Midterm Exam — Cumulative (Weeks 1–7) · Objectives 1–4

Course: Introduction to Statistics (MATH 11) · Silver Oak University (fictional sample) · Prof. Rivera
Scope: Cumulative — Weeks 1–7, Objectives 1–4 (foundations & data types · summarizing one variable · relationships between two variables · probability & random variables).
Format: 20 items, 100 points (5 each) · application-skewed · mixed item types (multiple-choice, multiple-answer, true/false, matching).
Points: 100 · Assignment group: Midterm (20% of the course grade) · Window: opens at the start of Module 8; due 6 days later.

This is the human-readable exam with its vetted answer key and one-line feedback. The import-ready Classic QTI 1.2 is in L-midterm-week-08-qti.xml (generated by a validated Python script — parses with 20 items, every single-answer item exactly one correct). The item-bank/coverage note and the Canvas placement block are at the bottom of this file.

This is the live exam. Its paired ungraded rehearsal — O-practice-exam-week-08.md — mirrors this blueprint with fresh variants and shares none of these items.

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Blueprint (items → objective)

Coverage is proportional to teaching time: Obj 1 ≈ 4 · Obj 2 ≈ 5 · Obj 3 ≈ 3 · Obj 4 ≈ 8. No trick questions; all arithmetic is pre-computed; every single-answer item has exactly one correct option.

#	Type	Concept	Objective	Week
1	Multiple choice	Population vs. sample (identify the sample)	1	1
2	Multiple choice	Parameter vs. statistic	1	1
3	Matching	Levels of measurement (NOIR)	1	1
4	Multiple choice	Observational study & correlation ≠ causation	1	1
5	Multiple choice	Relative frequency from a table	2	2
6	Multiple choice	Skew direction from a histogram	2	2
7	Multiple choice	Compute the median	2	3
8	Multiple choice	IQR from a five-number summary	2	3
9	Multiple choice	Compute a sample standard deviation	2	3
10	Multiple choice	Interpret a correlation r	3	4
11	Multiple choice	Conditional proportion (two-way table)	3	4
12	Multiple choice	Identify the lurking variable	3	4
13	Multiple choice	Complement rule	4	5
14	Multiple choice	Addition rule with overlap	4	5
15	Multiple choice	Conditional probability (two-way table)	4	5
16	Multiple choice	Multiplication rule & independence	4	5
17	Multiple choice	Compute an expected value E(X)	4	6
18	Multiple choice	Variance of a discrete random variable	4	6
19	Multiple choice	Binomial probability	4	7
20	Multiple answer	Binomial mean, SD & normal approximation	4	7

Objective totals: Obj 1 = 4 items (20 pts) · Obj 2 = 5 items (25 pts) · Obj 3 = 3 items (15 pts) · Obj 4 = 8 items (40 pts) → 20 items, 100 points.

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Questions, key, and feedback

Objective 1 — Foundations & Types of Data (Week 1)

Q1 (MC). The Silver Oak University library wants to estimate the average number of books checked out per semester by its 22,000 enrolled students. Staff record the checkouts of a randomly chosen 400 students. In this study, what is the sample?
- A. All 22,000 enrolled students
- B. The 400 students whose checkouts were recorded ✅
- C. The average number of books checked out
- D. Students who checked out at least one book
Feedback: The sample is the part actually measured — the 400 randomly chosen students. The 22,000 are the population.

Q2 (MC). A polling firm reports that 47% of all registered voters in a state approve of a new transit plan, a figure obtained from the complete state voter records. The value 47% is best described as a —
- A. Statistic
- B. Sample
- C. Parameter ✅
- D. Margin of error
Feedback: It describes the entire population (all registered voters), so it is a parameter. A statistic comes from a sample.

Q3 (Matching). Match each variable to the level of measurement (NOIR) at which it is recorded.
| Variable | Correct level |
|---|---|
| Movie star-rating (1, 2, 3, 4, or 5 stars) | Ordinal — ordered categories, unequal gaps |
| A runner's shirt number (e.g., bib #57) | Nominal — labels with no order |
| Temperature in degrees Fahrenheit | Interval — equal gaps but no true zero |
| Number of text messages a student sent today | Ratio — equal gaps and a true zero |
Feedback: Star-ratings are ordered but the gaps aren't equal (ordinal); a bib number is a label (nominal); °F has no true zero (interval); a count has a true zero (ratio).

Q4 (MC). A health team tracks 5,000 adults for ten years and notices that those who report eating more nuts tend to have fewer heart attacks. No one was assigned a diet. What is the most defensible conclusion?
- A. Eating nuts prevents heart attacks
- B. Heart attacks cause people to eat fewer nuts
- C. There is an association, but a confounding variable could explain both ✅
- D. Because the study is large, it proves cause and effect
Feedback: This is an observational study (no treatment assigned), so it shows association, not causation; a confounder such as overall lifestyle could drive both. Size doesn't fix that.

Objective 2 — Summarizing Data (Weeks 2–3)

Q5 (MC). A barista logs the number of espresso shots in 40 drink orders, grouped by count: 1 shot → 6 orders; 2 shots → 20; 3 shots → 10; 4 shots → 4. What share of orders had 3 or more shots?
- A. 10%
- B. 25%
- C. 35% ✅
- D. 65%
Feedback: Relative frequency = count ÷ total. 3-or-more = 10 + 4 = 14 orders; 14 ÷ 40 = 0.35 = 35%.

Q6 (MC). A histogram of the time customers wait in a pharmacy line has most of its tall bars bunched at the short waits on the left, with a long thin tail stretching right toward a few very long waits. This distribution is —
- A. Skewed left
- B. Skewed right ✅
- C. Symmetric
- D. Uniform
Feedback: Skew is named for the tail. The tail points toward the large values (right), so it is skewed right — even though the tall bulk sits on the left.

Q7 (MC). Six daily high temperatures (°F) were recorded: 58, 61, 64, 66, 70, 95. What is the median high temperature?
- A. 64
- B. 65 ✅
- C. 66
- D. 69
Feedback: With six (even) sorted values, the median is the average of the 3rd and 4th: (64 + 66) ÷ 2 = 65. The mean (≈ 69) is pulled up by the 95.

Q8 (MC). A set of quiz scores has the five-number summary Min = 12, Q1 = 28, Median = 35, Q3 = 44, Max = 60. What is the interquartile range (IQR)?
- A. 48
- B. 16 ✅
- C. 9
- D. 32
Feedback: IQR = Q3 − Q1 = 44 − 28 = 16 (the spread of the middle 50%). The range, Max − Min = 48, is the classic distractor.

Q9 (MC). Compute the sample standard deviation of these five values: 2, 4, 6, 8, 10. (Their mean is 6; the squared deviations 16, 4, 0, 4, 16 sum to 40.)
- A. √8 ≈ 2.83
- B. √10 ≈ 3.16 ✅
- C. 8
- D. 10
Feedback: Sample variance = sum of squared deviations ÷ (n − 1) = 40 ÷ 4 = 10; the SD is √10 ≈ 3.16. (Dividing by n = 5 gives variance 8 — the population formula, distractor A.)

Objective 3 — Exploring Relationships (Week 4)

Q10 (MC). For a sample of cars, the correlation between a car's weight and its highway fuel economy (mpg) is r = −0.78. The best description of this relationship is —
- A. A weak negative linear relationship
- B. A strong positive linear relationship
- C. A strong negative linear relationship ✅
- D. No linear relationship
Feedback: The size (0.78, close to 1) means strong; the minus sign means negative. Heavier cars tend to get fewer miles per gallon.

Q11 (MC). A survey of 300 commuters cross-classifies commute mode and on-time arrival. Among drivers (180 people), 153 arrived on time; among transit riders (120 people), 78 arrived on time. Among drivers, what proportion arrived on time?
- A. 153 / 300 = 0.51
- B. 153 / 180 = 0.85 ✅
- C. 78 / 120 = 0.65
- D. 180 / 300 = 0.60
Feedback: "Among drivers" is a conditional proportion — divide by the driver total (180), not the grand total: 153 ÷ 180 = 0.85.

Q12 (MC). Across many summer days, a city finds that ice-cream sales and the number of drownings rise and fall together. A blogger claims ice cream causes drownings. The most likely lurking variable is —
- A. The price of ice cream
- B. The outdoor temperature / hot weather ✅
- C. The number of ice-cream shops
- D. There is no lurking variable; ice cream causes drownings
Feedback: Hot weather drives both — people buy more ice cream and more people swim (so more drownings). Temperature is the lurking variable; the link isn't causal.

Objective 4 — Probability & Random Variables (Weeks 5–7)

Q13 (MC). An airline says the probability a particular flight departs on time is 0.86. Using the complement rule, what is the probability the flight does not depart on time?
- A. 0.86
- B. 0.14 ✅
- C. 0.50
- D. 1.14
Feedback: P(not A) = 1 − P(A) = 1 − 0.86 = 0.14. An event and its complement add to 1.

Q14 (MC). You draw one card from a standard 52-card deck. What is the probability the card is a Queen or a Heart? (There are 4 Queens, 13 Hearts, and 1 Queen of Hearts.)
- A. 17 / 52
- B. 16 / 52 ✅
- C. 13 / 52
- D. 4 / 52
Feedback: Subtract the overlap (the Queen of Hearts is in both): 4/52 + 13/52 − 1/52 = 16/52. Forgetting to subtract gives the 17/52 trap.

Q15 (MC). A clinic screens 500 patients. Of the 150 patients who actually have the condition, 135 tested positive. What is P(tests positive | the patient has the condition)?
- A. 135 / 500 = 0.27
- B. 150 / 500 = 0.30
- C. 135 / 150 = 0.90 ✅
- D. 15 / 150 = 0.10
Feedback: "Given the patient has the condition" restricts you to those 150 patients; 135 tested positive, so 135 ÷ 150 = 0.90 (the test's sensitivity).

Q16 (MC). On a quality line, a randomly chosen part passes inspection with probability 0.9, independently. If two parts are chosen independently, what is the probability that both pass?
- A. 1.8
- B. 0.18
- C. 0.81 ✅
- D. 0.95
Feedback: Independent events multiply: 0.9 × 0.9 = 0.81. (Adding, 1.8, is impossible — a probability can't exceed 1.)

Q17 (MC). A payout random variable X has the distribution X = 0 with probability 0.5, X = 10 with probability 0.4, X = 50 with probability 0.1. What is the expected value E(X)?
- A. 9 ✅
- B. 20
- C. 30
- D. 60
Feedback: E(X) = 0(0.5) + 10(0.4) + 50(0.1) = 0 + 4 + 5 = 9. It's the probability-weighted average, not the simple average of the payouts.

Q18 (MC). A discrete random variable X has mean E(X) = 2 and E(X²) = 5. What is the variance of X?
- A. 1 ✅
- B. 3
- C. 5
- D. 9
Feedback: Variance = E(X²) − [E(X)]² = 5 − 2² = 5 − 4 = 1. Forgetting to subtract the square of the mean leaves 5 — the classic slip.

Q19 (MC). A student guesses on 5 true/false questions, so the number correct X ~ Binomial(n = 5, p = 0.5). Using C(5, 5) = 1, what is the probability of getting all 5 correct, P(X = 5)?
- A. 1/5 = 0.20
- B. 1/32 = 0.03125 ✅
- C. 5/32 = 0.15625
- D. 0.5
Feedback: P(X = 5) = 1 × (0.5)⁵ = 1/32 = 0.03125. Each of the five independent guesses must be right.

Q20 (Multiple answer — select all that apply). A fair coin is flipped 100 times and X = the number of heads, so X ~ Binomial(n = 100, p = 0.5). Select all statements that are true. (Mean = np; SD = √[np(1 − p)].)
- A. The mean number of heads is 50 ✅
- B. The standard deviation is 5 ✅
- C. The standard deviation is 25
- D. The normal approximation is appropriate here (np ≥ 10 and n(1 − p) ≥ 10) ✅
- E. The mean number of heads is 100
Feedback: Mean = 100 × 0.5 = 50 (A). Variance = 100 × 0.5 × 0.5 = 25, so SD = √25 = 5 (B true; C is the variance, not the SD). np = 50 and n(1 − p) = 50 are both ≥ 10, so the normal approximation applies (D). E doubles the mean.

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Answer key (quick reference)

Q	Answer	Q	Answer
1	B	11	B (153/180 = 0.85)
2	C	12	B
3	Star-rating→Ordinal / Bib #→Nominal / °F→Interval / Texts→Ratio	13	B (0.14)
4	C	14	B (16/52)
5	C (35%)	15	C (135/150 = 0.90)
6	B	16	C (0.81)
7	B (65)	17	A (9)
8	B (16)	18	A (1)
9	B (√10 ≈ 3.16)	19	B (1/32)
10	C	20	A, B, D

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Quality gate (H5 — self-checked, computer-verified)

• Structure: 20 items, 5 points each, 100 points total; coverage Obj 1 = 4 · Obj 2 = 5 · Obj 3 = 3 · Obj 4 = 8 matches the blueprint exactly.
• Single-answer integrity: every multiple-choice and true/false item (Q1, Q2, Q4–Q19) has exactly one correct option; the matching item (Q3) pairs all four variables one-to-one; the multiple-answer item (Q20) keys A, B, D (and requires C and E to be left unselected).
• Arithmetic pre-computed and independently re-verified (Python verify_arithmetic.py): Q5 14/40 = 0.35; Q7 median (64+66)/2 = 65; Q8 IQR 44−28 = 16; Q9 sample SD √(40/4) = √10 ≈ 3.16; Q11 153/180 = 0.85; Q13 1−0.86 = 0.14; Q14 4/52+13/52−1/52 = 16/52; Q15 135/150 = 0.90; Q16 0.9×0.9 = 0.81; Q17 0+4+5 = 9; Q18 5−4 = 1; Q19 (0.5)⁵ = 1/32 = 0.03125; Q20 mean 50, var 25, SD 5, np = 50 ≥ 10. All checks PASS.
• QTI parse confirmation: L-midterm-week-08-qti.xml parses as imsqti_xmlv1p2 with 20 items; every single-answer respcondition sets SCORE = 100 on exactly one option; the matching item's four partial-credit blocks sum to 100.
• Integrity vs. the practice exam: 0 items are shared with O-practice-exam-week-08.md (verified by full stem-plus-options comparison; max overlap is a same-concept slot with different numbers/contexts).
• No content outside the Weeks 1–7 course definitions; no hallucinated facts.

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Item-bank & coverage note

All 20 items are fresh variants assembled from the Week 1–7 item banks per Prompt L (changed numbers and contexts to reduce answer-sharing with the weekly quizzes), tagged course=MATH11 · exam=midterm · weeks=1–7 · objectives=1–4 and deposited back into the banks for future per-term ($39) regenerations:

Objective	Drawn from banks	Items
1	Week 1 (Foundations & Types of Data)	Q1–Q4
2	Weeks 2–3 (Summarizing Data; Center & Spread)	Q5–Q9
3	Week 4 (Exploring Relationships)	Q10–Q12
4	Weeks 5–7 (Probability; Random Variables; Binomial & Normal)	Q13–Q20

Each term's update regenerates fresh midterm variants from these same banks; the paired practice exam is regenerated alongside and continues to share none of the live items.

Canvas placement block

canvas_object             = Quizzes::Quiz
title                     = "Midterm Exam — Cumulative (Weeks 1–7)"
assignment_group          = "Midterm"
points_possible           = 100
grading_type              = points
available_from_offset_days = 0        # opens at the start of Module 8
due_offset_days           = 6        # 6 days after module start
published                 = true
allowed_attempts          = 1
shuffle_answers           = true
provenance                = "~ Prof. Rivera's edition · Fall 2026 · built with thecoursemaker.com"