AP Statistics Exploring One-Variable Data — Worked Answer Explanations

Unit 1 · 12 questions explained

Below is a complete answer key for our AP Statistics Exploring One-Variable Data practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Exploring One-Variable Data practice test and come back here to review, or head back to the Exploring One-Variable Data unit overview.

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Question 1 · Easy
A data set of 12 exam scores has a mean of 78 and a median of 82. Which of the following best describes the shape of the distribution?
- A
  Symmetric
  Why not A: Symmetric distributions have mean ≈ median; here they differ significantly.
- B
  Skewed left (negatively skewed)Correct
- C
  Skewed right (positively skewed)
  Why not C: Right-skewed distributions pull the mean above the median, not below.
- D
  Bimodal
  Why not D: Bimodal shape is not determined by comparing mean and median alone.
Explanation
When the mean is less than the median, a few unusually low values pull the mean downward, indicating left skew. Right skew would make mean > median.

Key takeaway
Mean < median signals left (negative) skew; mean > median signals right skew.
Question 2 · Easy
A stemplot of hourly wages (in dollars) for 10 part-time workers is shown below. What is the median wage?

Stem Leaves
8 2 5
9 1 3 7
10 0 4 4 8
11 2 6
- A
  $9.70
  Why not A: This is the 5th value, not the average of the 5th and 6th.
- B
  $10.00
  Why not B: This is the 6th value, not the average of the 5th and 6th.
- C
  $10.02
  Why not C: Rounding error; the average of $9.70 an d$ 10.00 is exactly $9.85.
- D
  $9.85Correct
Explanation
With 10 data points, the median is the average of the 5th and 6th values. Ordered values: 8.2, 8.5, 9.1, 9.3, 9.7, 10.0, 10.4, 10.4, 10.8, 11.2. The 5th is 9.7 and the 6th is 10.0, so the median is (9.7 + 10.0)/2 = $9.85.

Key takeaway
For an even number of data points, the median is the average of the two middle values.
Question 3 · Easy
A boxplot shows the following five-number summary for a data set: Min = 10, Q1 = 22, Median = 30, Q3 = 45, Max = 70. What is the interquartile range (IQR)?
- A
  60
  Why not A: 60 is the range (Max − Min), not the IQR.
- B
  23Correct
- C
  8
  Why not C: 8 is the distance from Q1 to the median, not Q3 − Q1.
- D
  15
  Why not D: 15 is the distance from the median to Q3, not Q3 − Q1.
Explanation
IQR = Q3 − Q1 = 45 − 22 = 23. The IQR measures the spread of the middle 50% of the data and is resistant to outliers, unlike the range.

Key takeaway
IQR = Q3 − Q1; it measures the spread of the middle 50% of data.
Question 4 · Easy
A distribution of test scores is approximately normal with mean $μ = 72$ and standard deviation $σ = 8$ . Approximately what percentage of scores fall between 64 and 88?
- A
  68%
  Why not A: 68% covers scores within 1 standard deviation (64 to 80), not 64 to 88.
- B
  95%
  Why not B: 95% covers ±2 SD (56 to 88), not 64 to 88 which spans 1 SD below to 2 SD above.
- C
  81.5%Correct
- D
  99.7%
  Why not D: 99.7% covers ±3 SD from the mean.
Explanation
64 = 72 − 8 is 1 SD below the mean; 88 = 72 + 2(8) is 2 SD above. Using the 68-95-99.7 rule: the area from −1 SD to +2 SD = 34% + 47.5% = 81.5%.

Key takeaway
For non-symmetric normal intervals, add the separate half-rule percentages: 34% (−1 to mean) + 47.5% (mean to +2 SD) = 81.5%.
Question 5 · Easy
A student scored 85 on a test where the class mean was 78 and the standard deviation was 7. What is the student's standardized score ( $z$ -score)?
- A
  $z = 0.88$
  Why not A: Dividing by mean (78) instead of standard deviation.
- B
  $z = 1.00$ Correct
- C
  $z = - 1.00$
  Why not C: Sign error; the student scored above the mean, so z must be positive.
- D
  $z = 7.00$
  Why not D: Forgetting to divide by the standard deviation.
Explanation
$z = \frac{x - μ}{σ} = \frac{85 - 78}{7} = \frac{7}{7} = 1.00$ . A z-score of 1.00 means the student scored exactly one standard deviation above the class mean.

Key takeaway
$z = (x - \mu)/\sigma$; positive z means above the mean, negative z means below.
Question 6 · Easy
A histogram of household incomes in a neighborhood is strongly right-skewed. Which measure of center and spread is most appropriate to report?
- A
  Mean and standard deviation
  Why not A: Mean and SD are sensitive to outliers and skew; they are not resistant measures.
- B
  Median and IQRCorrect
- C
  Mean and IQR
  Why not C: Mixing a non-resistant center measure with a resistant spread measure is inconsistent.
- D
  Median and standard deviation
  Why not D: Mixing a resistant center with a non-resistant spread measure is inconsistent.
Explanation
For skewed distributions or those with outliers, the median (resistant to extreme values) and IQR (resistant spread) give a more accurate picture than the mean and standard deviation, which are both pulled by extreme values.

Key takeaway
For skewed data, use median + IQR; for symmetric data without outliers, mean + SD is appropriate.
Question 7 · Easy
Which of the following correctly identifies an outlier using the 1.5 × IQR rule? A data set has Q1 = 20, Q3 = 50, and contains the value 95.
- A
  95 is not an outlier because it is less than the maximum.
  Why not A: Being less than the maximum does not prevent a value from being an outlier.
- B
  95 is an outlier because it exceeds $Q 3 + 1.5 \times I QR = 95$ .
  Why not B: The upper fence is Q3 + 1.5×IQR = 50 + 1.5(30) = 95, so 95 is exactly at the fence, not beyond it.
- C
  95 is not an outlier because it does not exceed $Q 3 + 1.5 \times I QR = 95$ .Correct
- D
  95 is an outlier because it is more than 1.5 standard deviations above the mean.
  Why not D: The 1.5 × IQR rule uses quartiles, not standard deviations.
Explanation
IQR = Q3 − Q1 = 50 − 20 = 30. Upper fence = Q3 + 1.5 × IQR = 50 + 45 = 95. A value must be strictly greater than the fence to be classified as an outlier, so 95 falls exactly on the boundary and is not an outlier.

Key takeaway
Upper fence = Q3 + 1.5 × IQR; values strictly above this fence are outliers.
Question 8 · Easy
A cumulative relative frequency plot (ogive) shows that the 60th percentile of commute times is 35 minutes. Which of the following statements is correct?
- A
  60% of commute times are greater than 35 minutes.
  Why not A: Confusing percentile direction; the p-th percentile means p% of data fall at or below that value.
- B
  35% of commuters have a commute time below 60 minutes.
  Why not B: This reverses the roles of the percentile and the value.
- C
  60% of commute times are at or below 35 minutes.Correct
- D
  The median commute time is 35 minutes.
  Why not D: The median is the 50th percentile, not the 60th.
Explanation
By definition, the p-th percentile is the value at or below which p% of the data fall. The 60th percentile being 35 minutes means 60% of commute times are ≤ 35 minutes.

Key takeaway
The p-th percentile means p% of data values fall at or below that point.
Question 9 · Medium
A biology class recorded the number of seeds that germinated out of 20 seeds for each of 30 students. The data are roughly bell-shaped with mean $\overset{x}{ˉ} = 14.2$ and $s = 2.1$ . Using the normal model, approximately what proportion of students had fewer than 10 seeds germinate?
- A
  About 2.5%
  Why not A: 2.5% corresponds to z = −2; here z ≈ −2.0 so this is approximately correct but the calculated z is slightly different.
- B
  About 0.15%
  Why not B: 0.15% corresponds to z = −3, which would require 10 to be 3 SD below the mean.
- C
  About 2.3%Correct
- D
  About 16%
  Why not D: 16% corresponds to z = −1, but 10 is about 2 SDs below the mean.
Explanation
$z = \frac{10 - 14.2}{2.1} = \frac{- 4.2}{2.1} = - 2.0$ . Using the standard normal table, $P (Z < - 2.0) \approx 0.0228$ , or about 2.3%.

Key takeaway
Convert to a z-score, then use the standard normal table: $P(X < x) = P\left(Z < \frac{x-\mu}{\sigma}\right)$.
Question 10 · Medium
A dot plot shows the ages (in years) of 9 trees: 3, 5, 5, 7, 8, 9, 10, 12, 40. Which measure of center best represents the typical age, and why?
- A
  Mean, because it uses all values in the data set.
  Why not A: Using all values is a property, not a justification for best representation when an outlier is present.
- B
  Median, because it is resistant to the extreme value of 40.Correct
- C
  Mean, because it equals the median in symmetric distributions.
  Why not C: This distribution is not symmetric due to the outlier 40.
- D
  Mode, because 5 appears twice.
  Why not D: The mode (5) is the most frequent value but does not represent the center of the distribution.
Explanation
With the outlier 40, the mean is pulled to $(3 + 5 + 5 + 7 + 8 + 9 + 10 + 12 + 40) /9 \approx 11.0$ , well above most values. The median is 8, which better represents the typical age. The median is resistant to outliers.

Key takeaway
When outliers are present, the median is a more representative measure of center than the mean.
Question 11 · Hard
A set of exam scores has mean 70 and standard deviation 10. Each score is multiplied by 1.05 (a 5% curve), then 3 points are added. What are the new mean and standard deviation?
- A
  Mean = 73, SD = 10
  Why not A: Adding 3 to each score shifts the mean by 3 but ignores the multiplication step.
- B
  Mean = 76.5, SD = 10.5Correct
- C
  Mean = 76.5, SD = 10
  Why not C: Correctly computes the mean but forgets that multiplying by a constant also scales the SD.
- D
  Mean = 73, SD = 10.5
  Why not D: Applies the scaling to SD but not the mean, and misses the multiplication step for the mean.
Explanation
When every value is transformed as $y = 1.05 x + 3$ : the new mean is $1.05 (70) + 3 = 73.5 + 3 = 76.5$ and the new SD is $1.05 \times 10 = 10.5$ . Adding a constant shifts the mean but does not affect SD; multiplying by a constant scales both.

Key takeaway
For $y = ax + b$: new mean $= a\bar{x}+b$, new SD $= |a|s$. Adding a constant does not change spread.
Question 12 · Hard
A researcher says a score at the 90th percentile in a normal distribution with mean 50 and SD 6 corresponds to a score of about 57.7. Another student says the 90th percentile score is about 42.3. Who is correct, and why?
- A
  The student, because 42.3 is 1.28 standard deviations below the mean.
  Why not A: 42.3 is at the 10th percentile, not the 90th, which is below the mean.
- B
  Both are wrong; the 90th percentile requires the exact normal table value.
  Why not B: The researcher's value is correct using z* ≈ 1.28 for the 90th percentile.
- C
  The researcher, because $z^{*} \approx 1.28$ for the 90th percentile, giving $x = 50 + 1.28 (6) \approx 57.7$ .Correct
- D
  Neither; the correct value is 56.0, using $z^{*} = 1.00$ .
  Why not D: z* = 1.00 corresponds to the 84th percentile, not the 90th.
Explanation
The 90th percentile corresponds to $z^{*} \approx 1.28$ (from the standard normal table). Then $x = μ + z^{*} σ = 50 + 1.28 (6) \approx 57.68 \approx 57.7$ . The value 42.3 is at the 10th percentile ( $z^{*} = - 1.28$ ).

Key takeaway
To find a percentile value: look up $z^*$, then compute $x = \mu + z^* \sigma$. The 90th percentile uses $z^* \approx 1.28$.

Stem	Leaves
8	2 5
9	1 3 7
10	0 4 4 8
11	2 6