ACT Math study guide · Chapter 37
Part V Review
This 20-question set mixes the four core Part V chapters introduced first: Center and Spread, Study Design and Sampling, Two-Variable Data and Models, and Probability. The short addendum that follows adds Normal Distributions and Z-Scores and Statistical Inference so the complete Part V review covers all six chapters.
1. Find the mean of 6, 9, 13, 17, 20.
A) 11 B) 9 C) 13 D) 17
2. Find the median of 14, 3, 19, 8, 11.
F) 14 G) 11 H) 8 J) 19
3. Five numbers, 5, 6, 7, 8, 9, have a mean of 7. If the 9 is replaced with 40, find the new mean.
A) 13.2 B) 7 C) 9 D) 8
4. A box plot has Q1 = 12 and Q3 = 29. Find the IQR.
F) 12 G) 29 H) 41 J) 17
5. Find the range of 8, 14, 3, 21, 17.
A) 14 B) 18 C) 21 D) 3
6. A pollster calls randomly selected phone numbers from the entire city's phone directory. What kind of sample is this?
F) Convenience sample G) Voluntary response sample H) Random sample J) Cannot be determined
7. Researchers randomly assign patients to receive a new drug or a placebo, then compare outcomes. What kind of study is this, and what can it conclude?
A) Observational study; causation B) Observational study; association only C) Randomized experiment; association only D) Randomized experiment; causation
8. In a survey, 35 people under 25 and 20 people over 25 answered “Yes” to a question; 15 people under 25 and 30 people over 25 answered “No.” How many people under 25 answered “Yes”?
F) 50 G) 55 H) 35 J) 15
9. Using the survey from the previous question, find the overall proportion of all respondents who answered “Yes.”
A) 0.55 B) 0.50 C) 0.45 D) 0.70
10. A survey only samples people who volunteer to respond online. What is the main risk?
F) No risk, it is still random G) A systematic under-count of the population H) It is guaranteed to match the population J) Bias, since only motivated volunteers respond
11. A line of best fit is y = 4x − 7. Predict y when x = 9.
A) 25 B) 29 C) 33 D) 22
12. A line of best fit is y = 3x + 2. An actual data point is (6, 25). Find its residual.
F) 5 G) −5 H) 15 J) 25
13. A scatterplot shows points trending upward, tightly clustered around a line. This describes what kind of correlation?
A) Strong negative B) Weak positive C) Strong positive D) No correlation
14. Countries with more Nobel Prizes tend to have higher chocolate consumption per capita. What is the most likely explanation?
F) Chocolate consumption causes Nobel Prizes G) A third variable, such as national wealth, drives both H) Nobel Prizes cause chocolate consumption J) The correlation is a calculation error
15. A line of best fit for advertising spend versus sales is y = 2.5x + 1000. What does the slope represent?
A) Sales with zero ad spend B) The total ad spend C) The total number of sales D) The predicted increase in sales per additional unit of ad spend
16. Events A and B are independent, with P(A) = 0.45 and P(B) = 0.8. Find P(A and B).
F) 0.36 G) 1.25 H) 0.8 J) 0.45
17. Using the survey from questions 8-9, find the probability that a randomly chosen under-25 respondent answered “Yes.”
A) 0.55 B) 0.7 C) 0.35 D) 0.5
18. A game pays $40 with probability 0.15, pays $15 with probability 0.35, and costs $8 (a loss) with probability 0.5. Find the expected value.
F) $6.75 G) $7.25 H) $8.00 J) $5.50
19. How many ways can a committee of 3 be chosen from 6 people?
A) 20 B) 120 C) 15 D) 6
20. How many ways can 3 distinct offices be filled from 6 candidates, where order matters?
F) 120 G) 20 H) 15 J) 6
▸ Answers and Explanations
1: C. (6+9+13+17+20)/5=13. Choice A comes from an arithmetic slip while summing the values before dividing.
2: G. Ordered: 3, 8, 11, 14, 19. The middle value is 11. Choice F reports a value adjacent to the true median instead of the middle value itself.
3: A. Original sum = 7×5 = 35; replacing 9 with 40 gives a new sum of 35−9+40=66; new mean = 66/5=13.2. Choice B reports the original mean, forgetting to recompute after replacing 9 with 40.
4: J. IQR = Q3 − Q1 = 29 − 12 = 17. Choice F reports Q1 itself instead of subtracting it from Q3.
5: B. Range = maximum − minimum = 21 − 3 = 18. Choice A comes from an arithmetic slip while subtracting the minimum from the maximum.
6: H. Calling randomly selected numbers from the entire directory gives every listed resident an equal chance, making this a random sample. Choice F mistakes this genuinely random selection process for a convenience sample.
7: D. Random assignment to treatment groups is what allows this study to support a causal conclusion. Choice A correctly concludes that causation is supported but misidentifies the study as observational instead of a randomized experiment.
8: H. Directly given: 35 people under 25 answered Yes. Choice F adds the under-25 Yes and No counts together instead of reporting the Yes count alone.
9: A. Total Yes: 35+20=55; grand total: 100; 55/100=0.55. Choice B comes from an arithmetic slip while combining the two age groups' Yes counts.
10: J. Only people who choose to respond online are included, risking bias toward those with strong opinions. Choice F fails to recognize that voluntary participation is not the same as random selection.
11: B. 4(9) − 7 = 36 − 7 = 29. Choice A comes from an arithmetic slip while combining the slope and constant terms.
12: F. Predicted value: 3(6)+2=20; residual=actual−predicted=25−20=5. Choice G drops the sign while computing the residual, subtracting the actual value from the predicted value instead of the reverse.
13: C. An upward trend means positive correlation; tight clustering means the correlation is strong. Choice A mistakes the upward trend for a downward one, reversing the correlation's direction.
14: G. A plausible third variable, national wealth, could increase both Nobel Prize output and chocolate consumption, without either directly causing the other. Choice F assumes a direct causal link between the two measured variables instead of considering a plausible third variable.
15: D. In y=mx+b, the slope m represents the predicted change in y for each one-unit increase in x, here each additional unit of ad spend. Choice A confuses the slope with the y-intercept, which represents the value when x=0.
16: F. 0.45×0.8=0.36. Choice G adds the two probabilities instead of multiplying them, a common independent-events error.
17: B. Restricting to the under-25 group: 35 Yes out of 50 total, 35/50=0.7. Choice A reports the overall proportion instead of the proportion restricted to the under-25 group.
18: G. (40)(0.15)+(15)(0.35)+(−8)(0.5)=6+5.25−4=7.25. Choice F comes from an arithmetic slip while combining the three weighted outcomes.
19: A. Order does not matter, so this is a combination: 6C3=20. Choice B treats the selection as if order mattered, computing a permutation instead of a combination.
20: F. Order matters (distinct offices), so this is a permutation: 6P3=120. Choice G treats the selection as if order didn't matter, computing a combination instead of a permutation.
▸ Additional Part V Review
This short addendum gives extra mixed practice in Normal Distributions and Z-Scores and Statistical Inference, along with the general probability rules covered earlier in Part V, so those skills are tested alongside the rest of Part V.
1. A histogram is strongly right-skewed. Which relationship is most likely?
A) Mean < median B) Mean > median C) Mean = median always D) IQR = 0
2. A roughly normal distribution has mean 50 and standard deviation 6. What z-score corresponds to 62?
F) 1 G) 2 H) 6 J) 12
3. About what percent of a roughly normal distribution lies within 2 standard deviations of the mean?
A) 34% B) 68% C) 95% D) 99.7%
4. A poll reports 57% ±3%. Is 61% inside the reported plausible interval?
F) Yes G) No H) Only if n>1,000 J) Cannot be decided
5. Under a population claim, 1,000 simulations produce results at least as extreme as the observed result only 8 times. Which conclusion is best?
A) The claim is strongly supported B) The observed result is unusual under the claim and is evidence against it C) The claim is proven false D) Sample size is irrelevant
6. P(A)=0.55, P(B)=0.40, and P(A and B)=0.20. Find P(A or B).
F) 0.22 G) 0.75 H) 0.95 J) 1.15
7. P(A)=0.6 and P(B|A)=0.5. Find P(A and B).
A) 0.1 B) 0.3 C) 0.5 D) 1.1
8. Two models fit the same data. Model A systematically overpredicts high x-values; Model B has residuals scattered randomly around 0. Which model is more appropriate?
F) A, because its errors are patterned G) B, because random residual scatter suggests the model has captured the main pattern H) Both equally J) Neither can ever be used
▸ Answers and Explanations
1: B. In a right-skewed distribution, a long tail of high values pulls the mean upward more than the median, so the mean is typically greater than the median. Choice A describes the pattern for a left-skewed distribution instead.
2: G. z=(62−50)/6=12/6=2. Choice F reports the standard deviation's distance in the wrong units, treating the raw difference incompletely.
3: C. The empirical rule states that about 95% of values in a roughly normal distribution fall within 2 standard deviations of the mean. Choice B applies the 1-standard-deviation percentage instead of the 2-standard-deviation percentage.
4: G. The plausible interval is 57%−3%=54% to 57%+3%=60%. Since 61% is greater than the upper bound of 60%, it falls outside the interval. Choice F assumes 61% is automatically plausible without checking it against the computed interval.
5: B. A result this unusual under the claim (occurring in only 8 of 1,000 simulations) is evidence against the claim, supporting a decision to reject it under a reasonable threshold. Choice A misreads a rare, unusual outcome as support for the claim instead of evidence against it.
6: G. P(A or B)=P(A)+P(B)−P(A and B)=0.55+0.40−0.20=0.75. Choice F multiplies P(A) and P(B) instead of applying the addition rule, computing 0.55(0.40)=0.22 as if this were a P(A and B) calculation.
7: B. P(A and B)=P(A)·P(B|A)=0.6(0.5)=0.3. Choice A subtracts the two given probabilities instead of multiplying them, computing 0.6−0.5=0.1.
8: G. Residuals scattered randomly around 0 indicate the model has captured the underlying pattern, leaving only unpredictable noise. A systematic pattern in the residuals, as in Model A, signals that the model is missing part of the real relationship. Choice F treats a patterned residual as a strength instead of a warning sign.