StatsIQ · AP Statistics

Mock Exam 1 — Mid-Course Diagnostic (Lessons 1–20)

StatsIQ — AP Statistics Exam Prep · NEW 2026–27 Format

Coverage: Unit 1 (exploring & collecting data), Unit 2 (probability, random variables, binomial, continuous RVs, simulation, sampling distributions, bias vs. variability), and the START of Unit 3 (confidence intervals: the big idea + the one-proportion z-interval). This diagnostic does not include hypothesis tests, chi-square, inference for means, or regression — those come after Lesson 20.

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Calculator: A graphing calculator (TI-84 family) is allowed on every question in both sections.

Scoring note: There is no penalty for guessing — answer every multiple-choice question.

Student Instructions

You have about 90 minutes total. Work Section I first (45 min), then Section II (45 min). For the free-response questions, show all work and communicate your reasoning in context — full credit requires more than a final number. Use the PANIC structure (Parameter, Assumptions, Name the procedure, Interval, Conclusion) for any inference problem. When a question asks you to interpret a result, your sentence must (1) be about the population parameter, (2) be in context, and (3) include the confidence level where relevant. This is a self-assessment diagnostic; an approximate score guide appears at the very end.

SECTION I — Multiple Choice (21 questions · 45 minutes)

Directions: Each question has four answer choices. Choose the single best answer.

Question 1

A researcher records, for each car in a parking lot, its color, its number of doors, and its fuel efficiency in miles per gallon. Which classification of these three variables is correct?

(A) All three are quantitative.
(B) Color is categorical; number of doors and fuel efficiency are quantitative.
(C) Color and number of doors are categorical; fuel efficiency is quantitative.
(D) Color is categorical; number of doors and fuel efficiency are categorical.

Question 2

The histogram below summarizes the number of text messages sent in one day by 80 students. [GRAPH: Histogram. X-axis "Texts sent per day" with bins 0–20, 20–40, 40–60, 60–80, 80–100, 100–120. Y-axis "Number of students" (frequency). Bar heights: 0–20 → 6, 20–40 → 14, 40–60 → 26, 60–80 → 18, 80–100 → 10, 100–120 → 6. The distribution rises to a peak at the 40–60 bin and then tails off gradually to the right.] Which best describes the shape of this distribution?

(A) Roughly symmetric and bell-shaped
(B) Skewed to the right (positively skewed)
(C) Skewed to the left (negatively skewed)
(D) Uniform

Question 3

A dataset of daily high temperatures has a five-number summary of Min = 18, Q1 = 24, Median = 31, Q3 = 40, Max = 71 (degrees Fahrenheit). Using the 1.5 × IQR rule, which value(s) would be flagged as outliers?

(A) Only 18 (the minimum)
(B) Only 71 (the maximum)
(C) Both 18 and 71
(D) Neither value is an outlier

Question 4

SAT section scores are approximately Normal with mean μ = 500 and standard deviation σ = 100. A student scores 650. What is this student's z-score, and approximately what percent of students score below 650?

(A) z = 1.5; about 93% score below
(B) z = 1.5; about 7% score below
(C) z = 0.65; about 74% score below
(D) z = −1.5; about 7% score below

Question 5

The heights of adult women in a population are approximately Normal with mean 64 inches and standard deviation 2.5 inches. Using the 68–95–99.7 (empirical) rule, approximately what percent of women are between 59 and 69 inches tall?

(A) 68%
(B) 95%
(C) 99.7%
(D) 50%

Question 6

A school district has elementary, middle, and high schools. To study student opinions on a new schedule, researchers take a separate simple random sample from each of the three school levels and combine them. This sampling method is:

(A) cluster sampling
(B) systematic random sampling
(C) stratified random sampling
(D) convenience sampling

Question 7

A gardening company wants to know whether a new fertilizer increases tomato yield. They randomly assign 60 plants to receive either the new fertilizer or the standard fertilizer, grow them under identical conditions, and compare average yields. Which statement best describes the role of randomization here?

(A) It allows the results to be generalized to all tomato plants everywhere.
(B) It creates two groups that are roughly equivalent at the start, so a yield difference can be attributed to the fertilizer.
(C) It guarantees that the two groups have exactly equal average yields before treatment.
(D) It removes the need for a control or comparison group.

Question 8

An observational study finds that teenagers who drink more sugary soda tend to have lower grade-point averages. A news headline reads, "Soda lowers your grades." Why is this headline not justified?

(A) The sample size was probably too small.
(B) Because the data come from an observational study, a lurking variable could explain the association; a cause-and-effect claim requires a randomized experiment.
(C) Association between two variables is impossible without causation.
(D) The study should have used a voluntary response sample instead.

Question 9

Two side-by-side boxplots compare commute times (in minutes) for City A and City B. [GRAPH: Two horizontal boxplots sharing an X-axis "Commute time (minutes)" from 0 to 80. City A boxplot: Min = 10, Q1 = 20, Median = 28, Q3 = 38, Max = 55. City B boxplot: Min = 15, Q1 = 30, Median = 42, Q3 = 52, Max = 70. City B's box is shifted noticeably to the right of City A's box and is slightly wider.] Which statement is best supported by the boxplots?

(A) City A has a larger median commute time and more variability than City B.
(B) City B has a larger median commute time and greater variability (IQR) than City A.
(C) The two cities have the same median commute time.
(D) City A's maximum commute time is greater than City B's maximum.

Question 10

The two-way table classifies 200 survey respondents by gender and whether they support a proposed policy. | | Support | Oppose | Total | |---|---|---|---| | Male | 45 | 30 | 75 | | Female | 60 | 65 | 125 | | Total | 105 | 95 | 200 | Given that a randomly selected respondent is female, what is the probability that she supports the policy?

(A) 0.30
(B) 0.48
(C) 0.57
(D) 0.60

Question 11

For events A and B, P(A) = 0.50, P(B) = 0.30, and P(A and B) = 0.20. What is P(A or B), and are A and B independent?

(A) 0.60; independent
(B) 0.80; not independent
(C) 0.60; not independent
(D) 0.80; independent

Question 12

A small carnival game costs \$3 to play. With probability 0.20 you win \$10 (and otherwise win nothing). Let the random variable W be your net gain from one play (winnings minus the \$3 cost). What is the expected net gain E(W)?

(A) +\$2.00
(B) +\$0.50
(C) −\$1.00
(D) −\$3.00

Question 13

A discrete random variable X has the probability distribution below. | x | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(x) | 0.10 | 0.30 | 0.40 | 0.20 | What is the standard deviation of X? (Mean E(X) = 1.7.)

(A) 0.81
(B) 0.90
(C) 1.70
(D) 1.30

Question 14

A multiple-choice quiz has 12 questions, each with 4 options. A student guesses randomly on every question (so the probability of a correct guess is 0.25 on each). Let X be the number correct. What are the mean and standard deviation of X?

(A) mean = 3, SD = 1.5
(B) mean = 3, SD = 2.25
(C) mean = 4, SD = 1.5
(D) mean = 4, SD = 1.73

Question 15

Using the same setup as Question 14 (n = 12, p = 0.25), what is the probability that the student gets exactly 3 correct?

(A) 0.122
(B) 0.190
(C) 0.258
(D) 0.301

Question 16

A bus is equally likely to arrive at any moment during a 20-minute window after you reach the stop; that is, your waiting time X (in minutes) follows a uniform distribution on [0, 20]. What is the probability you wait more than 15 minutes?

(A) 0.15
(B) 0.25
(C) 0.50
(D) 0.75

Question 17

A teacher wants to estimate the probability that, in a class of 4 randomly chosen students, at least one was born in the summer (assume each student independently has probability 0.25 of a summer birthday). She assigns the digits 1 and 2 to "summer" and 3–9 and 0 to "not summer," then reads groups of 4 digits from a random digit table, recording whether each group of 4 contains at least one 1 or 2. After 50 repetitions, 34 of the groups contained at least one "summer" digit. Based on this simulation, the estimated probability is closest to:

(A) 0.25
(B) 0.50
(C) 0.68
(D) 0.86

Question 18

In a large population, 40% of adults have a particular gene. A random sample of n = 50 adults is taken, and p̂ is the sample proportion who have the gene. What is the standard deviation of the sampling distribution of p̂?

(A) 0.069
(B) 0.24
(C) 0.0048
(D) 0.40

Question 19

A population of account balances has mean μ = \$70 and standard deviation σ = \$12. A random sample of n = 36 accounts is selected. According to the Central Limit Theorem, the sampling distribution of the sample mean x̄ is approximately Normal with:

(A) mean \$70 and standard deviation \$12
(B) mean \$70 and standard deviation \$2
(C) mean \$70 and standard deviation \$0.33
(D) mean \$11.67 and standard deviation \$2

Question 20

A polling company wants its estimate of a population proportion to be precise (small variability from sample to sample) and unbiased. Which change to its sampling plan best reduces the variability of the estimate without introducing bias?

(A) Switch from a random sample to a larger voluntary-response sample.
(B) Increase the size of the random sample.
(C) Word the survey questions to encourage a particular answer.
(D) Keep the sample size the same but repeat the poll many times.

Question 21

A random sample of 300 voters finds that 180 support a ballot measure, giving a 95% confidence interval of approximately (0.544, 0.655) for the true proportion of all voters who support it. Which interpretation of the 95% confidence level is correct?

(A) 95% of all voters support the ballot measure.
(B) There is a 95% probability that the true proportion is exactly 0.60.
(C) If this sampling process were repeated many times, about 95% of the resulting intervals would capture the true population proportion.
(D) We are 95% sure that 95% of future samples will fall between 0.544 and 0.655.

SECTION II — Free Response (2 questions · ~45 minutes)

Directions: Show all work and write your answers in context. Communicate clearly — full credit requires correct reasoning, not just a final answer.

FRQ 1 (Full — 10 points) · Practices 1 & 2: Formulate Questions + Collect Data

A city's parks department wants to learn what proportion of the city's 24,000 registered library cardholders would use a proposed mobile e-book lending app. The department has a complete, up-to-date database of all 24,000 cardholders, including each cardholder's home branch (the city has 8 neighborhood library branches) and age group.

(a) State an appropriate statistical (investigative) question the parks department could answer with this study, and identify the population of interest. (2 points)

(b) Describe how the department could select a simple random sample (SRS) of 400 cardholders from the database. Be specific enough that another person could carry out your procedure. (3 points)

(c) A staff member instead proposes: "Put a banner ad inside the library's existing website for two weeks. Anyone who clicks it gets the survey, and we'll use all the responses we collect." Identify the sampling method this describes, name the most serious source of bias it introduces, and explain in context how that bias would likely affect the estimated proportion who would use the app. (3 points)

(d) Suppose the department carries out a proper SRS and finds that 35% of sampled cardholders would use the app. The director writes: "Therefore, 35% of all residents of the city would use the app." Explain why this conclusion goes beyond what the study supports — i.e., describe the correct scope of inference. (2 points)

MODEL RESPONSE — FRQ 1

(a) A reasonable statistical question: "What proportion of the city's registered library cardholders would use the proposed mobile e-book lending app?" The population of interest is all 24,000 registered library cardholders in the city (not all city residents). (The question must be answerable with data and must reference the cardholder population, not a single individual.)

(b)

1. Label every cardholder in the database with a distinct number, 1 through 24000.

2. Use a random device — on the TI-84, run randInt(1, 24000, 400) (generate a few extra numbers to cover duplicates), or use a random number generator / random digit table reading 5-digit groups and keeping 00001–24000.

3. Ignore any repeated labels (skip a number that has already appeared) and continue until 400 distinct cardholders are selected.

4. Survey those 400 cardholders about whether they would use the app.

(c) This is a voluntary response sample — cardholders self-select by choosing whether to click the banner and respond. The most serious source of bias is voluntary response bias (self-selection bias): people who feel strongly about the app — most likely tech-comfortable cardholders who are excited about an e-book app — are far more likely to click and respond than indifferent or less tech-savvy cardholders. As a result, the sample would overrepresent enthusiastic users, so the estimated proportion who would use the app would likely be biased too high (an overestimate of the true proportion). A large number of responses does not fix this.

(d) The sample was drawn from registered library cardholders only, so the results can be generalized only to the population of cardholders, not to all city residents. Residents who do not have a library card were never part of the sampling frame and may differ systematically (e.g., they may be less likely to use a library e-book app). Because the data come from a survey (an observational study) rather than a randomized experiment, the study can describe an association/proportion within the cardholder population but the director cannot extend "35%" to all residents. The correct scope: we can estimate the proportion of all city cardholders who would use the app, but we cannot generalize to non-cardholder residents.

POINT-BY-POINT RUBRIC — FRQ 1 (sums to 10)

Part (a) — 2 points

+1 States a valid statistical/investigative question that can be answered with data and concerns a proportion (or use) of the cardholder group.

+1 Correctly identifies the population as all registered library cardholders (the 24,000), not all city residents and not a single person.

Part (b) — 3 points

+1 Labels the 24,000 cardholders (e.g., 1–24000) using the database.

+1 Names a legitimate random device (randInt, random number generator, or random digit table) and applies it to the labels.

+1 States that duplicate labels are ignored until 400 distinct cardholders are obtained, and identifies them as the sample to survey.

Part (c) — 3 points

+1 Identifies the method as voluntary response (self-selected) sampling.

+1 Names voluntary response / self-selection bias as the source.

+1 Explains the mechanism and direction in context: enthusiastic/tech-comfortable cardholders respond more → estimate biased too high (and/or notes that large response count does not remove the bias).

Part (d) — 2 points

+1 Notes the sample represents only cardholders, so results generalize only to that population, not to all residents (non-cardholders were not in the frame).

+1 Correctly frames the scope of inference — we can generalize to the cardholder population (because the sample was random) but not to all city residents.

Where students lose points:

In (a), giving a question about one person, or a non-statistical yes/no question, or naming "all residents" as the population.

In (b), writing "pick 400 at random" with no labeling, no named random device, or no "ignore repeats / 400 distinct" step. Vague descriptions earn at most 1 of 3; the "ignore repeats" idea is the most commonly missed point.

In (c), naming the bias but giving no contextual mechanism (who responds and why) or no direction (too high).

In (d), saying the results can't be generalized to anyone (too strong — a random sample DOES generalize to the cardholder population), or failing to explain why non-cardholders are excluded.

FRQ 2 (Slightly simplified — 10 points) · Practices 3 & 4: Analyze + Interpret

A national streaming service believes that 45% of all its subscribers watch at least one show on mobile devices. A market-research team takes a random sample of 80 subscribers and will examine the sample proportion p̂ who watch on mobile. (Treat the company's 45% figure as the true population proportion p = 0.45 for parts (a)–(c).)

(a) Describe the shape, center, and standard deviation of the sampling distribution of p̂ for samples of size 80. Justify the shape by checking the Large Counts condition. (4 points)

(b) Find the probability that the sample proportion p̂ is greater than 0.55. Show your work. (3 points)

(c) In the actual sample, 44 of the 80 subscribers watched on mobile. Construct a 90% confidence interval for the true proportion of all subscribers who watch on mobile, and interpret it in context. (You may use the calculator; report the interval.) (3 points)

MODEL RESPONSE — FRQ 2

(a) Sampling distribution of p̂ when p = 0.45, n = 80:

Center (mean): μ_p̂ = p = 0.45.

Standard deviation: SD = sqrt( p(1 − p)/n ) = sqrt( 0.45 × 0.55 / 80 ) = sqrt( 0.2475 / 80 ) = sqrt(0.00309375) = 0.0556.

Shape: Approximately Normal, justified by the Large Counts condition: np = 80 × 0.45 = 36 ≥ 10 and n(1 − p) = 80 × 0.55 = 44 ≥ 10. Both are at least 10, so the Normal approximation is appropriate.

(b) We want P(p̂ > 0.55). Standardize using the mean 0.45 and SD 0.0556:

z = (0.55 − 0.45) / 0.0556 = 0.10 / 0.0556 = 1.80

P(p̂ > 0.55) = P(z > 1.80) = 1 − 0.9641 = 0.0359

Calculator check — normalcdf(lower = 0.55, upper = 1E99, μ = 0.45, σ = 0.0556) ≈ 0.0361. So P(p̂ > 0.55) ≈ 0.036. There is only about a 3.6% chance a random sample of 80 subscribers would have more than 55% watching on mobile.

(c) Now use the observed data: x = 44, n = 80, so p̂ = 44/80 = 0.55. (Conditions: random sample given; 80 is far less than 10% of all subscribers; Large Counts using p̂ → np̂ = 44 ≥ 10 and n(1 − p̂) = 36 ≥ 10 ✓.) One-sample z-interval for a proportion, 90% confidence (z* = 1.645):

SE = sqrt( 0.55 × 0.45 / 80 ) = sqrt(0.2475/80) = sqrt(0.00309375) = 0.0556

ME = 1.645 × 0.0556 = 0.0915

Interval: 0.55 ± 0.0915 → (0.458, 0.642)

Calculator check — 1-PropZInt, x = 44, n = 80, C-Level = 0.90 → (0.4585, 0.6415). ✓

Interpretation: We are 90% confident that the interval from 0.459 to 0.642 captures the true proportion of all of the streaming service's subscribers who watch at least one show on mobile devices.

POINT-BY-POINT RUBRIC — FRQ 2 (sums to 10)

Part (a) — 4 points

+1 Center: μ_p̂ = p = 0.45.

+1 Standard deviation: correctly uses sqrt(p(1−p)/n) and gets ≈ 0.0556.

+1 Shape: states approximately Normal.

+1 Justifies Normal shape with Large Counts showing numbers: np = 36 ≥ 10 and n(1−p) = 44 ≥ 10.

Part (b) — 3 points

+1 Standardizes correctly: z = (0.55 − 0.45)/0.0556 ≈ 1.80.

+1 Finds the correct upper-tail area (uses P(z > 1.80), not the lower tail).

+1 Correct probability ≈ 0.036 (accept 0.035–0.037 depending on rounding / calculator).

Part (c) — 3 points

+1 Correct p̂ = 0.55 and correct SE/critical value (z* = 1.645 for 90%), or correct 1-PropZInt setup (x = 44, n = 80, C = 0.90).

+1 Correct interval (0.459, 0.642) [accept (0.458, 0.642)].

+1 Interpretation: "90% confident" + interval + population proportion in context.

Where students lose points:

In (a), giving SD without the formula or computing variance instead of SD; forgetting to show the Large Counts numbers for the shape justification.

In (b), using the wrong tail (reporting 0.964 instead of 0.036), or using the wrong standard deviation (e.g., dividing by n instead of √ of p(1−p)/n).

In (c), using z* = 1.96 (that's 95%, not 90%), interpreting the interval as a probability about p, omitting the confidence level, or interpreting it as being about the sample rather than the population.

A subtle trap: part (b) uses the assumed p = 0.45 for the sampling-distribution SD, while part (c) uses the observed p̂ = 0.55 for the interval's SE. Mixing these up loses points.

🔑 Answer Key & Scoring Guide

Section I — Multiple Choice (recomputed independently)

1. (C) Color is categorical; number of doors (a count) and fuel efficiency (a measurement) are quantitative. — Distractor (B) wrongly treats "number of doors" as categorical-only; it is a numeric count. (A) ignores that color has no numeric meaning.

2. (B) Skewed right. The peak is at the 40–60 bin with a long tail extending toward the higher bins (80–120). The longer tail points to the right (positive direction). — (C) reverses the tail direction; (A) ignores the long right tail.

3. (B) IQR = Q3 − Q1 = 40 − 24 = 16; 1.5 × IQR = 24. Upper fence = 40 + 24 = 64, so 71 > 64 is an outlier. Lower fence = 24 − 24 = 0, and 18 > 0, so 18 is not an outlier. — (C) wrongly flags the minimum; (A)/(D) misapply the fences.

4. (A) z = (650 − 500)/100 = 1.5; from the Normal table, P(z < 1.5) ≈ 0.933, about 93% below. — (B) gives the upper-tail area (percent above, ≈ 7%) instead of below. (C) divides by the wrong number. (D) has the wrong sign.

5. (B) 59 is 2 SD below the mean (64 − 2×2.5 = 59) and 69 is 2 SD above (64 + 2×2.5 = 69). The empirical rule gives ≈ 95% within 2 SD. — (A) is the 1-SD interval (61.5–66.5); (C) is 3 SD.

6. (C) Stratified: the three school levels are the strata, and a separate SRS is drawn from each. — (A) cluster would sample all of a few groups; (B) needs a fixed interval; (D) has no randomness.

7. (B) Random assignment balances the groups on both known and unknown variables, so a difference in yield can be causally attributed to the fertilizer. — (A) confuses random assignment (causation) with random sampling (generalization). (C) is too strong ("exactly equal"). (D) is false — a comparison group is still needed.

8. (B) Observational study → a lurking variable (e.g., overall diet, activity level) could drive both. Causation requires a randomized experiment. — (C) is false: association does not require causation. (A) and (D) are not the core flaw.

9. (B) City B's median (42) > City A's median (28), and City B's IQR (52 − 30 = 22) > City A's IQR (38 − 20 = 18), so City B has greater variability too. — (A) reverses the medians; (C)/(D) contradict the values shown.

10. (B) P(Support | Female) = 60/125 = 0.48. — (C) is P(Female | Support) = 60/105 ≈ 0.57 (the reversed conditional — the classic trap). (D) 0.60 uses the wrong denominator, e.g. 60/100 or treats the female-support count (60) as a probability out of 100; (A) 0.30 = 60/200 uses the overall total as the denominator instead of conditioning on female.

11. (C) P(A or B) = P(A) + P(B) − P(A and B) = 0.50 + 0.30 − 0.20 = 0.60. Independence check: P(A)·P(B) = 0.50 × 0.30 = 0.15, which does not equal P(A and B) = 0.20, so A and B are not independent. Correct choice: 0.60; not independent. — (A) has the right union but wrongly calls them independent; (B)/(D) use 0.80 (this forgets to subtract the overlap, giving 0.50 + 0.30).

12. (C) E(W) = (0.20)(\$10) + (0.80)(\$0) − \$3 cost = \$2 − \$3 = −\$1.00. (Equivalently, net is +\$7 with prob 0.20 and −\$3 with prob 0.80: 0.2×7 + 0.8×(−3) = 1.4 − 2.4 = −1.) — (A) forgets to subtract the cost; (D) ignores winnings.

13. (B) Var(X) = Σ(x − 1.7)²P(x) = (1.7²)(0.1) + (0.7²)(0.3) + (0.3²)(0.4) + (1.3²)(0.2) = 0.289 + 0.147 + 0.036 + 0.338 = 0.81; SD = √0.81 = 0.90. — (A) reports the variance, not the SD; (C) is the mean.

14. (A) Binomial: mean = np = 12 × 0.25 = 3; SD = √(np(1−p)) = √(12 × 0.25 × 0.75) = √2.25 = 1.5. — (B) reports the variance as the SD; (C)/(D) use the wrong mean.

15. (C) P(X = 3) = C(12,3)(0.25)³(0.75)⁹ = 220 × 0.015625 × 0.0751 ≈ 0.258. Calculator: binompdf(12, 0.25, 3) ≈ 0.258. — (D) ≈ 0.301 is P(X = 2); the others are nearby wrong values.

16. (B) For a uniform distribution on [0, 20], P(X > 15) = (20 − 15)/20 = 5/20 = 0.25. — (A) reports the length (5) over 100; (D) is P(X < 15).

17. (C) Estimated probability = 34/50 = 0.68. (This is close to the true value 1 − 0.75⁴ ≈ 0.684.) — (A) is the probability for a single student; (B)/(D) misread the simulation count.

18. (A) SD of p̂ = √(p(1−p)/n) = √(0.40 × 0.60 / 50) = √(0.24/50) = √0.0048 = 0.069. — (C) is the variance (0.0048, forgot the square root); (B) is √(p(1−p)) without dividing by n; (D) is p itself.

19. (B) Mean of x̄ = μ = \$70; SD of x̄ = σ/√n = 12/√36 = 12/6 = \$2. — (A) forgets to divide by √n; (C) divides by n instead of √n; (D) corrupts the mean.

20. (B) A larger random sample reduces the standard deviation of the estimator (less variability) while keeping it unbiased. — (A) introduces bias (voluntary response); (C) introduces response/wording bias; (D) repeating a same-size poll does not reduce the per-estimate variability.

21. (C) Correct interpretation of a confidence level: in repeated sampling, about 95% of such intervals capture the true p — the confidence is in the long-run success of the method. — (A) confuses the level with the parameter; (B) wrongly assigns a probability to a fixed value; (D) is nonsense about future samples.

Section II — FRQ Rubrics (restated)

FRQ 1 (10 pts):

- (a) 2 pts — valid statistical question (+1); population = all cardholders (+1).

- (b) 3 pts — label 1–24000 (+1); named random device applied (+1); ignore repeats → 400 distinct (+1).

- (c) 3 pts — voluntary response (+1); voluntary-response/self-selection bias (+1); mechanism + direction (too high) in context (+1).

- (d) 2 pts — sample represents cardholders only / non-cardholders not in frame (+1); correct scope of inference, generalize to cardholders but not all residents (+1).

FRQ 2 (10 pts):

- (a) 4 pts — center 0.45 (+1); SD ≈ 0.0556 via √(p(1−p)/n) (+1); shape Normal (+1); Large Counts with numbers np = 36, n(1−p) = 44 (+1).

- (b) 3 pts — z ≈ 1.80 (+1); correct upper tail (+1); P ≈ 0.036 (+1).

- (c) 3 pts — correct p̂ = 0.55 with z* = 1.645 / correct 1-PropZInt setup (+1); interval (0.459, 0.642) (+1); interpretation with "90% confident" + context (+1).

This diagnostic is half-length (21 MC instead of 42; 2 FRQs instead of 4). On the real exam, Section I and Section II each count for 50% of the score. To approximate a composite for this diagnostic, weight the two sections equally:

Section I score = (MC correct / 21) × 50

Section II score = (total FRQ points / 20) × 50

Composite (0–100 scale) = Section I score + Section II score

| Composite (approx.) | Rough AP band (self-assessment) |

|---|---|

| 80–100 | On track for a 4–5 |

| 65–79 | On track for a 3–4 |

| 50–64 | Borderline 3 — review weak areas |

| Below 50 | Below passing — revisit Units 1–2 fundamentals |

Important caveats: These bands are approximate and for self-assessment only — they are not official College Board cut scores, which are set by equating each year and are not published as fixed percentages. Because this is a mid-course diagnostic covering only Lessons 1–20, a low score on topics you haven't mastered yet is expected; use the per-item review to target your studying. Pay special attention to any interpretation items you missed (Questions 8, 20, 21 and FRQ interpretations) — interpretation in context is where the most AP points are won and lost.

StatsIQ · Mock Exam 1 (Mid-Course Diagnostic, Lessons 1–20) · Units 1–2 + early Unit 3

This mock exam aligns to the NEW 2026–27 AP Statistics Course and Exam Description (first administered May 2027). It uses the new format: 4-choice multiple choice and 10-point point-based free-response. AP® is a trademark registered by the College Board, which is not affiliated with and does not endorse this product.

Statistical-accuracy note: Every multiple-choice answer and every free-response value was independently recomputed and verified (Normal areas via standard Normal cdf; binomial via binompdf; sampling-distribution SDs via √(p(1−p)/n) and σ/√n; the one-proportion interval via 1-PropZInt). Key checks: Q3 fences 0 and 64; Q4 z = 1.5, P ≈ 0.933; Q11 P(A or B) = 0.60 with not-independent confirmed (0.50×0.30 = 0.15 ≠ 0.20); Q13 SD = 0.90; Q15 binompdf(12,0.25,3) ≈ 0.258; Q18 SD = 0.069; FRQ 2(b) P(p̂ > 0.55) ≈ 0.036; FRQ 2(c) 90% interval (0.459, 0.642). Reviewed for statistical accuracy by Isaac, retired actuary.

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