AP Statistics Quiz: Summary Statistics For A Quantitative Variable

What this quiz covers

This quiz focuses on Summary Statistics For A Quantitative Variable, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The IQR is $33-15=18$ cm, so the middle 50% of fish lengths span 18 cm.
2. A typical fish length is $4.0$ cm because the SD equals the typical length.
3. The middle 50% of fish lengths are between 22 cm and 27 cm, so the IQR is 5 cm, and the distribution appears roughly symmetric because the mean and median are close and the tails are similar. (correct answer)
4. Because the median is 24 cm, about half the fish are between 15 cm and 24 cm.
5. Since $Q_1=22$ cm, 25% of fish are longer than 22 cm.

Explanation: This AP Statistics question tests interpreting quantitative variable summaries, including spread and shape. Choice C correctly notes middle 50% from 22 to 27 (IQR=5) and rough symmetry as mean (24.5) ≈ median (24.0) with similar tails (15-22=7, 27-33=6). Distractor E misstates Q1 (22), saying 25% longer than 22, but actually 75% ≥22. Lesson: Close mean-median suggests symmetry; IQR for central spread. SD (4.0) indicates low variability. Five-number summary shows balanced tails. Distribution appears nearly symmetric with moderate spread.

Question 2

1. The middle 50% of waiting times are between 8 and 160 minutes because those are the minimum and maximum.
2. The distribution is likely right-skewed, with a typical wait near 49 minutes, because the mean exceeds the median and the upper tail is much longer. (correct answer)
3. The IQR equals $160-8=152$ minutes, so half of waits vary by 152 minutes.
4. A typical waiting time is 30 minutes because that is the SD.
5. Because $Q_1=32$, 25% of patients waited less than 8 minutes.

Explanation: In AP Statistics, this tests understanding quantitative variable summaries, emphasizing shape and center. Choice B correctly describes right-skewness with typical wait near 49 (median), as mean (58) > median and upper tail (70 to 160) much longer than lower (8 to 32). Distractor A confuses middle 50% with min-max (8 to 160) instead of Q1-Q3. Key: For skew, median is robust center; compare tails in five-number summary. IQR (70-32=38) measures central spread. SD (30) shows high variability. Indicates right-skewed waits with long upper extremes.

Question 3

1. Because the maximum is much larger than $Q_3$ and the mean is greater than the median, the distribution is likely skewed right due to a few very large fish. (correct answer)
2. The IQR of 150 g means the heaviest fish is 150 g heavier than the lightest fish.
3. The standard deviation of 120 g means the typical fish weighs 120 g.
4. Since $Q_3=470$ g, about 75% of fish weigh more than 470 g.
5. Because the median is 395 g, exactly half the fish weigh 395 g.

Explanation: This question assesses recognition of right-skewed distributions in biological data. The mean (410 g) being greater than the median (395 g) suggests right skew, where the distribution has a tail extending toward heavier fish. The maximum value (920 g) being much larger than Q3 (470 g) strongly confirms this - there are a few unusually large fish pulling the mean upward. The distractors contain typical errors: IQR represents Q3-Q1, not max-min; standard deviation measures variability around the mean, not a typical weight; Q3=470 g means 75% of fish weigh 470 g or less, not more; and the median indicates that half the fish weigh less than and half weigh more than 395 g, not that half weigh exactly 395 g.

Question 4

1. Because the mean is greater than the median and the maximum is far above $Q_3$, the distribution is likely skewed right due to a few unusually high-visit days. (correct answer)
2. An IQR of 33 visits means the number of visits typically equals 33.
3. The standard deviation of 26 visits means the range of the data is 26 visits.
4. Since the median is 105, exactly 105 visits occurred on half of the days.
5. Because $Q_1=92$ and $Q_3=125$, about 75% of days had between 92 and 125 visits.

Explanation: This question assesses recognition of right-skewed distributions through summary statistics. The mean (112 visits) exceeding the median (105 visits) is a classic indicator of right skew, where the distribution has a tail extending toward higher values. Additionally, the maximum value (210) being far above Q3 (125) confirms this interpretation - there are a few days with unusually high emergency room visits pulling the mean upward. The incorrect options reflect common misunderstandings: IQR represents the spread of the middle 50% of data, not a typical value; standard deviation measures variability, not range; the median tells us that half the days had fewer than 105 visits, not that exactly 105 visits occurred; and the middle 50% of days (not 75%) had between Q1 and Q3 visits.

Question 5

1. Because the mean and median are essentially equal and the five-number summary is fairly balanced, the distribution is likely roughly symmetric. (correct answer)
2. The IQR of 6.0 miles means the range of route distances is 6.0 miles.
3. The standard deviation of 4.2 miles means the typical route distance is 4.2 miles.
4. Since $Q_3=21.0$ miles, about 75% of routes are longer than 21.0 miles.
5. Because the median is 18.0 miles, exactly half the routes are 18.0 miles long.

Explanation: This question tests recognition of symmetric distributions through summary statistics. The mean (18.1 miles) and median (18.0 miles) being essentially equal is a strong indicator of a roughly symmetric distribution. The five-number summary also shows good balance - the distances from median to quartiles are equal (3.0 miles each), and the extremes are roughly equidistant from their nearest quartiles. The incorrect options demonstrate common errors: IQR represents the spread of the middle 50% of data, not the total range; standard deviation measures typical deviation from the mean, not a typical value itself; Q3=21.0 means 75% of routes are 21.0 miles or shorter, not longer; and the median indicates that half the routes are shorter and half are longer than 18.0 miles, not that half are exactly 18.0 miles.

Question 6

1. Because the IQR is 3, the number of books checked out ranged from 2 to 5 books.
2. Because the mean is slightly larger than the median and the maximum is well above $Q_3$, the distribution is likely right-skewed, with a few patrons checking out many books. (correct answer)
3. Because the SD is 2.2, most patrons checked out about 2.2 books.
4. Because the median is 3, about 75% of patrons checked out fewer than 3 books.
5. Because the IQR is 3, the range is 3 books (from 0 to 3).

Explanation: This question examines mild right skewness in count data. The mean (3.6 books) being slightly larger than the median (3 books) suggests right skewness, confirmed by the maximum (12 books) being well above Q3 (5 books). This indicates a few patrons checking out many books pull the mean upward. The IQR of 3 correctly measures the spread between Q1=2 and Q3=5, representing the middle 50% of the data. The standard deviation of 2.2 books measures variability, not that patrons typically checked out 2.2 books. The range is actually 12 books (from 0 to 12), not 3 books. Understanding these distinctions helps correctly interpret summary statistics for count data.

Question 7

1. Because the mean is close to the median and the tails are similar in length, the distribution is likely roughly symmetric, with the middle 50% between 25 and 36 minutes. (correct answer)
2. Because the SD is 8.5 minutes, the typical commute time is 8.5 minutes.
3. Because the IQR is 11 minutes, the range of commute times is 11 minutes.
4. Because the median is 30 minutes, about 75% of workers commute less than 30 minutes.
5. Because the maximum is 49 minutes, at least half the commutes are longer than 49 minutes.

Explanation: This question tests recognition of a roughly symmetric distribution. The mean (31.2) and median (30) are very close, suggesting symmetry. The five-number summary shows similar tail lengths: the minimum (14) is 16 minutes below the median, while the maximum (49) is 19 minutes above it. The IQR of 11 minutes correctly identifies that the middle 50% of commute times fall between Q1=25 and Q3=36 minutes. The standard deviation of 8.5 minutes measures spread, not a typical commute time. Remember that the median divides the data in half - 50% below and 50% above, not 75% below as one distractor suggests.

Question 8

1. The distribution is likely symmetric because $Q_1$ and $Q_3$ are equally spaced from the median.
2. The IQR is $6.0-4.0=2.0$ days, meaning the middle 50% of delivery times span about 2 days. (correct answer)
3. A typical order took $2.4$ days to arrive because the SD is $2.4$ days.
4. The IQR is $18-2=16$ days, which shows the middle 50% is very spread out.
5. Because the mean is greater than the median, at least 50% of orders took more than 5.6 days.

Explanation: This AP Statistics question evaluates summary statistics for quantitative variables, particularly spread. Choice B accurately computes IQR as 6-4=2 days, spanning middle 50%. Distractor D miscalculates IQR as range (18-2=16), ignoring actual quartiles. Lesson: IQR resists outliers, ideal for skew; here mean (5.6) > median (5.0) suggests right-skew despite close Q1/median/Q3. SD (2.4) quantifies deviation, not typical time. Five-number summary shows long upper tail. Distribution is right-skewed with low central spread but outliers.

Question 9

1. The middle 50% of times are between $60.6$ s and $63.4$ s, so the IQR is $2.8$ s. (correct answer)
2. The standard deviation of $2.3$ seconds means most athletes ran exactly $2.3$ seconds.
3. Because the mean is slightly greater than the median, the distribution must be extremely right-skewed.
4. The IQR is $67.0-57.5=9.5$ seconds because IQR is the same as the range.
5. Since the maximum is $67.0$ s, at least 75% of athletes ran slower than $67.0$ s.

Explanation: This question assesses IQR and quartile interpretation for run times. Q1 = 60.6 s and Q3 = 63.4 s mean middle 50% between them, with IQR = 2.8 s, as choice A states. Choice D is a distractor, confusing IQR with range (9.5 s). Mini-lesson: IQR = Q3 - Q1 measures central spread; quartiles divide data into quarters. Slight mean (62.1) > median (61.8) suggests mild right skew, and standard deviation (2.3 s) quantifies deviation from mean, not implying 'most' values are exactly that.

Question 10

1. Because the mean is less than the median, the distribution is likely skewed left, possibly due to a few students with very low sleep times. (correct answer)
2. The IQR of 75 minutes means the range of sleep times is 75 minutes.
3. The standard deviation of 55 minutes means most students slept about 55 minutes total.
4. Since $Q_3=465$ and $Q_1=390$, about 75% of students slept between 390 and 465 minutes.
5. The median of 435 minutes means exactly half the students slept 435 minutes.

Explanation: This question tests understanding of how the relationship between mean and median indicates skewness in a distribution. When the mean (420 minutes) is less than the median (435 minutes), this suggests the distribution is skewed left, with a tail extending toward lower values. This occurs because the mean is pulled down by a few students who slept much less than typical. The other options contain common misconceptions: IQR measures the spread of the middle 50% of data, not the total range; standard deviation measures typical deviation from the mean, not total sleep time; the middle 50% of students slept between Q1 and Q3, not 75%; and the median indicates that half slept less than and half slept more than 435 minutes, not that half slept exactly 435 minutes.

Question 11

1. The distribution is likely left-skewed because the mean is less than the median. (correct answer)
2. The IQR of $17$ ppb means the ozone values range from $12$ to $29$ ppb.
3. A typical ozone concentration is $12$ ppb because the standard deviation is $12$.
4. Since the median is $41$ ppb, exactly 75% of days had ozone at or below $41$ ppb.
5. Because the maximum is $60$ ppb, the IQR must be $60-12=48$ ppb.

Explanation: This question focuses on shape interpretation using summary statistics for ozone concentrations. The mean of 38 ppb is less than the median of 41 ppb, indicating left skewness, as correctly noted in choice A. Choice D distracts by saying exactly 75% are at or below the median (41 ppb), but median is the 50th percentile. Mini-lesson: When mean < median, left skew is likely due to low outliers pulling the mean down, supported here by longer left tail (min to Q1 = 18 ppb vs. Q3 to max = 13 ppb). The IQR (17 ppb) measures central spread, while standard deviation (12 ppb) shows overall variability from the mean.

Question 12

1. Because the mean exceeds the median and the maximum is much larger than $Q_3$, the distribution is likely skewed right due to a few very expensive homes. (correct answer)
2. The IQR of 160 means the typical home price is $160{,}000$.
3. The standard deviation of 110 means home prices range from $365-110$ to $365+110$ with no exceptions.
4. Since $Q_1=260$ and $Q_3=420$, about 75% of homes sold between $260{,}000$ and $420{,}000$.
5. The median of 340 means exactly half the homes sold for $340{,}000$.

Explanation: This question assesses understanding of right-skewed distributions in real estate data. The mean ($365,000) substantially exceeding the median ($340,000) is a strong indicator of right skew, where expensive homes pull the mean upward. The maximum value ($820,000) being much larger than Q3 ($420,000) confirms the presence of outliers on the high end - luxury homes that are much more expensive than typical. Common misinterpretations in other options include: IQR represents the spread of the middle 50%, not a typical price; standard deviation doesn't define absolute boundaries for all data; the middle 50% (not 75%) of homes sold between Q1 and Q3; and the median indicates that half sold for less than and half for more than $340,000, not that half sold for exactly that amount.

Question 13

1. Because the mean and median are nearly equal and the five-number summary is fairly balanced, the distribution is likely roughly symmetric. (correct answer)
2. An IQR of 15 means the range is 15 emails.
3. A standard deviation of 11 means half the employees received between 46 and 68 emails.
4. Since $Q_1=50$, about 75% of employees received fewer than 50 emails.
5. The median of 57 means all employees received 57 emails.

Explanation: This question tests understanding of symmetric distributions and their characteristics. The mean (58) and median (57) being nearly equal is a key indicator of a roughly symmetric distribution. Additionally, the five-number summary shows reasonable balance - the distances from the median to Q1 and Q3 are similar (7 and 8 respectively), and the extremes are roughly equidistant from the quartiles. Common misconceptions in other options include: IQR measures the spread of the middle 50%, not the total range; standard deviation doesn't define exact boundaries for data; Q1=50 means 25% (not 75%) received fewer than 50 emails; and the median indicates the middle value when data is ordered, not that all employees received that exact number.

Question 14

1. Because the IQR is 7.2, the difference between the most expensive and least expensive cars is 7.2 thousand dollars.
2. Because the SD is 6.1, most cars sold for about 6.1 thousand dollars.
3. Because the mean exceeds the median and the upper tail (to 39.5) is longer, the selling prices are likely right-skewed, with some high-priced cars pulling the mean up. (correct answer)
4. Because the median is 17.2, about 75% of cars sold for less than 17.2 thousand dollars.
5. Because $Q_3$ is 21.0, exactly 21 cars sold for 21.0 thousand dollars or more.

Explanation: This question tests interpretation of right-skewed car price data. The mean ($18.9k) exceeding the median ($17.2k) indicates right skewness, confirmed by the long upper tail - the maximum ($39.5k) is $18.5k above Q3 ($21.0k), while the minimum ($7.5k) is only $6.3k below Q1 ($13.8k). The IQR of $7.2k measures the spread of the middle 50% of prices between Q1 and Q3, not the total range which is $32k. The standard deviation of $6.1k measures variability, not that cars sold for $6.1k. Q3 being $21.0k means 75% of cars sold for less than this amount, not that exactly 21 cars sold for this price or more.

Question 15

1. Because the SD is 8.9 years, most students are about 8.9 years old.
2. Because the IQR is 9 years, the ages range from 21 to 30 years and no students are outside that range.
3. Because the mean is larger than the median and the maximum is far above $Q_3$, the age distribution is likely right-skewed due to some older students. (correct answer)
4. Because the median is 24, about half of the students are between 24 and 58 years old.
5. Because the IQR is 9 years, the difference between the oldest and youngest student is 9 years.

Explanation: This question tests recognition of strong right skewness in age data. The mean (27.8 years) substantially exceeding the median (24 years) indicates right skewness, dramatically confirmed by the maximum (58 years) being 28 years above Q3 (30 years). This shows some much older students in the evening program pull the mean upward. The IQR of 9 years correctly identifies that the middle 50% of students are between 21 and 30 years old. The total age range is 40 years (58-18), not 9 years. The standard deviation of 8.9 years measures spread around the mean, not that students are typically 8.9 years old. Remember that 50% of students are below 24 years and 50% above, not that half are between 24 and 58.

Question 16

1. The distribution is strongly right-skewed because the mean is slightly larger than the median.
2. The middle 50% of beetle lengths fall between $17.0$ mm and $19.4$ mm. (correct answer)
3. About half of the beetles have lengths within $1.9$ mm of the mean because the standard deviation covers the middle half.
4. The IQR is $22.1-13.5=8.6$ mm because IQR is the distance from minimum to maximum.
5. Because the median is $18.0$ mm, exactly half of the beetles are longer than $19.4$ mm.

Explanation: This question evaluates understanding summary statistics for beetle lengths, particularly the meaning of quartiles. The five-number summary indicates Q1 = 17.0 mm and Q3 = 19.4 mm, so the middle 50% of lengths are between these values. Choice B is correct in this interpretation. A distractor is choice A, which overstates 'strongly' right-skewed based on a slight mean (18.2 mm) > median (18.0 mm) difference, but the distribution is fairly symmetric with balanced tails. Mini-lesson: Quartiles divide data into quarters; the IQR (2.4 mm here) is Q3 - Q1, not the full range (8.6 mm). Use mean vs. median for shape: small differences suggest near symmetry, and standard deviation (1.9 mm) measures spread, not the middle 50%.

Question 17

1. The IQR of $23$ means households spend between $0$ and $23$ dollars per month.
2. The distribution is likely right-skewed because the mean is greater than the median and the upper tail extends far beyond $Q_3$. (correct answer)
3. A typical household spends $18.6$ dollars per month because the standard deviation is a typical spending amount.
4. Because the median is $22$, the range must be $22$ dollars.
5. Since $Q_1=12$, about 12% of households spend $12$ dollars or less.

Explanation: Interpreting summaries for spending involves skewness from centers and tails. Mean 27.4 > median 22.0, with upper tail (max - Q3 = 60) longer than lower (Q1 - min = 12), indicates right skewness, correctly in choice B. Choice E distracts, saying about 12% spend ≤ Q1 (12), but it's 25%. Mini-lesson: Right skew when mean > median due to high values; five-number summary shows asymmetry. IQR (23) is robust spread of middle 50%, unlike standard deviation (18.6) affected by extremes.

Question 18

1. Because the SD is $9.6$, most students took about $9.6$ minutes to finish.
2. The typical completion time is about $40$ minutes, and the distribution is likely right-skewed since the mean is greater than the median and the upper tail extends farther. (correct answer)
3. The middle 50% of times span about $78-22=56$ minutes, so the IQR is $56$ minutes.
4. The median must be larger than the mean because the maximum is much larger than the minimum.
5. About half of the students finished between $22$ and $35$ minutes because $Q_1=35$.

Explanation: This question assesses understanding of summary statistics for a quantitative variable, focusing on center, spread, and shape in AP Statistics. The correct interpretation is choice B, which accurately states that the typical completion time is about 40 minutes (the median) and infers a right-skewed distribution since the mean (42.8) exceeds the median (40.0), with the upper tail extending farther (from 47 to 78) compared to the lower tail (from 22 to 35). A common distractor is choice C, which mistakenly calculates the IQR as 78-22=56 instead of the correct Q3-Q1=47-35=12, confusing range with interquartile range. In interpreting summaries, remember that for skewed distributions, the median better represents the center than the mean, as the mean is pulled toward the tail. The standard deviation (9.6) measures variability around the mean, but doesn't indicate typical values directly. Shape can be inferred by comparing mean and median, and examining whisker lengths in the five-number summary. Overall, these statistics paint a picture of a right-skewed distribution with moderate spread.

Question 19

1. The middle 50% of pumpkin weights are between $6.0$ lb and $44.0$ lb.
2. The IQR is $44.0-6.0=38.0$ lb, so half the pumpkins vary by about 38 lb.
3. The distribution is likely right-skewed, with a typical weight near $16.9$ lb, because the mean exceeds the median and the upper tail is much longer. (correct answer)
4. Since the SD is $7.4$ lb, most pumpkins weigh about $7.4$ lb.
5. Because $Q_3=21.0$ lb, only 25% of pumpkins weigh less than 21.0 lb.

Explanation: This AP Statistics item focuses on interpreting summary statistics for a quantitative variable, covering shape and center. Choice C correctly identifies right-skewness with typical weight near 16.9 lb (median), as mean (18.6) > median and upper tail (21.0 to 44.0) is longer than lower (6.0 to 13.2). Distractor D misinterprets SD (7.4) as meaning most pumpkins weigh 7.4 lb, confusing variability with center. Key lesson: in skewed distributions, median represents center better; mean is influenced by outliers. Five-number summary reveals shape via tail lengths. IQR (21.0-13.2=7.8) shows central spread. This indicates right-skewed weights with heavier outliers.

Question 20

1. The distribution is likely right-skewed because the maximum is far above $Q_3$ and the mean is slightly greater than the median. (correct answer)
2. The IQR equals $165-70=95$ minutes, so the middle 50% of movie lengths span 95 minutes.
3. A typical movie length is $18.5$ minutes because that is the SD.
4. Since the median is 101 minutes, about 75% of movies are shorter than 92 minutes.
5. Because $Q_1=92$ and $Q_3=112$, exactly half of the movies are shorter than 92 minutes.

Explanation: Assessing AP Statistics skills in summary statistics for quantitative variables, this question highlights shape inference. Choice A is right, noting right-skewness due to mean (104) slightly > median (101) and long upper tail (112 to 165) versus lower (70 to 92). A distractor, choice B, wrongly uses range (165-70=95) for IQR instead of 112-92=20. Mini-lesson: Compare mean-median for skew direction; examine five-number summary for tail asymmetry. SD (18.5) measures spread from mean, not typical length. Median is preferred center for skew. Overall, mild right-skew with moderate variability.