AP Prep — Statistics Practice

Question 1 of 10

MCQU2Topic 2.3Medium Calc Word

To estimate the probability that at least two of four randomly chosen people were born in the same calendar month, a student assigned each person a random integer from 1 to 12 and repeated the experiment for 200 independent trials. In 128 of the trials at least two of the four people shared a birth month. Based on this simulation, what is the best estimate of the probability?

A0.32, found by dividing 128 by 400

B0.36, the proportion of trials in which no two people shared a birth month

CAbout 0.427, the exact value computed with the multiplication rule

D0.64

Explanation

A simulation-based estimate is the observed relative frequency of the event: 128/200 = 0.64. Computing an exact theoretical value with the multiplication rule is a different quantity from the estimate the question asks for; 72/200 = 0.36 counts the complementary trials, and dividing by 400 doubles the number of trials.

Question 2 of 10

MCQU1Topic 1.3Easy Calc Word

A regional drone-delivery hub classified each of its 400 deliveries from one week by final outcome. The frequency table below summarizes the results.

Outcome        Count
On time         246
Delayed          92
Damaged          34
Returned         28
Total           400

What proportion of the week's deliveries had an outcome other than "On time"?

A0.615

B0.155, since only the Damaged and Returned deliveries represent outcomes in which the package failed to reach the customer as intended.

C0.230

D0.385

Explanation

A relative frequency for one categorical variable is a category count divided by the table total. The deliveries that were not "On time" are Delayed, Damaged, and Returned: 92 + 34 + 28 = 154, and 154/400 = 0.385. Equivalently, 1 − 246/400 = 1 − 0.615 = 0.385. The value 0.230 counts only the Delayed deliveries and ignores the other two non-on-time categories; 0.615 is the relative frequency of the "On time" category itself, which is the complement of what was asked; and 0.155 drops Delayed on the incorrect grounds that a late package is still a successful delivery — the question asks about every outcome other than on time.

Question 3 of 10

MCQU4Topic 4.1Medium Calc Word

The number of minutes drivers spend at a toll plaza has a right-skewed distribution with mean μ = 27 minutes and standard deviation σ = 8 minutes. A transportation analyst takes a random sample of n = 50 drivers and computes the sample mean x̄. Approximately what is P(x̄ > 29 minutes)?

AAbout 0.0386

BAbout 0.4013

CAbout 0.5987

DAbout 0.9614

Explanation

Because n = 50 is large, the Central Limit Theorem makes the sampling distribution of x̄ approximately Normal with mean 27 and standard deviation σ/√n = 8/√50 ≈ 1.131. Then z = (29 − 27)/1.131 ≈ 1.77, so P(x̄ > 29) ≈ 0.0386. The value 0.4013 comes from using the population SD of 8 directly (z = 0.25), which ignores that the CLT describes the spread of x̄, not of individual drivers; that variability shrinks by √n.

Question 4 of 10

MCQU3Topic 3.3Medium Calc Word

An online retailer samples 200 recent orders at random and finds that 52 of them included a gift-wrapping request. Assuming the conditions for inference are met, which of the following is the correct 95% confidence interval for the true proportion of all orders that request gift wrapping? (Use z* = 1.96.)

A(0.209, 0.311)

B(0.199, 0.321)

C(0.256, 0.264)

D(0.189, 0.331)

Explanation

With p̂ = 52/200 = 0.26, the standard error is √(p̂(1−p̂)/n) = √(0.26·0.74/200) ≈ 0.0310. The margin of error is 1.96·0.0310 ≈ 0.0608, giving 0.26 ± 0.0608 = (0.199, 0.321). The interval (0.209, 0.311) uses z* = 1.645 (a 90% critical value) instead of 1.96. The narrow interval (0.256, 0.264) comes from dividing by n rather than √n in the standard error. The interval (0.189, 0.331) drops the (1−p̂) factor, using √(p̂/n) and inflating the margin.

Question 5 of 10

MCQU2Topic 2.2Medium Calc Word

A city surveyed 300 workers about their commute mode and whether they arrived on time.

            On time   Late    Total
Drive         120      60      180
Transit        96      24      120
Total         216      84      300

What is the conditional relative frequency of arriving late among workers who drive?

A0.20, the proportion of all workers who both drove and were late

B0.3333

CAbout 0.714, since 60 of the 84 late workers drove

D0.28, since 84 of the 300 workers arrived late

Explanation

Condition on the 180 drivers and find how many were late: 60/180 = 0.3333. The value 84/300 = 0.28 is the marginal relative frequency of being late; 60/84 reverses the condition (of late workers, the fraction who drove); 60/300 is the joint relative frequency of driving and being late.

Question 6 of 10

MCQU5Topic 5.3Medium Calc Word

A running coach recorded, for six athletes, x = average weekly training distance (miles) and y = finishing time in a 5-km race (minutes):

x: 10, 15, 20, 25, 30, 35 y: 27.5, 26.1, 24.8, 23.9, 22.6, 21.8

The least-squares regression line is ŷ = 29.58 - 0.228x. Which statement correctly interprets the slope?

AEach additional weekly mile of training is associated with a predicted increase of about 0.228 minutes in 5-km time.

BThe correlation between weekly training distance and 5-km finishing time is about -0.228, indicating a weak relationship.

CEach additional weekly mile of training is associated with a predicted decrease of about 0.228 minutes in 5-km time.

DFor each additional minute in 5-km time, an athlete is predicted to train about 0.228 fewer miles per week.

Explanation

A negative slope means the predicted response decreases as x increases: here each extra weekly mile lowers predicted 5-km time by about 0.228 minutes. The sign must be read as a decrease, so calling it an increase is wrong. The slope is 'change in y per unit change in x' (time per mile), not the reverse (miles per minute), and the slope is not the same quantity as the correlation r, so quoting -0.228 as r confuses slope with correlation.

Question 7 of 10

MCQU4Topic 4.3Medium Calc Word

From a random sample of 15 espresso shots, a barista computes a mean extraction time of x̄ = 48.2 seconds with s = 5.6 seconds. Conditions for inference are met, and the 95% confidence interval for the true mean extraction time is about (45.1, 51.3) seconds. The machine's manufacturer claims the true mean is 50 seconds. Which statement is best supported by this interval?

ABecause 50 lies inside the interval, there is a 95% probability that the population mean equals exactly 50 seconds.

BWe are 95% confident that 95% of all individual espresso shots have extraction times between 45.1 and 51.3 seconds.

CThe interval shows the population mean must equal the sample mean of 48.2 seconds.

DA mean of 50 seconds is plausible, since 50 lies within the interval of about 45.1 to 51.3.

Explanation

A confidence interval gives the plausible values of the population mean. Because 50 falls inside (45.1, 51.3), the data are consistent with the manufacturer's claim; we cannot rule it out at the 5% level. Choice A misreads the confidence level as covering 95% of individual observations rather than expressing confidence about the mean. Choice B treats a fixed parameter as if it had a probability of equaling a value, and choice C confuses the point estimate with the parameter. The margin here is t*(s/√n) = t.ppf(0.975, 14)·5.6/√15 ≈ 3.10.

Question 8 of 10

MCQU3Topic 3.2Medium Calc Word

A city recycling office believes that p = 0.40 of households actually rinse their containers before recycling. An auditor takes a simple random sample of 50 households from the city's very large population. Assuming p = 0.40, and treating the sampling distribution of the sample proportion p̂ as approximately Normal, what is the approximate probability that p̂ is at least 0.50?

A0.925

B0.0745

C0.419

D0.149

Explanation

The sampling distribution of p̂ is approximately Normal with mean p = 0.40 and standard deviation √(p(1−p)/n) = √(0.40·0.60/50) ≈ 0.0693. Standardize 0.50: z = (0.50 − 0.40)/0.0693 ≈ 1.443, and P(p̂ ≥ 0.50) = P(Z ≥ 1.443) ≈ 0.0745. The value 0.419 comes from forgetting to divide by n inside the square root (using √(p(1−p)) as the standard deviation), which wrongly shrinks the z-score. The value 0.925 is the lower-tail probability P(p̂ ≤ 0.50), the wrong tail, and 0.149 is the result of doubling the tail as if a two-sided calculation were needed.

Question 9 of 10

MCQU1Topic 1.5Medium Calc Word

A ceramics studio recorded the number of minutes each of 12 kiln loads took to cool to a safe handling temperature. The stemplot below displays the data (stem = tens, leaf = ones; 1 | 4 means 14 minutes).

1 | 4 7 9 2 | 2 3 3 6 8 3 | 0 1 5 4 | 2

What is the median cooling time?

A24.5 minutes

B26 minutes, the value of the leaf that sits at the physical center of the printed stemplot when the rows are counted.

C25.83 minutes

D23 minutes

Explanation

A stemplot preserves every individual value, so read them off in order: 14, 17, 19, 22, 23, 23, 26, 28, 30, 31, 35, 42. With n = 12 (an even count) the median is the average of the 6th and 7th ordered values: (23 + 26)/2 = 24.5 minutes. Reporting 23 takes only the 6th value (and coincides with the most repeated leaf, which is the mode, not the median); 25.83 is the mean, which is pulled upward by the value 42 and is not the median; and 26 takes only the 7th value. With an even n, neither single middle value is the median on its own.

Question 10 of 10

MCQU5Topic 5.2Medium Calc Word

A botanist recorded x = stem length (cm) and y = flower mass (g) for five plants:

x: 3, 5, 6, 8, 9 y: 2.1, 3.0, 3.4, 4.1, 4.6

She then decides to report the stem lengths in millimeters, so she multiplies every x-value by 10 (1 cm = 10 mm). What happens to the correlation r between the two variables?

AThe value of r is divided by 10, since larger x-values shrink the correlation.

BThe value of r is multiplied by 10, because every explanatory value was multiplied by 10.

CThe value of r changes because the slope of the least-squares line changes when the units of x change.

DThe value of r stays exactly the same.

Explanation

Correlation r is unit-free and is unaffected by any linear rescaling (multiplying or adding a constant to) either variable, because r is computed from standardized values (z-scores). Multiplying every x by 10 multiplies both the deviations and the standard deviation of x by 10, so the standardized x-values are unchanged, leaving r identical. The slope b = r·(sy/sx) does change with the units of x, but that change in the slope does not change r itself.

Free Response 1 · Section II

FRQMulti-Focus on Practices 1 and 2U1 Calc

The parks department of Marlow, a mid-sized city, is planning trail improvements and wants to understand how much residents use the public trail system. The city is divided into three residential districts. The number of adult residents in each district is:

  District      Adult residents
  ------------  ---------------
  North           4,200
  Central         2,800
  South           3,000
  ------------  ---------------
  Total          10,000

(a) State a clearly defined statistical question the parks department could investigate about trail use, and identify the population and the population parameter of interest.

(b) An intern proposes standing at the busiest trailhead on a sunny Saturday morning and asking every adult who walks by how many hours per week they use the trails, until 150 people have answered. Name the sampling method this describes and identify one specific source of bias. State the likely DIRECTION of that bias (would the estimated mean be too high or too low) and explain why.

(c) The department decides instead to take a stratified random sample of 150 adults, using the three districts as strata and allocating the sample proportionally to district size. Determine how many adults should be sampled from each district. Show the calculation for at least one district.

(d) In a pilot run, 10 randomly selected residents reported these weekly trail-use hours:

2, 5, 3, 8, 1, 6, 4, 3, 7, 2

Compute the sample mean and the sample standard deviation of these pilot values. Report the mean to one decimal place and the standard deviation to two decimal places, with units. Briefly state one reason a pilot sample like this is useful before running the full study.

Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.

Model answer

(a) Statistical question: "What is the mean number of hours per week that adult residents of Marlow use the public trail system?" The population is all adult residents of Marlow (the 10,000 adults). The parameter of interest is the population mean weekly trail-use hours, μ.

(b) This is a convenience sample (interviewing whoever happens to walk by the trailhead). One source of bias: only people who are AT the trail on a sunny Saturday morning can be selected, so frequent/enthusiastic trail users are massively overrepresented and non-users have essentially no chance of being chosen (undercoverage / selection bias). Direction: the estimated mean would be TOO HIGH, because the sample is dominated by heavy users while light users and non-users (who pull the mean down) are almost entirely excluded.

(c) Proportional allocation: sample fraction = 150 / 10,000 = 0.015.
  North:   4,200 × 0.015 = 63 adults
  Central: 2,800 × 0.015 = 42 adults
  South:   3,000 × 0.015 = 45 adults
These total 63 + 42 + 45 = 150. Within each district, select that many adults at random.

(d) Data: 2, 5, 3, 8, 1, 6, 4, 3, 7, 2. Sum = 41, n = 10. Mean x̄ = 41 / 10 = 4.1 hours per week. Sample standard deviation s ≈ 2.33 hours per week (using ddof = 1). A pilot sample is useful because it gives a rough estimate of the variability (s) in trail-use hours, which lets the department judge whether a sample of 150 is large enough to estimate μ with the desired precision, and it lets them test the survey wording before the full study.

Scoring · 10 points

1. (a) Poses a clearly worded statistical question about trail use that admits variability (e.g., mean weekly hours).

2. (a) Identifies the population as all adult residents of Marlow.

3. (a) Identifies the parameter as the population mean μ of weekly trail-use hours (not a statistic).

4. (b) Names the method as a convenience (or voluntary) sample rather than a random sample.

5. (b) Identifies a specific, valid source of bias (undercoverage/selection: heavy users overrepresented, non-users excluded).

6. (b) States the direction as an overestimate AND ties the direction to who is over/under-represented.

7. (c) Uses the correct sampling fraction 150/10,000 = 0.015 (or equivalent proportional reasoning) with work shown for at least one stratum.

8. (c) Reports the correct allocation North 63, Central 42, South 45 (summing to 150).

9. (d) Correct sample mean 4.1 hours AND correct sample standard deviation ≈ 2.33 hours, with units.

10. (d) Gives a valid reason a pilot is useful (estimate variability for sample-size planning, or test the instrument).

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