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Week 7 · Lecture outline

Week 7 — Lecture Outline · Binomial & Normal Models

Introduction to Statistics · MATH 11 Fall 2026 · Prof. Rivera Fictional sample

Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
Objectives covered: Objective 4 — Apply basic probability rules, including conditional probability, and work with random variables. · Objective 5 — Use normal and sampling distributions to reason about variability.
SLOs touched: A (reason quantitatively from data) · B (communicate results to a non-technical audience)
Meeting pattern: 2 sessions × 75 min = 150 min. Segment minutes below total ~150; scale to your own pattern.

Week at a Glance
The week's big question "When a yes/no thing happens over and over, how do we predict the count — and when can the bell curve do the heavy lifting for us?"
By the end of the week, students can… (1) check whether a situation is a binomial setting using the four conditions (binary outcome, fixed n, independent trials, same p); (2) compute a binomial probability for a small case by hand; (3) find the mean (np) and standard deviation (√(np(1−p))) of a binomial count; (4) decide when the normal approximation applies (the np ≥ 10 and n(1−p) ≥ 10 check) and explain why it works.
Key vocabulary binomial setting, trial, success/failure, binary outcome, fixed number of trials (n), independent trials, probability of success (p), binomial random variable X, binomial probability, the count vs. the proportion, mean of a binomial (np), variance (np(1−p)), standard deviation (√(np(1−p))), normal approximation to the binomial, the np ≥ 10 / n(1−p) ≥ 10 rule, "success" is just a label
Materials slides (Deck 7), the week's readings + video links, a spreadsheet (Google Sheets or Excel) for =BINOM.DIST, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and the tutorial. A four-coin demo (real coins or a =RANDBETWEEN column) is handy.
Timing note 8 segments, ~150 min total. Session 1 = Segments 1–4 (~75). Session 2 = Segments 5–8 (~75).

Builds on Week 6 (Random Variables). Last week we met a random variable X, its probability distribution, its expected value (the mean of a RV) and its variance/SD. This week is one specific, hugely useful random variable — the binomial count — where those same ideas have a clean shortcut formula. Re-state the Week-6 definitions briefly whenever you lean on them.

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

Hook. Pull four coins out of your pocket (or put a =RANDBETWEEN(0,1) × 4 row on the board). "I'm going to flip four coins. Before I do — how many heads will I get?" Let them shout guesses. Flip. Then: "You can't know the exact number — but you can know the odds of each number, and that turns out to be one of the most useful tricks in all of statistics."
- Same shape shows up everywhere: how many of 4 free throws drop, how many of 50 emails get opened, how many of 100 voters say "yes." A yes/no thing, repeated a fixed number of times.
- "Today we name that pattern — the binomial setting — learn to compute its odds, and then meet the shortcut that lets the bell curve answer the big-n versions for us."

The promise (write it on the board): "By the end of this week you can look at a 'how many out of n' question, decide in seconds whether it's a binomial, compute the small ones by hand, and know exactly when you're allowed to hand the big ones to the normal curve."

Why it matters line (memory hook): "Count the successes, name the four conditions, and the formula does the rest — until n gets big, and then the bell curve takes over."

Segment 2 — The Binomial Setting: The Four Conditions (22 min)

Plain language first. A binomial setting is the most common "repeated yes/no" situation in statistics. You do the same little experiment a fixed number of times, each time it either works ("success") or doesn't ("failure"), and you count the successes. Four things must all be true — miss one and it is not binomial.

The four conditions (put each on a slide, one at a time):
1. Binary outcome — every trial has exactly two outcomes we collapse to success and failure. ("Success" is just a label for the outcome we're counting — it can be "gets a defect.")
2. Fixed number of trials, n — you decide n in advance. (Flip exactly 4 times. Not "flip until you get a head.")
3. Independent trials — one trial's result doesn't change the next one's odds. (Coins don't remember.)
4. Same probability of success, p — p is identical on every trial. (A fair coin is p = 0.5 every single flip.)

Memory hook (say it twice): B-I-N-S — Binary, Independent, N fixed, Same p. "If all four hold, it's a BINS — a binomial."

One fully worked example — does it qualify? (do the check out loud):

Setup: A basketball player who makes 50% of her free throws shoots 3 of them. Let X = the number she makes.
- Binary? Yes — each shot is make / miss. ✓
- Fixed n? Yes — exactly 3 shots, decided ahead. ✓
- Independent? We'll assume one shot doesn't change the next's odds. ✓
- Same p? Yes — p = 0.5 every shot. ✓
All four hold → X is binomial with n = 3, p = 0.5. Write it: X ~ B(n = 3, p = 0.5).

A non-example (this is where the learning is):

Setup: "Deal 3 cards from one deck; count the hearts." Binary (heart / not)? Yes. Fixed n = 3? Yes. Independent? No — after you remove a heart, fewer hearts remain, so p changes (13/52, then 12/51…). Dealing without replacement breaks condition 3. Not binomial.

Land the key idea: the four conditions are a checklist you run before any formula. If "same p" or "independent" fails, the binomial formula gives a confidently wrong answer.

Segment 3 — Computing a Binomial Probability for Small Cases (25 min)

Plain language first. To get the chance of exactly k successes in n trials, you need two ingredients:
- How many ways you can arrange k successes among n trials — written C(n, k) ("n choose k"), the number of orderings.
- The chance of any one such arrangement — multiply p for each success and (1−p) for each failure: pᵏ (1−p)ⁿ⁻ᵏ.

Multiply them: P(X = k) = C(n, k) · pᵏ · (1−p)ⁿ⁻ᵏ.

Note on scope: we keep n tiny and compute C(n, k) by listing or with the small numbers below — no heavy counting machinery (that's beyond Objective 4). The values you need are pre-listed so the arithmetic stays clean.

The little "choose" values you'll use (memorize the small ones):
| C(n, k) | k=0 | k=1 | k=2 | k=3 | k=4 |
|---|---|---|---|---|---|
| n = 2 | 1 | 2 | 1 | — | — |
| n = 3 | 1 | 3 | 3 | 1 | — |
| n = 4 | 1 | 4 | 6 | 4 | 1 |

One fully worked example — every step out loud (n = 3, p = 0.5 free throws):

X ~ B(3, 0.5). Find P(X = 2) — she makes exactly 2 of 3.
- C(3, 2) = 3 (the 2 makes can be in shots {1,2}, {1,3}, or {2,3}).
- Each such arrangement: p² (1−p)¹ = (0.5)²(0.5)¹ = (0.125).
- P(X = 2) = 3 × 0.125 = 0.375 = 3/8.
Say it in words: "There's about a 37.5% chance she makes exactly two of her three free throws."

The whole tiny distribution (show it — it's the payoff slide):

X ~ B(3, 0.5): P(0) = 1/8 = 0.125, P(1) = 3/8 = 0.375, P(2) = 3/8 = 0.375, P(3) = 1/8 = 0.125. They sum to 1 ✓ (a quick sanity check students can always run).

A second worked example, p ≠ 0.5 (n = 2, p = 0.3 — a part fails inspection 30% of the time):

X ~ B(2, 0.3), X = number defective in 2 parts.
- P(X = 0) = C(2,0)·(0.3)⁰·(0.7)² = 1 × 1 × 0.49 = 0.49.
- P(X = 1) = C(2,1)·(0.3)¹·(0.7)¹ = 2 × 0.3 × 0.7 = 0.42.
- P(X = 2) = C(2,2)·(0.3)²·(0.7)⁰ = 1 × 0.09 × 1 = 0.09.
Sum = 0.49 + 0.42 + 0.09 = 1.00 ✓.

Land it: the formula is count the orderings × chance of one ordering. When the numbers are friendly, you can check yourself with the "must sum to 1" rule.

Segment 4 — Misconceptions + Quick Interaction (20 min) · Session 1 closes (~75)

Name the misconceptions out loud, then cure each:

  • ❌ "Any yes/no situation is binomial."
    ✅ Cure: all four conditions must hold. "Draw cards without replacement" is yes/no but the trials aren't independent and p changes — not binomial. Run the BINS checklist every time.
  • ❌ "'Success' means something good."
    ✅ Cure: success is a neutral label for the outcome you're counting. If you count defective parts, a defect is a "success." Don't let the everyday word mislead you.
  • ❌ "Independent and mutually exclusive are the same thing."
    ✅ Cure: different idea. Independent = one trial doesn't change another's odds (Week-6 language). Mutually exclusive = can't both happen on the same trial. Binomial needs independence across trials, not exclusivity.
  • ❌ "P(X = 2) and P(at least 2) are the same."
    ✅ Cure: "exactly 2" is one term; "at least 2" adds P(2) + P(3) + …. Read the wording: exactly, at least, at most point to different sums.
  • ❌ "You flip until you get a head — that's still binomial."
    ✅ Cure: that has no fixed n (you stop on the first head). Fixed-n is condition 2; "keep going until…" breaks it.

Interaction — "Binomial or not?" rapid-fire (Think-Pair-Share, ~10 min):
Put 5 scenarios on a slide; students decide solo (30 sec), compare with a neighbor (1 min), then vote thumbs up/down for "binomial."
1. Flip a fair coin 10 times, count heads. (Binomial ✓ — n=10, p=0.5.)
2. Roll a die until you get a 6; count the rolls. (Not — no fixed n.)
3. Ask 4 randomly chosen students "do you have a campus job?"; count "yes." (Binomial ✓ — n=4, same p, independent enough with random selection.)
4. Deal 5 cards from a deck; count aces. (Not — without replacement, p changes, trials dependent.)
5. A factory line where each item is defective with probability 0.02, inspect 50; count defects. (Binomial ✓ — n=50, p=0.02.)
Debrief the two that split the room: #2 (no fixed n) and #4 (dependence). These are the classic traps.

Segment 5 — Mean and Standard Deviation of a Binomial (25 min) · Session 2 opens

Hook back in: "Last session we computed exact odds for small n. But if I flip a coin 100 times, you don't want to add up 101 terms. Good news — the center and spread of a binomial have one-line formulas."

Plain language first — and connect to Week 6. Last week, the mean of any random variable was its expected value — the long-run average. For a binomial, that average has a beautiful shortcut:
- Mean (expected count): μ = np. Intuition: if each of n trials succeeds with probability p, you "expect" np of them. (Flip a coin 100 times, expect 50 heads.)
- Variance: σ² = np(1−p).
- Standard deviation: σ = √(np(1−p)). This is the typical distance of the actual count from np.

Memory hook: "Mean is np — successes-per-trial times trials. Spread is √(np(1−p)) — and it's biggest when p = ½."

One fully worked example (n = 100, p = 0.5 — engineered to land clean):

A fair coin is flipped 100 times. X = number of heads, X ~ B(100, 0.5).
- Mean: μ = np = 100 × 0.5 = 50 heads.
- Variance: σ² = np(1−p) = 100 × 0.5 × 0.5 = 25.
- SD: σ = √25 = 5 heads.
Say it in words: "Expect about 50 heads, give or take about 5." So ~45–55 is a typical range.

A second worked example (n = 64, p = 0.5 — also clean):

X ~ B(64, 0.5). μ = 64 × 0.5 = 32. σ² = 64 × 0.5 × 0.5 = 16. σ = √16 = 4. Center 32, spread 4.

A third worked example, p ≠ 0.5 (n = 50, p = 0.2 — opening rates):

50 emails, each opened with probability 0.2. X = number opened, X ~ B(50, 0.2).
- μ = np = 50 × 0.2 = 10 opens.
- σ² = np(1−p) = 50 × 0.2 × 0.8 = 8. σ = √8 ≈ 2.83 opens.
"Expect about 10 opens, give or take roughly 2.8."

Land it: for ANY binomial, you get center and spread with two products — no listing terms. (Tie back: Week 6's "variance of a random variable," now with a formula.)

Segment 6 — The Normal Approximation to the Binomial (22 min)

Plain language first. When n is large, the binomial's bar chart of probabilities gets tall, dense, and — here's the magic — bell-shaped. So instead of summing dozens of terms, we can lay a normal curve with the same mean and SD over the bars and read off areas. (We'll do the actual normal-curve calculations formally in Week 9 — this week is the setup and, crucially, when you're allowed to do it.)

Why it works (one sentence + a picture): add up many independent yes/no trials and the pile-up of outcomes smooths into a bell. Picture: draw the B(10, 0.5) bars on the board — they already look like a bell; now imagine B(100, 0.5), even smoother.

The rule for "is it bell-shaped enough?" (write it big):

Use the normal approximation only when np ≥ 10 AND n(1−p) ≥ 10. Both must clear 10. (When p is far from ½, you need a bigger n to pass.)

One fully worked check — PASSES (n = 100, p = 0.5):

np = 100 × 0.5 = 50 ≥ 10 ✓ and n(1−p) = 100 × 0.5 = 50 ≥ 10 ✓. Both pass → normal approximation is fine. Use a normal curve centered at μ = np = 50 with σ = √(np(1−p)) = 5.

One fully worked check — FAILS (n = 20, p = 0.02 — a rare defect):

np = 20 × 0.02 = 0.4, which is far below 10 ✗. Fails. With so few expected successes the distribution is lopsided (piled at 0), not bell-shaped — the normal curve would lie. Stick with the exact binomial here.

One borderline check (n = 50, p = 0.2):

np = 50 × 0.2 = 10 ✓ and n(1−p) = 50 × 0.8 = 40 ✓. Both clear 10 → approximation OK (just barely on the np side). μ = 10, σ = √8 ≈ 2.83.

Memory hook: "Big n, p not too extreme → the binomial wears a bell. Check np ≥ 10 and n(1−p) ≥ 10 before you trust it."

Callback: this is the same mean (np) and SD (√(np(1−p))) from Segment 5 — the normal curve just borrows them. Point back explicitly.

Segment 7 — Putting It Together: A Real Scenario, Start to Finish (16 min)

Plain language first. Walk one scenario through all four skills so students see the pipeline: check conditions → (small) compute or (large) summarize → decide on the normal curve.

Worked example (n = 200, p = 0.1 — clicks on an ad):

An ad is shown to 200 randomly chosen users; each clicks with probability 0.1, independently. X = number of clicks.
1. Binomial? Binary (click / no), fixed n = 200, independent (random users), same p = 0.1 → yes, X ~ B(200, 0.1). ✓
2. Center & spread: μ = np = 200 × 0.1 = 20 clicks. σ² = np(1−p) = 200 × 0.1 × 0.9 = 18. σ = √18 ≈ 4.24 clicks.
3. Normal approximation? np = 20 ≥ 10 ✓ and n(1−p) = 180 ≥ 10 ✓ → yes, model X by a normal curve centered at 20 with SD ≈ 4.24.
4. Interpret (SLO B): "We expect about 20 clicks, typically give or take about 4; results from roughly 16 to 24 clicks would be unsurprising."

Contrast case — same idea, approximation NOT allowed (n = 30, p = 0.05):

X ~ B(30, 0.05). μ = np = 30 × 0.05 = 1.5. Since np = 1.5 is well under 10, the normal curve is off the table — use the exact binomial. Lesson: a perfectly good binomial isn't always normal-approximable; check first.

Mini-debate (genuinely arguable, ~3 min): "A pollster says, 'We surveyed 9 people; 6 said yes — by the normal curve that's 67% ± a bit.' Do you trust the normal step?" Surface that n = 9 gives np ≈ 4.5 < 10 → the normal approximation isn't licensed; the exact binomial (or a bigger sample) is needed.

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

Technology workflow — binomial probabilities in a spreadsheet (exact steps):
1. To get P(X = k) exactly, use =BINOM.DIST(k, n, p, FALSE). The FALSE means "exactly k" (the probability mass).
- Example to verify by hand: =BINOM.DIST(2, 3, 0.5, FALSE) returns 0.375 — matches our 3/8 from Segment 3. ✓
2. To get P(X ≤ k) (cumulative, "at most k"), use TRUE: =BINOM.DIST(k, n, p, TRUE).
- =BINOM.DIST(2, 3, 0.5, TRUE) returns 0.875 = P(0)+P(1)+P(2) = 0.125+0.375+0.375. ✓
3. For mean and SD, just type the formulas: =n*p and =SQRT(n*p*(1-p)). For B(100, 0.5): =100*0.5 → 50, =SQRT(100*0.5*0.5) → 5.
- Google Sheets and Excel both have BINOM.DIST; older sheets may use BINOMDIST (no dot).

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

Paste this to an approved chatbot: "A basketball player makes 50% of her free throws and shoots 3. What's the probability she makes exactly 2? Also, for 100 fair coin flips, what's the mean and standard deviation of the number of heads?"
Then check its work against what we did in class. The exact answers are 0.375 (= 3/8) and mean 50, SD 5. Chatbots usually nail these — but they sometimes flip "exactly 2" into "at least 2," quietly switch to the proportion instead of the count, or apply a normal approximation where np < 10 isn't met. Your job all semester: the tool drafts, you judge — and you already know the right numbers.

Callback + tease:
- Callback: "The binomial is last week's random variable with a name and a formula — mean np, SD √(np(1−p)) — and when n is big it borrows the bell curve."
- Tease next week: "Week 8 is midterm review and the midterm — we'll pull together everything from Weeks 1–7. Then Week 9 we give the normal distribution its own full treatment: density curves and z-scores, the machinery the approximation was leaning on."

Hand-off (the week's graded work):
- Lecture Tutorial 7 (AI tutor, share-link submission) — the four conditions, small binomial probabilities, mean & SD, and the normal-approximation check.
- Quiz 7 (end of week), Discussion 7 ("Is it binomial? Could a normal model approximate it?"), and Assignment 7 (four problems, adaptive coach). Note: the midterm is next week — this week's work doubles as review.

Instructor FAQ — Common Stumbles

Student says / does Quick cure
"It's yes/no, so it's binomial, right?" Run all four conditions (BINS). Yes/no is only condition 1. "Without replacement" or "until I get a…" breaks the others.
"Why isn't 'draw 5 cards, count aces' binomial?" Without replacement, removing a card changes p and links the trials → not independent, p not constant. Binomial needs both.
"Does 'success' have to be good?" No — it's just the outcome you're counting. Counting defects? A defect is a "success." It's a neutral label.
Computes P(X = 2) but the problem said "at least 2." "At least 2" = P(2) + P(3) + …; "exactly 2" is one term. Underline the word exactly / at least / at most before computing.
Mixes up the count X and the proportion. This week X is the count of successes (mean np). The proportion is X/n (comes later). Keep them separate.
Forgets the (1−p) exponent, e.g., writes C(n,k)·pᵏ. Every trial is accounted for: k successes contribute pᵏ, the other n−k failures contribute (1−p)ⁿ⁻ᵏ. Exponents must add to n.
Uses √(np) instead of √(np(1−p)) for the SD. SD = √(np(1−p)) — the (1−p) factor is essential. √(np) is the mean's square root, not the SD.
Applies the normal curve to small n (e.g., n = 8, p = 0.5). Check np ≥ 10 and n(1−p) ≥ 10 first. n = 8 gives np = 4 < 10 → not licensed; use the exact binomial.
"Both checks are np ≥ 10 — why two?" One guards the success side (np), the other the failure side (n(1−p)). With p far from ½, one side can fail even when the other passes.
Thinks "the sum doesn't have to be 1." A full binomial distribution's probabilities always sum to 1 — a free self-check on small cases (we used it in Segment 3).

Scope flag

This outline stays within Objectives 4 & 5. We compute C(n, k) only for tiny n by listing — no general combinatorics (deliberately out of scope per the spine's depth decisions). The formal normal-curve area calculations (z-scores, probabilities) are previewed only; they are taught fully in Week 9 — this week stops at setting up the approximation and the np ≥ 10 / n(1−p) ≥ 10 license. Poisson is not covered (omitted by design). Cut the Segment-7 contrast case for a leaner 60-minute version.

~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com