A teacher defines X as the number of questions a student answers correctly on a 10-question quiz selected at random from a large bank. Over many quizzes, the distribution of X has mean μX=7.1 and standard deviation σX=1.3. Which interpretation of the standard deviation is correct?
Answer: A typical quiz score is about 1.3 questions away from 7.1.
This question tests interpretation of standard deviation for quiz scores. The standard deviation σ_X = 1.3 measures how much a typical quiz score varies from the mean of 7.1 questions correct. Choice C correctly states that "a typical quiz score is about 1.3 questions away from 7.1." This properly interprets standard deviation as measuring typical deviation from the mean. Choice A wrongly creates rigid boundaries (5.8 to 8.4), choice B confuses standard deviation with range, choice D incorrectly treats 1.3 as an exact score, and choice E wrongly relates standard deviation to the mean value. Standard deviation helps us understand how consistent or variable the student's performance is across different quizzes.
In a game, a player earns a random number of tokens X each round (a discrete random variable). Over many rounds, the distribution of X has mean μX=7.4 tokens and standard deviation σX=2.1 tokens. Which interpretation of the mean is correct?
Answer: In the long run, the player's average number of tokens per round will be about 7.4.
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.
A movie theater tracks X, the number of minutes of previews shown before a randomly selected showing. The distribution of X has mean μX=12.6 minutes and standard deviation σX=3.0 minutes. Which interpretation of the mean is correct?
Answer: In the long run, the average preview time per showing is about 12.6 minutes.
This question examines understanding of the mean for movie preview times. The mean μ_X = 12.6 minutes represents the long-run average preview time across many showings. Choice B correctly states that "in the long run, the average preview time per showing is about 12.6 minutes." This properly interprets the mean as an expected value over many observations. Choice A wrongly suggests exactly 12.6 minutes in most showings, choice C incorrectly creates rigid boundaries (9.6 to 15.6 minutes), choice D misinterprets standard deviation as a maximum duration, and choice E incorrectly claims 13 minutes is impossible. The mean tells us what to expect on average, while individual showings may have various preview lengths.
A streaming service looks at how many episodes a randomly selected viewer watches in one sitting. Let X be this number, a discrete random variable with μX=2.6 episodes and σX=1.4 episodes. Which interpretation of the mean is correct?
Answer: In repeated sittings across many viewers, the average number of episodes watched will be about 2.6.
This question asks about interpreting the mean of a discrete random variable. The mean μ = 2.6 episodes represents the expected value or long-run average number of episodes watched per sitting. Choice B correctly states that across many viewers and sittings, the average will be about 2.6. Choice A incorrectly suggests viewers watch exactly 2.6 episodes (impossible since episodes are discrete). Choice C confuses the mean with defining a range. Choice D describes the standard deviation, not the mean. Choice E makes an unfounded claim about specific values. The mean is the theoretical average that emerges when we observe the random variable many times.
A call center records the number of calls an operator handles in an hour. Let X be the number of calls for a randomly selected operator-hour; X is discrete. The distribution of X has mean μX=22 calls and standard deviation σX=5 calls. Which interpretation of the mean is correct?
Answer: In the long run, the average number of calls handled per operator-hour is about 22.
This question assesses mean interpretation for a discrete random variable in AP Statistics. The mean μ_X = 22 is the long-run average calls handled per operator-hour, correctly in choice C. Over numerous hours, the average would be about 22. Distractor A confuses mean with mode, suggesting 22 is most common, but mean is an average, not necessarily the most frequent. Choice D misuses SD to set hard limits of 17 to 27, whereas SD measures typical deviation. In a mini-lesson, the mean is the expected value E(X) = Σ x P(x) for discrete variables, representing the center of mass and long-term average outcome.
A parking garage counts the number of cars entering between 8:00 and 9:00 a.m. each weekday. Let X be the number of cars on a randomly selected weekday; X is discrete. The distribution of X has mean μX=120 cars and standard deviation σX=15 cars. Which interpretation of the standard deviation is correct?
Answer: The number of cars is typically about 15 cars away from 120.
This question focuses on standard deviation interpretation for a discrete random variable in AP Statistics. The σ_X = 15 indicates the number of cars typically deviates from 120 by about 15, as in choice A. This captures the average spread from the mean. Distractor C incorrectly asserts a strict range of 105 to 135, but SD doesn't guarantee all values within mean ± SD. Choice B treats SD as an exact value like a mode, which it's not. In a mini-lesson, SD is the square root of variance, where variance averages the squared deviations from the mean, providing a unit-matched measure of variability in the distribution.
A manufacturing plant defines X as the number of defective items found in a random sample of 200 items from one day's production. Over many days, the distribution of X has mean μX=4.8 defects and standard deviation σX=1.6 defects. Which interpretation of the mean is correct?
Answer: In the long run, the average number of defects per 200-item sample is about 4.8.
This question tests understanding of the mean in quality control. The mean μ_X = 4.8 defects represents the long-run average number of defects found in 200-item samples across many days. Option A correctly states this interpretation - it's about the average over many samples, not any single sample. Option B incorrectly suggests exactly 4.8 defects occur each time (impossible since defects must be whole numbers), while option D wrongly claims the mean sets a lower bound. The mean of 4.8 tells us what to expect on average when we examine many such samples over time.
A discrete random variable K represents the number of students absent from a particular class on a randomly chosen day. The distribution of K has mean μK=1.1 and standard deviation σK=1.3. Which interpretation of the standard deviation is correct?
Answer: A typical day's number absent differs from 1.1 by about 1.3 students.
This question focuses on interpreting standard deviation in context. The standard deviation σ = 1.3 measures the typical deviation of daily absences from the mean μ = 1.1. Choice C correctly states "a typical day's number absent differs from 1.1 by about 1.3 students." This properly interprets standard deviation as a measure of typical variability around the mean. Choice A incorrectly treats the standard deviation as an upper bound for the data. Note that the standard deviation (1.3) is actually larger than the mean (1.1), which is common for discrete distributions with small means. This indicates high variability relative to the average.
In a large school, let X be the number of minutes a randomly selected student spends waiting in the cafeteria line on a randomly selected day. The distribution of X has mean μX=6.8 minutes and standard deviation σX=2.4 minutes. Which interpretation of the mean is correct?
Answer: In the long run, the average waiting time is about 6.8 minutes.
This question tests understanding of the mean in a continuous context (waiting time). The mean μ_X = 6.8 minutes represents the long-run average waiting time across many students and many days. Option C correctly states this interpretation - it's about the average over many observations, not any specific instance. Option A incorrectly suggests every student waits exactly this amount, while option D wrongly claims the mean is the median (50th percentile). The mean is the balance point of the distribution and represents what we expect on average across many randomly selected student-days.
A commuter records X, the number of red lights encountered on the drive to work. The distribution of X has mean μX=5.8 and standard deviation σX=1.9. Which interpretation of the mean is correct?
Answer: In the long run, the commuter's average number of red lights per drive is about 5.8.
This question assesses understanding of the mean for red lights encountered while commuting. The mean μ_X = 5.8 represents the long-run average number of red lights per drive. Choice B correctly interprets this as "in the long run, the commuter's average number of red lights per drive is about 5.8." This properly treats the mean as an expected value over many drives. Choice A wrongly suggests exactly 5.8 red lights on most drives, choice C incorrectly creates rigid boundaries (3.9 to 7.7), choice D misinterprets the standard deviation as a maximum, and choice E incorrectly claims 6 red lights are impossible. Since red lights are discrete, individual drives will have whole numbers of red lights, but the average over many drives can be 5.8.
A website tracks the number of items a customer places in their cart per visit. Let X be the number of items for a randomly selected visit; X is discrete. The distribution of X has mean μX=4.0 items and standard deviation σX=2.5 items. Which interpretation of the standard deviation is correct?
Answer: The number of items in the cart is typically about 2.5 items away from 4.0.
This question evaluates standard deviation for a discrete random variable in AP Statistics. The σX=2.5 means the number of items typically is about 2.5 away from the mean 4.0, correctly in choice B. This reflects typical deviation magnitude. Distractor C suggests a fixed range of 1.5 to 6.5, but SD doesn't impose such limits. Choice A confuses SD with the mode or exact value. In a mini-lesson, calculate SD as ∑(x−μ)2P(x), which quantifies spread and helps compare variability across similar distributions.
At a clinic, let X be the number of phone calls received between 9:00 and 10:00 AM on a randomly selected weekday. The distribution of X has mean μX=31.4 calls and standard deviation σX=5.6 calls. Which interpretation of the standard deviation is correct?
Answer: About 5.6 calls is the typical amount by which the number of calls in an hour differs from the mean of 31.4 calls.
This question asks about interpreting standard deviation in a healthcare setting. The standard deviation σ_X = 5.6 calls measures the typical variability in hourly call volume. Option A correctly identifies this as the typical amount by which the number of calls differs from the mean of 31.4. Option B incorrectly interprets the parameters as defining a range, while option C confuses the standard deviation with the mean. The standard deviation tells us about the consistency of call volume - on a typical hour, the actual number of calls will be about 5.6 calls away from the average of 31.4.
A discrete random variable T is the number of tardy students in a school's first-period class on a randomly chosen day. The distribution of T has mean μT=2.0 and standard deviation σT=0.6. Which interpretation of the mean is correct?
Answer: Over many days, the average number of tardy students is about 2.0.
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.
A student records the random variable X, the number of text messages received in an hour. From a long period of observation, the distribution of X has mean μX=3.2 and standard deviation σX=1.6. Which interpretation of the mean is correct?
Answer: In the long run, the average number of texts received per hour will be about 3.2.
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.
Let X be the discrete random variable representing the number of text messages a student receives during a 50-minute class period. For this student, μX=12.0 and σX=4.0. Which interpretation of the standard deviation is correct?
Answer: In the long run, the number of messages typically differs from 12 by about 4 messages.
This question focuses on interpreting the standard deviation of a discrete random variable. The standard deviation σ_X = 4.0 measures the typical deviation from the mean of 12 messages. Option A incorrectly describes a deterministic outcome. Option B wrongly suggests the range is always μ ± σ, which isn't guaranteed. Option D misinterprets the mean as a certain outcome. Option E confuses the standard deviation with the mode or most frequent value. The correct interpretation is that in the long run, the number of messages typically varies from the mean of 12 by about 4 messages, measuring the spread of the distribution.
Let X be the discrete random variable representing the number of heads in 10 flips of a particular coin. Suppose the distribution of X has mean μX=5.6 and standard deviation σX=1.5. Which interpretation of the standard deviation is correct?
Answer: In repeated sets of 10 flips, the number of heads typically differs from 5.6 by about 1.5 heads.
This question examines standard deviation interpretation for coin flips. The standard deviation σ_X = 1.5 measures the typical variation in the number of heads from the mean of 5.6. Option A describes an impossible exact fractional outcome. Option B incorrectly treats μ ± σ as absolute bounds for all outcomes. Option D confuses the standard deviation value with a probability or outcome. Option E misinterprets standard deviation as the range. The correct interpretation is that in repeated sets of 10 flips, the number of heads typically varies from the mean of 5.6 by about 1.5 heads, measuring the spread of outcomes.
Let X be the discrete random variable representing the number of days in a week that a person exercises. In a large group, the distribution of X has mean μX=3.9 days and standard deviation σX=1.6 days. Which interpretation of the mean is correct?
Answer: In the long run, the average number of exercise days per week in this group is about 3.9.
This question tests understanding of the mean for a discrete distribution. The mean μ_X = 3.9 represents the average number of exercise days per week across the entire group. Option A incorrectly suggests each individual exercises exactly 3.9 days. Option C confuses the mean with the mode (most frequent value). Option D misinterprets the standard deviation as defining absolute bounds for all individuals. Option E incorrectly relates the mean to a specific probability. The correct interpretation is that when averaging across all people in the group, the mean number of exercise days per week is approximately 3.9.
A discrete random variable N represents the number of times a website user refreshes a page during a 5-minute visit. The distribution of N has mean μN=2.6 and standard deviation σN=2.0. Which interpretation of the mean is correct?
Answer: Over many 5-minute visits, the average number of refreshes is about 2.6.
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.
A bookstore tracks X, the number of items purchased in a randomly selected transaction. The distribution of X has mean μX=2.4 items and standard deviation σX=1.1 items. Which interpretation of the standard deviation is correct?
Answer: A typical transaction differs from 2.4 items by about 1.1 items.
This question tests interpretation of standard deviation for bookstore transactions. The standard deviation σ_X = 1.1 items measures how much a typical transaction differs from the mean of 2.4 items. Choice B correctly states that "a typical transaction differs from 2.4 items by about 1.1 items." This properly captures standard deviation as a measure of typical variability around the mean. Choice A incorrectly treats 1.1 as an exact transaction amount, choice C wrongly limits possible values to just two numbers, choice D confuses standard deviation with range, and choice E incorrectly relates standard deviation to the mean value. Standard deviation quantifies the spread of transaction sizes around their average.
A delivery company records X, the number of packages delivered by a driver in one day. For a randomly selected driver-day, the distribution of X has mean μX=52.0 packages and standard deviation σX=8.5 packages. Which interpretation of the standard deviation is correct?
Answer: About 8.5 packages is the typical distance between a day's package count and the mean of 52.0 packages.
This question asks about interpreting the standard deviation in context. The standard deviation σ_X = 8.5 packages measures the typical variability in the number of packages delivered. Option B correctly identifies this as the typical distance between a day's package count and the mean of 52.0 packages. Option A incorrectly confuses standard deviation with range, while option C misinterprets it as a typical value rather than a measure of spread. The standard deviation quantifies how much individual observations typically deviate from the mean, giving us a sense of the variability in the data.
