Week 7 — Quiz (auto-graded) · The Binomial Distribution

Course: Introduction to Statistics (18-week generic edition)
Objective tested: Objective 4 — the binomial model: recognizing the setting, computing probabilities, mean & SD, technology, shape.
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Week 7 · Closed to AI.

This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in F-quiz-week-07-qti.xml; the reusable item-bank entries and the Canvas placement block are at the bottom of this file.

Blueprint

No trick questions; distractors target the Week 7 misconceptions named in the lecture outline (the forgotten ways factor, exactly-vs-at-least, p vs. 1−p, "due" streaks, np as a guarantee, cumulative misread).

Questions, key, and feedback

Q1 (MC). Which of the following situations can be modeled by a binomial distribution?
- A. Counting the number of customers who happen to enter a store during one busy afternoon hour
- B. Counting how many of 15 sunflower seeds germinate, when each germinates with probability 0.7 independently ✅
- C. Counting how many of 4 names drawn without replacement from a 30-name class list belong to seniors
- D. Recording how many minutes it takes each of 15 sunflower seeds to sprout after planting
Feedback: B passes every B·I·N·S check: two outcomes, independent trials, n = 15 fixed in advance, same p = 0.7. A has no fixed number of trials; C's without-replacement draws aren't independent (p drifts); D measures time instead of counting successes.

Q2 (MC). An email platform reports that 20% of recipients click the link in a promotional message. A boutique sends the message to 30 customers, and X = the number who click. The binomial parameters are —
- A. n = 30, p = 0.8
- B. n = 20, p = 0.30
- C. n = 30, p = 0.2 ✅
- D. n = 6, p = 0.2
Feedback: n counts the trials (30 messages sent); p is the chance one trial succeeds (a click, 0.2). Option A swaps in the failure probability; B swaps the roles of the two given numbers; D confuses n with the mean np = 6.

Q3 (MC). A basketball player makes 80% of her free throws, each attempt independent. She takes 3 attempts. The probability she makes exactly 2 of the 3 is —
- A. 0.128
- B. 0.512
- C. 0.640
- D. 0.384 ✅
Feedback: P(X = 2) = C(3, 2) × 0.8² × 0.2¹ = 3 × 0.64 × 0.2 = 0.384. Option A (0.64 × 0.2 = 0.128) forgot the ways factor C(3, 2) = 3; B is 0.8³ = P(all three); C is 0.8² with the miss ignored.

Q4 (MC). Each of the 4 basil seeds in a starter kit germinates with probability 0.9, independently. The probability that at least one germinates is —
- A. 0.9999 ✅
- B. 0.6561
- C. 0.0001
- D. 0.9000
Feedback: At least one = 1 − P(none) = 1 − 0.1⁴ = 1 − 0.0001 = 0.9999. Option B (0.9⁴ = 0.6561) is P(all four germinate); C is P(none); D is just the single-seed probability.

Q5 (MC). A player makes 80% of her free throws. Over a stretch of 25 independent attempts, the mean and standard deviation of the number of makes are —
- A. mean 20, SD 4
- B. mean 20, SD 2 ✅
- C. mean 20, SD 4.47
- D. mean 12.5, SD 2
Feedback: Mean = np = 25 × 0.8 = 20; SD = √(np(1−p)) = √(25 × 0.8 × 0.2) = √4 = 2. Option A reports the variance (4) as the SD; C is √20 (the (1−p) factor dropped); D used p = 0.5.

Q6 (Matching). Match each binomial quantity to its formula.
| Quantity | Correct formula |
|---|---|
| Mean (expected number of successes) | n × p |
| Standard deviation of the count | √( n × p × (1 − p) ) |
| P(exactly k successes) | C(n, k) × p^k × (1 − p)^(n − k) |
| P(at least one success) | 1 − (1 − p)^n |
Feedback: The mean is the one-multiplication shortcut; the SD adds the (1−p) factor under the root; "exactly k" needs the ways factor; "at least one" is the complement of "none."

Q7 (Multiple answer — select all that apply). Which of the following are required for a count X to be binomial?
- A. A fixed number of trials, set before the trials begin ✅
- B. Trials that are independent of one another ✅
- C. A success probability of exactly 0.5 on each trial
- D. Knowing in advance exactly how many successes will occur
- E. The same probability of success on every trial ✅
Feedback: The checklist is B·I·N·S: binary outcomes, independent trials, fixed n, same p. The value of p is free to be anything (C is not required), and the number of successes is the random variable itself — never known in advance (D).

Q8 (True / False). A student guesses blindly on every question of a true/false quiz, each guess independent with probability 0.5 of being correct. "After three wrong guesses in a row, the next guess is more likely than 0.5 to be correct, because a right answer is due."
- True
- False ✅
Feedback: False. Independent trials don't remember: the probability is 0.5 on every guess regardless of any streak. "Due" thinking is the gambler's fallacy — the I in B·I·N·S rules it out.

Q9 (MC). In a spreadsheet, which formula returns P(X ≤ 3) for a binomial random variable with n = 10 trials and success probability p = 0.2?
- A. =BINOM.DIST(3, 10, 0.2, FALSE)
- B. =BINOM.DIST(10, 3, 0.2, TRUE)
- C. =BINOM.DIST(0.2, 10, 3, TRUE)
- D. =BINOM.DIST(3, 10, 0.2, TRUE) ✅
Feedback: Argument order is (k, n, p, cumulative), and P(X ≤ 3) is cumulative — so the last argument must be TRUE. Option A returns exactly-3 only; B and C scramble the argument order.

Q10 (MC). A binomial histogram with p = 0.1 and a small number of trials is strongly right-skewed. Which change would bring its shape closest to a symmetric bell?
- A. Increasing the number of trials n substantially ✅
- B. Decreasing the success probability even further below 0.1
- C. Reducing the number of trials all the way down to n = 2
- D. Nothing — binomial histograms are always strongly skewed
Feedback: As n grows, binomial histograms smooth out and approach a bell shape even when p is far from 0.5 — the bridge to next week's normal distribution. Pushing p toward 0 or shrinking n makes the skew worse, and D contradicts what the p = 0.5 histogram already shows.

Answer key (quick reference)

Quality gate (self-checked): each single-answer item has exactly one correct option; every numeric answer and distractor re-verified by script (0.384 = 3 × 0.64 × 0.2; 0.9999 = 1 − 0.1⁴; 20 and 2 from n = 25, p = 0.8; distractors 0.128, 0.512, 0.640, 0.6561, 0.0001, 4, 4.47 each trace to a named error); no positional pattern in the MC key (B C D A B D A) and no length giveaway (options within each item are comparable lengths); no item asserts a fact outside the Week 7 course definitions; no scenario reuses the tutorial, practice, chapter, lab, or assignment surfaces (quiz surfaces: sunflower seeds, boutique email, 3- and 25-attempt free-throw stretches, basil starter kit, true/false guessing streak — all fresh numbers).

Item-bank entries (for variants + the midterm/final)

All ten items are tagged week=7 · objective=4 · topic=binomial-distribution and deposited in Item Bank: Week 7 — The Binomial Distribution with idents w07q1–w07q10. The midterm (Week 9), the final (Week 18), and per-term variant updates draw fresh variants from this bank's concepts — never these live stems. (Tags: w07q1 binomial-setting, w07q2 parameters-n-p, w07q3 binomial-formula, w07q4 at-least-one-complement, w07q5 mean-sd, w07q6 formula-matching, w07q7 binomial-conditions, w07q8 independence-gamblers-fallacy, w07q9 technology-binomdist, w07q10 shape-normal-bridge.)

Canvas placement block

canvas_object    = Quizzes::Quiz
title            = "Week 7 Quiz — The Binomial Distribution"
assignment_group = "Quizzes"
points_possible  = 10
grading_type     = points
due_offset_days  = 6        # end of the module's week
published        = true
shuffle_answers  = true