). SLOs touched: A (reason quantitatively from data) · B (communicate results to a non-technical audience) Meeting pattern: planned as 2 sessions × ~75 min = ~150 min. Segment minutes below total ~150; scale them to your own pattern.

Week at a Glance

The week's big question | "When an outcome is uncertain but has numbers attached — a prize, a payout, a count — what is that uncertain number actually worth, and how wildly does it swing?"
By the end of the week, students can… | (1) recognize a random variable and tell a discrete one from a continuous one; (2) check whether a table of values-and-probabilities is a legitimate probability distribution and read probabilities from it; (3) compute and interpret the expected value E(X) as a long-run average; (4) compute the variance and standard deviation of a discrete random variable; (5) predict what happens to the mean and SD under a linear transformation (a + bX); (6) explain why, for a continuous random variable, probability is area under a density curve and any single exact value has probability 0.
Key vocabulary | random variable, discrete vs. continuous random variable, probability distribution, legitimate distribution (each probability in [0,1]; total exactly 1), expected value E(X) = μ_X, variance σ²_X, standard deviation σ_X, linear transformation (a + bX), density curve, uniform density
Materials | slides (Deck 6), the Week 6 chapter, the week's readings + video links, a spreadsheet (Google Sheets or Excel), the student's chatbot for the AI-critique moment and the tutorial
Timing note | 8 segments, ~150 min total. Session 1 = Segments 1–4 (~75). Session 2 = Segments 5–8 (~75).

Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "An arcade claw machine costs $1 a play. Most plays win nothing. Sometimes it drops a $1 keychain, occasionally a $5 plush, and once in a blue moon a $10 headset. Would you play? How would you even decide?" Take a quick show of hands — play / don't play / need more information. The "need more information" people are already doing statistics.

- Last week we could answer "how likely is a win?" This week we answer the sharper question: "what is a play worth?" — because the outcomes aren't just events anymore, they're events with dollar amounts attached.
- "By the end of this week, you'll compute exactly what one play of that machine is worth — and you'll be able to do the same for a warranty, a raffle ticket, and any other deal where chance carries a number."

The promise (write it on the board): "By the end of this week you can take any uncertain number — a payout, a count, a score — write down its possible values with their chances, and answer the two questions that matter: what's typical (the expected value) and how far off typical should I expect to be (the standard deviation)?"

Why it matters line (memory hook): "A random variable is just a number that hasn't happened yet — and we can still say what it's worth."

Segment 2 — Random Variables & Discrete Distributions (22 min)

Plain language first. - A random variable is a numeric outcome of a chance process — a number whose value is decided by chance. Before the claw drops, "the cash value of my prize" is a random variable. After it drops, it's just data. - We write random variables with capital letters (X, Y) and their possible values with small letters (x). "P(X = 5)" reads "the probability that X turns out to be 5." - Two kinds, and the split echoes Week 1's variable types: - Discrete — the possible values are separate, countable points (0, 1, 2, …). Counts of things. You can list them. - Continuous — the possible values fill an entire interval; between any two values there are infinitely more. Measurements. You can't list them. - Memory hook: "Discrete you count, continuous you measure" — the same instinct as NOIR in Week 1. - The probability distribution of a discrete random variable is the complete table of its possible values with the probability of each. Two rules make a table legitimate: 1. every probability is between 0 and 1, and 2. the probabilities add up to exactly 1 (something must happen).

One fully worked example (build the table live).

The claw machine from the hook. X = the cash value of the prize from one $1 play. The arcade's own service data:

x (prize value) | $0 | $1 | $5 | $10
P(X = x) | 0.70 | 0.20 | 0.08 | 0.02

- Legitimacy check, out loud: every entry is between 0 and 1 ✓, and 0.70 + 0.20 + 0.08 + 0.02 = 1.00 ✓. This is a legitimate distribution.
- Read from it: P(X = 0) = 0.70 — most plays win nothing. P(win something) = P(X ≥ 1) = 0.20 + 0.08 + 0.02 = 0.30. The complement rule from Week 5 still works: 1 − 0.70 = 0.30. Same machinery, new packaging.

Land the key idea: the distribution table is the entire personality of a discrete random variable — everything we compute this week comes from those two rows.

Segment 3 — Expected Value: What a Chance Is Worth (23 min)

Plain language first. If you played the claw machine thousands of times, your winnings per play would settle down to one number — that's the expected value. It is a long-run average, computed by weighting each value by its probability:

E(X) = Σ x · P(x) — multiply each value by its probability, add them all up. Also written μ_X (it is the mean of the random variable).

One fully worked example (do every step out loud).

The claw machine again. E(X) = 0(0.70) + 1(0.20) + 5(0.08) + 10(0.02) = 0 + 0.20 + 0.40 + 0.20 = $0.80. - Say it in words: "Over many, many plays, the machine pays out about 80 cents per play, on average." - The play costs $1.00. So per play, the average player loses 1.00 − 0.80 = $0.20 — the arcade's edge. Play 100 times: spend $100, expect about $80 back. - Now the punchline: $0.80 is not a possible prize. No single play ever pays 80 cents. The expected value is what your average approaches, not a value you should "expect" on any play.

Callback (Week 5): this is the long-run relative frequency idea wearing a new coat — in the Week 5 lab, the proportion of sixes wobbled early and settled near 1/6. An average settles the same way. Chance is wild in the short run, dependable in the long run.

Land the key idea (memory hook): "Expected value is what you'd average, not what you'd expect."

Segment 4 — Misconceptions + Quick Interaction (22 min) · Session 1 closes (~75)

Name the misconceptions out loud, then cure each:

- ❌ "The expected value is the most likely outcome." ✅ Cure: the claw machine's most likely outcome is $0 (probability 0.70); its expected value is $0.80 — a value it never produces. E(X) is a weighted average, not a prediction for one play.
- ❌ "E(X) must be one of the possible values of X." ✅ Cure: averages routinely land between the values being averaged ("2.3 people per household"). If your computed E(X) happens to be impossible as a single outcome, that is not an error.
- ❌ "Any table of values and probabilities is a distribution." ✅ Cure: run the two legitimacy rules first — each probability in [0,1], total exactly 1. If the total is 0.9 or 1.1, the table is broken before any computation.
- ❌ "A random variable is the same thing as the data we collected." ✅ Cure: the random variable is the number before it happens (with chances attached); data are what you get after. The distribution plays the role the population played in Week 1 — the truth behind the numbers we see.

Interaction — Think-Pair-Share (rapid-fire, ~12 min): Six quick prompts on a slide; solo answer (30 sec), compare with a neighbor (1 min), fingers vote. Suggested items: 1. X = the number of typos on a page you just typed — discrete or continuous? 2. X = the exact time you wait at a crosswalk — discrete or continuous? 3. A table lists probabilities 0.6, 0.3, 0.2 — legitimate? 4. A table lists probabilities 0.5, 0.25, 0.25 — legitimate? 5. X takes values 1 and 2, each with probability 0.5. E(X) = 1.5 — but 1.5 is impossible. Is the computation wrong? 6. Every value of X gets 10 added to it. Does the spread change?

(Answers: discrete · continuous · not legitimate — sums to 1.1 · legitimate · not wrong — E(X) is a long-run average, impossible values are fine · no — shifting every value moves the center, not the spread.) Debrief items 5 and 6 — they preview Segments 5–6 and are the week's two favorite traps.

Segment 5 — Variance & SD of a Random Variable (25 min) · Session 2 opens

Hook back in: "Last session: E(X) tells you what's typical in the long run. But two deals can share an average and feel completely different — a steady $2 versus a coin flip between $0 and $4. The missing number is the spread."

Plain language first. In Week 3 the standard deviation measured how far data values typically sit from their mean. Same idea here, but the averaging is weighted by probability:

Var(X) = σ²_X = Σ (x − μ)² · P(x) — each squared distance from the mean, weighted by its probability. SD(X) = σ_X = √Var(X) — back in the variable's own units.

One fully worked example (grind through every step — this is the week's longest computation).

A parking office tracks X = the number of vehicles a randomly chosen household in one neighborhood keeps registered:

x | 0 | 1 | 2 | 3
P(X = x) | 0.10 | 0.30 | 0.40 | 0.20

Step 1 — the mean: E(X) = 0(0.10) + 1(0.30) + 2(0.40) + 3(0.20) = 0 + 0.30 + 0.80 + 0.60 = 1.7 vehicles. Step 2 — squared deviations, each weighted: - (0 − 1.7)² = 2.89 → 2.89 × 0.10 = 0.289 - (1 − 1.7)² = 0.49 → 0.49 × 0.30 = 0.147 - (2 − 1.7)² = 0.09 → 0.09 × 0.40 = 0.036 - (3 − 1.7)² = 1.69 → 1.69 × 0.20 = 0.338 Step 3 — add: Var(X) = 0.289 + 0.147 + 0.036 + 0.338 = 0.81 vehicles². Step 4 — square root: SD(X) = √0.81 = 0.9 vehicles. Say it in words: "a randomly chosen household typically sits about 0.9 vehicles away from the mean of 1.7."

The trap to name (σ vs. σ²): variance comes out in squared units (vehicles²!) — nobody thinks in squared vehicles. Report the SD unless someone specifically asks for variance. If a computed "SD" looks suspiciously large, ask whether it's actually the variance still waiting for its square root.

Memory hook: "Variance is the math's scratch work; the SD is the answer you say out loud."

Segment 6 — Linear Transformations: Shifts & Stretches (18 min)

Plain language first. Real questions rarely stop at X. Fees get subtracted, rates get multiplied, units get converted. If Y = a + bX, you do not need to rebuild the distribution — two rules carry everything:

Mean: E(a + bX) = a + b·E(X) — the average goes through the same arithmetic. SD: SD(a + bX) = |b|·SD(X) — only the multiplier touches the spread. Adding a constant slides every outcome by the same amount; the outcomes don't spread out. (Variance picks up b²: Var(a + bX) = b²·Var(X).)

Memory hook: "Adding shifts the center. Multiplying stretches both. A constant can't stretch anything."

One fully worked example (do every step out loud).

A phone-repair kiosk's service records: X = the number of screen repairs in a day, with mean E(X) = 4 and SD(X) = 1.5. Each repair brings in $60, and the kiosk pays $50 a day for its mall stall. Daily profit: Y = 60X − 50. - E(Y) = 60(4) − 50 = 240 − 50 = $190. - SD(Y) = 60 × 1.5 = $90. The $50 stall fee does not appear — subtracting the same $50 every day shifts profits down but doesn't make them any more or less variable. - Quick check on the trap: if the stall fee rises to $60, E(Y) drops $10 to $180 — and SD(Y) is still $90.

Misconception + cure: - ❌ "Subtracting the $50 fee reduces the spread too." ✅ Cure: every possible day moves down by the same $50 — the gaps between good days and bad days are untouched. Only multiplication (the $60 rate) rescales the spread.

Segment 7 — Continuous Random Variables: Probability as Area (20 min)

Plain language first. A discrete variable's table lists each value with its chunk of probability. A continuous variable can't have a table — between any two values sit infinitely many more, so no single value can hold a chunk. Instead we draw a density curve and make one trade:

Probability = area under the density curve over an interval. The total area under any density curve is exactly 1. And a single exact value is an interval of width zero — so P(X = any exact value) = 0. Probability questions about continuous variables are always interval questions.

One fully worked example (the friendliest density there is).

Glance at a wall clock at a random moment. X = where the second hand points, measured in seconds from 0 up to 60. Every position is equally likely, so the density is a flat rectangle from 0 to 60. - The rectangle's total area must be 1, and its base is 60 — so its height is 1/60 (≈ 0.0167). The height is not a probability — it's just the level that makes the area work. - P(X ≤ 15) — the second hand is in the first quarter — = area = 15 × (1/60) = 0.25. Matches intuition: a quarter of the clock face. - P(20 ≤ X ≤ 50) = 30 × (1/60) = 0.50. - P(X = exactly 30) = a sliver of width 0 = 0. You can land near 30 easily; exactly 30.000000… never gets its own chunk of chance.

Misconception + cure: - ❌ "P(X = a) is the height of the curve at a." ✅ Cure: height is not probability — area is. A tall thin curve and a short wide one can enclose the same area. (No width, no area, no probability.) - Note the payoff of P(X = a) = 0: for continuous variables, P(X ≤ a) and P(X < a) are the same — the boundary point carries no probability.

Tease inside the segment: next week, one famous discrete variable gets a name (the binomial); the week after, the most famous density curve in the world (the normal). Both run on exactly this week's machinery.

Segment 8 — Technology Workflow + AI-Critique, Callback & Hand-off (12 min) · Session 2 closes (~75)

Technology workflow — E(X) and SD(X) in a spreadsheet (exact steps): 1. Put the values in A2:A5 (0, 1, 5, 10) and the probabilities in B2:B5 (0.70, 0.20, 0.08, 0.02) — the claw machine. 2. Expected value in one cell: =SUMPRODUCT(A2:A5,B2:B5) → 0.8. (SUMPRODUCT multiplies the columns pairwise and adds — exactly the E(X) formula.) 3. Sanity cell: =SUM(B2:B5) → must show exactly 1. If it doesn't, the distribution is broken; fix it before trusting anything. 4. Spread with a helper column: in C2, =(A2-0.8)^2*B2, fill down; Var = =SUM(C2:C5); SD = =SQRT(<that cell>). Google Sheets and Excel are identical here. 5. Preview of Data Lab 6: =RANDBETWEEN(1,100) will simulate draws from a distribution, and you'll watch the simulated average settle toward E(X). (Results are random — expect "close to," never "exactly.")

AI-critique moment (students verify, not consume):

Paste this to your chatbot: "X takes the values 0, 1, 5, 10 with probabilities 0.70, 0.20, 0.08, 0.02. Find the expected value and interpret it." Then check its work against SUMPRODUCT. Two classic chatbot stumbles: it averages the four values ignoring the probabilities — (0 + 1 + 5 + 10) ÷ 4 = 4, wildly wrong — or it computes 0.80 correctly and then interprets it as "you will most likely win about 80 cents," which is exactly this week's misconception. The tool drafts, you judge — same habit as every week.

Callback + tease: - Callback: "Week 5 gave chances to events. This week attached numbers to those chances and asked what they're worth. That completes the probability toolkit." - Tease next week: "One special random variable is so common it gets its own name, its own formula, and its own spreadsheet function: the number of successes in n tries — the binomial. Free-throws, click-throughs, guessing on a quiz — one distribution rules them all."

Hand-off (the week's work): - Chapter 6 (the primary reading) if they haven't read it yet — then Lecture Tutorial 6 (AI tutor, share-link submission) — random variables, E(X), SD, transformations, density intro. - Data Lab 6 (build and simulate a raffle in your spreadsheet) · Quiz 6 (end of week) · Discussion 6 (the extended-warranty question) · Assignment 6 (AI-coached).

Instructor FAQ — Common Stumbles

Student says / does | Quick cure
"How can the expected value be 0.8 if you can never win 80 cents?" | E(X) is the number your average over many plays homes in on — not a possible outcome. "Expected value is what you'd average, not what you'd expect."
Computes E(X) as the plain average of the values. | That silently assumes all values are equally likely. Each value must be weighted by its probability — that's the whole formula. (Chatbots make this exact error; see Segment 8.)
"The probabilities sum to 0.98 — close enough?" | Never. Legitimacy is exact: total = 1. A near-miss usually means a value is missing or a typo — find it.
Reports the variance when asked for the SD. | Variance is in squared units (vehicles², dollars²). Take the square root and report the SD in real units. If a "SD" looks huge, ask if it's an un-rooted variance.
"Subtracting the fee makes profits less variable." | Adding or subtracting a constant slides every outcome equally — spread untouched. Only the multiplier rescales SD (and variance gets b²).
"P(X = 30) should be 1/60 for the clock." | Single exact values have width 0, so area 0, so probability 0 for continuous X. Only intervals carry probability; 1/60 is the density's height, not a probability.
Confuses P(X ≥ 1) with P(X = 1). | "At least one" bundles several rows of the table (or use the Week 5 complement: 1 − P(X = 0)). Have them mark which rows the event includes before adding.
Writes probabilities for a continuous variable in a table. | If you can't list all the values, there is no table — that's what makes it continuous. The density curve replaces the table; areas replace the entries.

Scope flag

This outline stays within Objective 4. The "house edge" gambling framing and the long-run-average reprise of Week 5's law-of-large-numbers demo are added context (kept because they make E(X) stick); the continuous-variable treatment is deliberately qualitative + uniform-density only — general density calculations, z-scores, and the normal model are Week 8's job, and the binomial shortcut formulas are Week 7's. Cut the second worked check in Segment 6 for a leaner session.