AP Statistics Quiz: Mean And Standard Deviation

What this quiz covers

This quiz focuses on Mean And Standard Deviation, 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 student will score exactly 8.9 correct on most quizzes.
2. The standard deviation 1.4 means the score is usually within about 1.4 questions of 8.9 correct. (correct answer)
3. The standard deviation 1.4 is the difference between the highest and lowest possible quiz scores.
4. The mean number correct is 1.4.
5. The score must always be between 7.5 and 10.3 because $8.9\pm1.4$.

Explanation: This question focuses on interpreting standard deviation in a quiz score context. The standard deviation σ = 1.4 measures the typical deviation of quiz scores from the mean μ = 8.9. Choice B correctly states "the standard deviation 1.4 means the score is usually within about 1.4 questions of 8.9 correct." This properly interprets standard deviation as describing typical variability around the mean. Choice E incorrectly treats μ ± σ as absolute bounds (7.5 to 10.3), when scores can fall outside this range. Standard deviation describes typical variation, not the range of all possible scores. For a 12-question quiz, scores could range from 0 to 12, well beyond one standard deviation from the mean.

Question 2

1. Most students are absent exactly 2 days.
2. The number of absences is usually between 3 and 7 days because $\sigma_X=2$.
3. A typical student's number of absences differs from 5 days by about 2 days. (correct answer)
4. The standard deviation 2 means the range of absences is 2 days.
5. Exactly half of students have 5 or fewer absences.

Explanation: This question evaluates understanding the standard deviation of a discrete random variable in AP Statistics. The SD σ_X = 2 means a typical student's absences deviate from the mean of 5 by about 2 days, as accurately stated in choice C. This reflects the average distance from the mean in terms of root-mean-square. Distractor B incorrectly suggests SD defines a usual range like mean ± SD/2 or something similar, but it doesn't specify bounds that way. Choice A confuses SD with the mode, implying most have exactly 2 absences, which isn't what SD measures. In a mini-lesson, standard deviation is sqrt(Var(X)), where Var(X) = E[(X - μ)^2], providing a scale for how spread out the values are around the mean for discrete distributions.

Question 3

1. A typical student's number of books checked out differs from 3.2 by about 1.5 books. (correct answer)
2. Exactly 1.5 books are checked out by most students.
3. The number of books checked out ranges from 1.7 to 4.7 because the standard deviation is 1.5.
4. About 1.5% of students check out 3.2 books.
5. The standard deviation 1.5 means the maximum number of books any student can check out is 1.5.

Explanation: This question assesses the interpretation of the standard deviation for a discrete random variable in AP Statistics. The standard deviation σ_X = 1.5 indicates that the number of books checked out typically deviates from the mean of 3.2 by about 1.5 books, as correctly captured in choice A. This means, on average, the root-mean-square deviation from the mean is 1.5, providing a sense of typical variability. A distractor like choice C incorrectly assumes the range is strictly mean ± SD, but standard deviation doesn't guarantee all values fall within that interval, especially without knowing the distribution shape. Choice B is wrong because it treats SD as a mode or exact value rather than a measure of spread. In a mini-lesson, standard deviation quantifies variability: for discrete X, it's the square root of the variance, where variance is the expected value of (X - μ)^2, helping understand how much values fluctuate around the mean.

Question 4

1. Most shifts result in exactly 14 orders.
2. In the long run, drivers complete about 14 orders per 4-hour shift on average. (correct answer)
3. Drivers complete between 10 and 18 orders in every shift because $\sigma_X=4$.
4. The probability a driver completes 14 orders is 0.14.
5. Half of shifts have fewer than 14 orders and half have more than 14 orders.

Explanation: This question focuses on interpreting the mean of a discrete random variable in AP Statistics. The mean μ_X = 14 signifies the long-run average number of orders completed per 4-hour shift, correctly interpreted in choice B. Over many shifts, the average would stabilize around 14 orders. Choice E is a distractor that mistakes the mean for the median, which isn't necessarily true unless the distribution is symmetric. Choice C wrongly uses the standard deviation to imply a fixed range of 10 to 18, but SD measures average deviation, not absolute limits. In a mini-lesson, the mean E(X) for a discrete random variable is the sum of x * P(X=x) over all possible x, representing the center of the probability distribution and the expected outcome in repeated trials.

Question 5

1. Over many days, the average number of tardy students is about 2.0. (correct answer)
2. Exactly 2 students are tardy on a typical day.
3. The number tardy cannot be more than 2.0.
4. The number tardy is always between 1.4 and 2.6 because $\sigma_T=0.6$.
5. Half the days have 2.0 or fewer tardy students.

Explanation: This question tests the interpretation of mean for a discrete random variable. The mean μ = 2.0 represents the expected or average number of tardy students over many days. Choice A correctly states "over many days, the average number of tardy students is about 2.0." This is the fundamental interpretation of expected value - it's the long-run average. Choice B incorrectly suggests exactly 2 students are tardy on a typical day, confusing the mean with the mode. The mean tells us about the center of the distribution over many observations, not necessarily what happens most frequently. For discrete distributions, the mean often isn't even a possible value of the variable.

Question 6

1. In the long run, the average number of skipped songs is 3.0.
2. A typical session's number of skipped songs is about 3.0 away from the mean of 7.5. (correct answer)
3. The number of skipped songs is always between 4.5 and 10.5.
4. Most sessions have exactly 7.5 skipped songs.
5. The standard deviation 3.0 is the maximum possible number of skipped songs.

Explanation: This question assesses understanding of standard deviation interpretation. The standard deviation σ = 3.0 measures how much the number of skipped songs typically varies from the mean μ = 7.5. Choice B correctly states "a typical session's number of skipped songs is about 3.0 away from the mean of 7.5." This properly captures that standard deviation measures typical distance from the mean. Choice C incorrectly treats μ ± σ as absolute bounds (4.5 to 10.5), when values can certainly fall outside this range. Standard deviation describes typical variability, not the range of all possible values. Understanding this distinction is crucial for proper statistical interpretation.

Question 7

1. Over many 10-game stretches, the player averages about 6.7 games with at least one 3-point shot. (correct answer)
2. In exactly 6.7 of the 10 games, the player will make at least one 3-point shot.
3. The player's number of such games is always within 1.5 of 6.7.
4. Half of all 10-game stretches have 6.7 or fewer such games.
5. The standard deviation is 6.7 because that is the typical number of such games.

Explanation: This question tests understanding of the mean in a binomial-like context. The mean μ = 6.7 represents the expected number of games with at least one 3-point shot over many 10-game stretches. Choice A correctly states "over many 10-game stretches, the player averages about 6.7 games with at least one 3-point shot." This is the proper interpretation of expected value - it's the long-run average. Choice B incorrectly suggests exactly 6.7 games, which is impossible since the number of games must be a whole number. The mean represents what happens on average across many repetitions of the 10-game experiment, not what happens in any single stretch.

Question 8

1. In the long run, the average number of texts per class period is 2.1.
2. The number of texts is always between $3.4-2.1$ and $3.4+2.1$.
3. A typical class period's number of texts differs from 3.4 by about 2.1 texts. (correct answer)
4. Most class periods have exactly 3.4 texts.
5. The standard deviation 2.1 is the difference between the maximum and minimum possible number of texts.

Explanation: This question focuses on interpreting the standard deviation of a discrete random variable. The standard deviation σ = 2.1 measures the typical deviation from the mean μ = 3.4. Choice C correctly states that "a typical class period's number of texts differs from 3.4 by about 2.1 texts." This captures the essence of standard deviation - it quantifies how spread out the data is from the mean. Choice B incorrectly claims all values fall within one standard deviation of the mean, which is false. Standard deviation describes typical variability, not absolute bounds. Understanding that standard deviation measures typical distance from the mean is crucial for statistical interpretation.

Question 9

1. The company will receive exactly 1.8 complaints on most days.
2. The number of complaints per day is typically between 0.6 and 3.0 because the standard deviation is 1.2.
3. In the long run, the average number of complaints per day is about 1.8. (correct answer)
4. Half of all days have 1.8 complaints or fewer.
5. The maximum possible number of complaints in a day is 1.8.

Explanation: This question tests understanding of the mean of a discrete random variable. The mean μ = 1.8 represents the long-run average or expected value of the number of complaints per day. Choice C correctly states that "in the long run, the average number of complaints per day is about 1.8." This is the fundamental interpretation of expected value - it's what we expect on average over many observations. Choice A incorrectly suggests the company receives exactly 1.8 complaints most days, which is impossible since complaints must be whole numbers. The mean tells us about the center of the distribution over many days, not what happens on any single day.

Question 10

1. The user refreshes the page exactly 2.6 times in a typical 5-minute visit.
2. Over many 5-minute visits, the average number of refreshes is about 2.6. (correct answer)
3. The number of refreshes is usually between 0.6 and 4.6 because the standard deviation is 2.0.
4. The maximum possible number of refreshes in 5 minutes is 2.6.
5. Half of all visits have 2.6 or fewer refreshes.

Explanation: This question tests the interpretation of mean for a discrete random variable. The mean μ = 2.6 represents the expected or average number of page refreshes over many 5-minute visits. Choice B correctly states "over many 5-minute visits, the average number of refreshes is about 2.6." This is the fundamental meaning of expected value - it's what we expect on average in the long run. Choice A incorrectly suggests a typical visit has exactly 2.6 refreshes, which is impossible since refreshes must be whole numbers. The mean is a theoretical average that emerges over many observations, not a value that occurs frequently in practice.

Question 11

1. A typical street segment has exactly 4.1 potholes reported in a month.
2. The typical distance between a segment's pothole count and the mean is 4.1 potholes.
3. In the long run, the average number of potholes reported per street segment per month will be about 4.1. (correct answer)
4. Most street segments have between $4.1-2.3$ and $4.1+2.3$ potholes reported.
5. The mean of 4.1 is the smallest possible number of potholes that can be reported.

Explanation: This question evaluates the correct interpretation of the mean for a discrete distribution of potholes reported per street segment. The mean μ_X = 4.1 potholes means that, in the long run, the average number reported per segment per month would be about 4.1. Choice C accurately captures this long-term average. Distractors like Choice D misinterpret SD as defining 'most' values in a simple range, which isn't accurate for all distributions. Mini-lesson: The mean is the expected value over infinite trials and is central for budgeting repairs in city planning. It's not an exact count (Choice A), deviation (Choice B), or minimum (Choice E). Discrete distributions often have non-integer means, reflecting probabilistic averages.

Question 12

1. In repeated samples of 20 items, the average number of defectives is about 1.1. (correct answer)
2. Every sample of 20 items will contain either 1 or 2 defectives because $\mu_X=1.1$.
3. The probability of getting exactly 1 defective is 1.1.
4. Most samples will have 1.1 defectives, since the mean is the most common count.
5. The mean 1.1 means the number of defectives cannot be 0.

Explanation: This question tests understanding of the mean in a quality control context. The mean μ_X = 1.1 represents the expected or average number of defective items in samples of 20. Option B incorrectly suggests the outcome must be 1 or 2, ignoring that 0, 3, or more defectives are possible. Option C confuses the mean with a probability. Option D wrongly claims the mean is the most common count (mode). Option E incorrectly states that a positive mean eliminates the possibility of zero defectives. The correct interpretation is that over many samples of 20 items, the average number of defectives will be approximately 1.1.

Question 13

1. The number of items purchased is usually between 1.1 and 2.6 items.
2. The typical distance between a visit's number of items and the mean number of items is about 1.1 items. (correct answer)
3. The standard deviation means the maximum possible number of items is about 1.1.
4. About 1.1 items is the average number of items purchased per visit.
5. Half the visits have purchases within 1.1 items of 2.6 items.

Explanation: This question focuses on interpreting the standard deviation in a discrete distribution of items purchased from a vending machine. The standard deviation σ_X = 1.1 items measures the typical deviation or spread of the number of items from the mean of 2.6, meaning purchases usually vary by about 1.1 items around the average. Choice B accurately describes this as the typical distance from the mean. A distractor like Choice A incorrectly uses the SD as a range boundary without reference to the mean, while Choice D confuses SD with the mean itself. Mini-lesson: Standard deviation quantifies variability; a smaller SD like 1.1 indicates less spread, whereas a larger one shows more variation. It's the square root of the variance and isn't about maxima (as in Choice C) or medians (Choice E). For discrete distributions, SD helps understand consistency in counts like purchases.

Question 14

1. Most players use exactly 1.30 hints.
2. In repeated play across many players, the average number of hints used will be about 1.30. (correct answer)
3. About 1.30 hints is the typical distance from the mean.
4. A mean of 1.30 hints means players use between 0.13 and 1.30 hints.
5. A mean of 1.30 hints means the range of hints is 1.30.

Explanation: This question tests understanding of the mean for a discrete probability distribution. The mean μ = 1.30 hints represents the expected value or long-run average number of hints used per player. Choice B correctly states that across many players, the average number of hints will be about 1.30. Choice A incorrectly claims most players use exactly 1.30 hints (impossible for discrete data). Choice C describes standard deviation, not the mean. Choice D incorrectly creates a range. Choice E incorrectly equates the mean with the range. The mean is the weighted average of all possible values, representing what we expect on average over many observations.

Question 15

1. A typical box's number of defective bulbs is about 0.8.
2. In the long run, the number of defective bulbs per box will be exactly 0.9.
3. The number of defective bulbs varies from box to box by about 0.8 around the mean of 0.9. (correct answer)
4. The number of defective bulbs must be between 0.1 and 1.7 because $0.9\pm0.8$.
5. The standard deviation 0.8 is the largest possible number of defective bulbs.

Explanation: This question assesses understanding of standard deviation interpretation. The standard deviation σ = 0.8 measures how much the number of defective bulbs typically varies from the mean μ = 0.9. Choice C correctly states "the number of defective bulbs varies from box to box by about 0.8 around the mean of 0.9." This captures both aspects: variability between boxes and the typical magnitude of that variation. Choice D incorrectly treats μ ± σ as absolute bounds, when in reality values can fall outside this range. Standard deviation describes typical variability, not the range of all possible values. For discrete distributions, especially those that can take value 0, values often extend beyond one standard deviation from the mean.

Question 16

1. About 14.2 customers enter in every 10-minute interval from 8:00 to 8:10.
2. In the long run, the average number of customers entering from 8:00 to 8:10 is about 14.2. (correct answer)
3. About 14.2 is the most common number of customers entering from 8:00 to 8:10.
4. Most days, the number of customers entering is between $14.2-3.1$ and $14.2+3.1$.
5. The number of customers entering cannot be less than 14.2 because 14.2 is the mean.

Explanation: This question tests understanding of what the mean represents in a statistical distribution. The mean μ_X = 14.2 represents the long-run average number of customers entering during the 8:00-8:10 time period across many weekdays. Option B correctly states this interpretation - it's not about any single day, but rather the average over many observations. Option A incorrectly suggests this happens every interval, while option C confuses the mean with the mode (most common value). The mean tells us the center of the distribution when we average across many days, not what happens on any particular day.

Question 17

1. About 0.7 defectives is the typical distance between a box's defective count and 0.9 defectives. (correct answer)
2. About 0.7 defectives means most boxes have between 0.7 and 1.6 defectives.
3. About 0.7 defectives is the probability a box has 1 defective.
4. About 0.7 defectives is the range of defective counts across boxes.
5. About 0.7 defectives means the average number of defectives is 0.7.

Explanation: This question asks about interpreting standard deviation. The standard deviation σ = 0.7 defectives measures the typical distance between a box's actual defective count and the mean of 0.9 defectives. Choice A correctly identifies this interpretation - it's the typical distance from the mean. Choice B incorrectly creates a range by adding the standard deviation to itself. Choice C confuses standard deviation with probability. Choice D incorrectly defines it as the range. Choice E confuses standard deviation with the mean. Standard deviation quantifies variability, telling us how spread out the defective counts typically are around the mean value.

Question 18

1. The player will earn exactly 7.4 tokens in most rounds.
2. In the long run, the player's average number of tokens per round will be about 7.4. (correct answer)
3. The player's tokens per round will usually be between 5.3 and 9.5 because the standard deviation is 2.1.
4. The mean of 7.4 implies the maximum possible tokens in a round is 7.4.
5. Because the standard deviation is 2.1, the range of possible tokens per round is 2.1.

Explanation: This question tests understanding of the mean of a discrete random variable. The mean μ_X = 7.4 represents the long-run average or expected value of tokens earned per round, not a value that occurs in any single round. Choice B correctly states that "in the long run, the player's average number of tokens per round will be about 7.4." Choice A incorrectly suggests the player earns exactly 7.4 tokens in most rounds, which is impossible since tokens are discrete (whole numbers). Choice C misinterprets the standard deviation, and choices D and E make false claims about maximum values and ranges. The mean is a measure of center that tells us what to expect on average over many repetitions of the random process.

Question 19

1. The driver delivers exactly 8 packages on most days.
2. The driver delivers 52 packages every day.
3. A typical day's package count is about 8 packages away from 52. (correct answer)
4. The number of packages delivered must be between 44 and 60.
5. The range of possible daily package counts is 8.

Explanation: This question tests interpretation of standard deviation in a delivery context. The standard deviation σ_X = 8 packages measures how much the daily package count typically varies from the mean of 52 packages. Choice C correctly states that "a typical day's package count is about 8 packages away from 52." This properly interprets standard deviation as a measure of typical deviation from the mean. Choice A incorrectly treats 8 as an exact count, choice B misunderstands the mean, choice D wrongly creates rigid boundaries (44 to 60), and choice E confuses standard deviation with range. Standard deviation helps us understand the consistency or variability in the number of packages delivered daily.

Question 20

1. The student will receive exactly 3.2 texts every hour.
2. In the long run, the average number of texts received per hour will be about 3.2. (correct answer)
3. Most hours will have between 1.6 and 3.2 texts because the standard deviation is 1.6.
4. The maximum possible number of texts in an hour is 3.2.
5. The range of texts per hour is 1.6.

Explanation: This question assesses understanding of the mean in the context of text messages received per hour. The mean μ_X = 3.2 represents the long-run average number of texts per hour, not a value that occurs in any specific hour. Choice B correctly interprets this as "in the long run, the average number of texts received per hour will be about 3.2." Choice A wrongly suggests 3.2 texts occur exactly in individual hours, choice C misuses the standard deviation to create boundaries, choice D incorrectly treats the mean as a maximum, and choice E confuses standard deviation with range. The mean tells us what to expect on average if we observe many hours and calculate the average number of texts.