Statistics
Practice Test 37 for Statistics: real questions and explanations from the Varsity Tutors practice-test pool.
0%
0 / 25 answered
Question 1 of 25
A manufacturer wants to estimate the mean lifetime (in hours) of all light bulbs produced on a particular day. From that day’s production, the quality team randomly selects 50 bulbs and measures their lifetimes. Which statement correctly describes the population and the sample?
Question Navigator
A manufacturer wants to estimate the mean lifetime (in hours) of all light bulbs produced on a particular day. From that day’s production, the quality team randomly selects 50 bulbs and measures their lifetimes. Which statement correctly describes the population and the sample?
Explanation: Statistics involves using samples to make inferences about populations. The population here is all light bulbs produced on that particular day - this is the complete group the manufacturer wants to understand. The sample consists of the 50 randomly selected bulbs that were actually tested for their lifetimes. The parameter being estimated is the mean lifetime of ALL bulbs from that day's production, while the sample statistic is the mean lifetime of just the 50 tested bulbs. Random sampling is crucial because it helps ensure the sample represents the population without systematic bias. Based on the sample mean, we can estimate the population mean, though the true value may differ somewhat. A common misconception is confusing which group is the population versus the sample - remember to ask 'Who do we want to know about?' (all bulbs from that day) versus 'Who did we measure?' (the 50 tested bulbs).
A histogram shows the distribution of package delivery times (in days) for 30 packages. Most packages are between 2 and 5 days, but there is a single package that took 14 days.
Histogram counts by bin: 2–3 days: 9 packages 3–4 days: 11 packages 4–5 days: 8 packages 5–6 days: 1 package 13–14 days: 1 package
Which statement is best supported by the histogram?
Explanation: This involves interpreting a histogram for shape, center, and outlier effects in delivery times. The histogram shows right-skewed shape, with tall bars at 2–5 days and a tail to 14 days. Typical time is around 3–4 days, using median for skew. The 14-day outlier increases mean more than median. Choice B is supported, as bars cluster left with isolated right bar, showing skew and mean pull. Error: calling symmetric from middle tall bar, ignoring tail. Examine tails in histogram first, then use median for center in skew.
A school counselor wants to estimate the average number of hours of sleep students at Central High get on school nights. From the student roster, the counselor uses a random number generator to select 120 students and emails them a short questionnaire asking, “How many hours did you sleep last night?” No changes are made to students’ routines. Which type of study is described, and what does the randomization (if any) allow the counselor to do?
Explanation: In statistics, we distinguish between types of studies—such as sample surveys, observational studies, and experiments—based on how data are collected, and randomization plays key roles in supporting generalizations or causal conclusions. Experiments are unique because researchers actively assign participants to receive different treatments or conditions. Random sampling involves selecting participants randomly from a larger population, which supports generalizing results from the sample to that population. Random assignment, on the other hand, randomly allocates participants to treatment groups, which helps establish cause-and-effect relationships by balancing out confounding factors. In this scenario, the counselor conducted a sample survey by randomly selecting students to ask about their sleep without imposing any changes, allowing generalization of the average sleep estimate to all Central High students but not causation. A common misconception is that random sampling alone enables causal claims, but it does not—it only aids generalization, unlike random assignment in experiments. To classify future studies, ask: 'Did researchers assign a treatment?' (no, so not an experiment) and 'Was the sample randomly selected?' (yes, supporting generalization).
A factory makes two types of phone chargers: Standard (80% of daily output) and Fast (20%). A quick screening test flags a charger as “likely defective.” The test correctly flags 90% of truly defective chargers (sensitivity 90%) and incorrectly flags 5% of non-defective chargers (false positive rate 5%). Historical defect rates differ by type: 1% of Standard chargers are defective, and 4% of Fast chargers are defective.
A shipment is flagged by the screening test, and the label shows it is a Standard charger. The quality manager must choose a policy for what to do next for flagged Standard chargers. The goal is to minimize unnecessary destructive teardown while still catching most truly defective units.
Which strategy is most reasonable given the probabilities?
Policy options:
Explanation: This question involves the skill of analyzing decisions using probability to choose the best policy for handling flagged Standard chargers. The decision goal is to minimize unnecessary destructive teardowns while still catching most truly defective units. The relevant probabilities show that for Standard chargers with a 1% defect rate, the screening test has a 90% sensitivity and 5% false positive rate, leading to a positive predictive value of about 15%, meaning most flagged units are not defective. Policy 2 best aligns with these probabilities by using a confirmatory test with 95% sensitivity and 1% false positive rate, which reduces unnecessary teardowns on false positives while catching nearly all true defects through the two-step process. In contrast, Policy 1 involves tearing down every flagged charger, which ignores the low positive predictive value and results in many unnecessary teardowns, a flaw known as base rate neglect. Remember that probabilities inform the likelihood of defects but do not guarantee any specific outcome for an individual charger. To apply this strategy elsewhere, first identify the decision goal, then compare the probabilities of true positives versus false positives across options to balance risks effectively.
A class has 10 students. Each student is equally likely to be selected at random.
The students are:
Let event A be “the selected student plays soccer” and event B be “the selected student plays a sport (soccer or basketball).”
Using the list above, what is P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question tests conditional probability from a model of equally likely student selections. 'Given B' means we restrict our attention to only the outcomes in event B, which is 'the selected student plays a sport.' There are 7 outcomes in B: the 4 soccer players and 3 basketball players. Of these 7, 4 are also in A, which are the soccer players. Therefore, the correct fraction 4/7 represents P(A|B), as it shows the proportion of sport-playing students who play soccer. A common mistake is using the total 10 students as the denominator, leading to 4/10 instead of focusing on B. To avoid this, first list or circle the outcomes in B, then count how many of those are in A.
In a factory, event A is “a randomly selected item is defective,” and event B is “the item comes from Machine 1.” Suppose P(B)=0.25 and P(A∩B)=0.02. Based on the information given, what is P(A∣B)? Give your answer as a fraction in simplest form.
Explanation: We need to find P(A|B), the probability that an item is defective given that it comes from Machine 1. This conditional probability means 'among items from Machine 1, the fraction that are defective.' Using the formula P(A|B) = P(A∩B)/P(B) = 0.02/0.25 = 2/25. This tells us that 2/25 of items from Machine 1 are defective. A common error would be to use P(A∩B) = 0.02 = 2/100 as the answer, but that's the joint probability among all items, not the conditional probability among Machine 1 items only. Remember: the given condition (Machine 1) becomes your denominator group, then you find what fraction of that specific group is defective.
A gym wants to estimate the mean number of minutes its members spend exercising per visit. It randomly selects 40 member visits from all visits in January and records the exercise minutes for each selected visit. Why is random sampling important in this situation?
Explanation: Statistics uses samples to make inferences about populations. Here, the population is all member visits to the gym in January, and the sample is the 40 randomly selected visits. The parameter being estimated is the mean exercise duration across ALL January visits. Random sampling is crucial because it helps ensure each visit has an equal chance of being selected, reducing selection bias. This makes the sample more likely to represent the variety of exercise patterns in the population - capturing both short and long workouts proportionally. Without random sampling, we might accidentally oversample certain times or types of members, leading to biased estimates. A common misconception is that random sampling proves causation or guarantees perfect accuracy - it doesn't. Random sampling simply helps us get unbiased estimates, not exact values or causal relationships.
A district wants to estimate the proportion of families who prefer email over paper for school announcements. From a list of all families in the district, they take a random sample of 400 families and send a questionnaire asking their preference. Which type of study is described?
Explanation: To classify studies, we must identify whether researchers assign treatments or simply collect information. A sample survey gathers data from a subset to estimate population characteristics without manipulating variables, while experiments assign treatments. Random sampling ensures each population member has a known selection probability, supporting generalization to the whole population. Random assignment allocates treatments in experiments to enable causal conclusions. Here, the district is estimating a population proportion (families preferring email) by asking a question—they're not assigning families to prefer one method over another. Using a random sample from the complete family list means results can generalize to all district families with known sampling error. This is a classic sample survey: no treatment assignment, just measurement of an existing preference using random sampling for valid inference. The common misconception that sending a questionnaire is a 'treatment' misunderstands that treatments must attempt to change outcomes, not merely measure them.
A gym uses the linear model y=12x+35 to predict total monthly cost. Here, x is the number of personal training sessions in a month (sessions) and y is the total monthly cost (dollars). What does the slope represent in this context? Include units in your interpretation.
Explanation: In this linear model, we're interpreting the slope of the equation y = 12x + 35, where x represents personal training sessions and y represents total monthly cost in dollars. The slope is 12, which means for each 1 additional training session, the total monthly cost increases by $12. This makes sense because the slope tells us the rate of change - specifically, how much y changes when x increases by 1 unit. The units are crucial here: the slope is 12 dollars per session, showing the cost increase for each additional session. The y-intercept of 35 represents the base monthly cost when x = 0 (no training sessions), likely a membership fee. A common mistake is flipping the interpretation, thinking the slope means sessions per dollar instead of dollars per session. To avoid this error, always label your axes clearly: x-axis is sessions, y-axis is dollars, so slope is dollars/session.
A warehouse is deciding how to screen incoming shipments for damaged items. Historically, 4% of items arrive damaged.
Two screening strategies are available:
Each flagged item is pulled for manual inspection, and the warehouse wants to minimize the chance that an item released without manual inspection is actually damaged, even if that means inspecting more items overall.
Which strategy is most reasonable given the probabilities?
Explanation: This problem involves analyzing decisions using probability to select the optimal screening strategy for damaged items. The decision goal is to minimize the probability that an item released without inspection is actually damaged. The key probabilities include a 4% base rate of damage, Strategy 1's 70% detection rate and 8% false positive rate, and Strategy 2's 90% detection rate and 18% false positive rate. Strategy 2 best aligns with the goal because its higher detection rate results in a lower conditional probability of damage given release (about 0.5% versus 1.34% for Strategy 1), prioritizing fewer missed damages over more inspections. One incorrect option is C, which commits the base rate fallacy by downplaying the detection difference due to rarity without calculating the actual impact on released items. Remember, these probabilities guide the decision to minimize risk but do not guarantee that no damaged items will ever be released. To transfer this strategy, identify the goal, then compare the conditional probabilities like P(damaged | released) that matter most.
A streaming service surveys users. Event A is “a user watches comedy,” and event B is “a user watches documentaries.” The survey finds P(A)=0.46, P(B)=0.38, and P(A∩B)=0.16. What is the probability that a user watches comedy or documentaries (inclusive)?
Explanation: This question tests the Addition Rule for finding the probability of watching comedy or documentaries (inclusive). Simply adding P(A) and P(B) would double-count users who watch both. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.46 + 0.38 - 0.16 = 0.68. This works because it includes all comedy watchers, all documentary watchers, but counts those who watch both only once. A common mistake is forgetting to subtract, which would give 0.84, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
A phone battery is draining at a nearly constant rate while playing music. The battery percentage is recorded below.
Time (hours): 0, 1, 2, 3, 4, 5 Battery (%): 100, 92, 84, 76, 68, 60
Which type of function best models the relationship between time and battery percentage?
Explanation: When modeling data, constant first differences suggest linear, constant ratios exponential, and curved shapes quadratic. The battery percentages show constant differences: 92-100=-8, 84-92=-8, 76-84=-8, 68-76=-8, 60-68=-8, all equal. This steady decline fits a constant drain rate while playing music. Linear is the best model for uniform depletion processes. A misconception is assuming exponential decay due to percentage context, but ratios aren't constant (e.g., 92/100=0.92, 84/92≈0.913, varying slightly). Always compare differences and ratios to avoid mixing linear with exponential decay.
A password is formed by arranging 4 different letters chosen from the 7 distinct letters {A, B, C, D, E, F, G}, with no repeats. Order matters because different sequences are different passwords. There are 7P4 equally likely passwords. What is the probability that the password starts with A?
Explanation: This problem requires counting the ways to form four-letter passwords from seven distinct letters (no repeats) to find the probability it starts with A. Since sequences matter, order matters—ABCD differs from ACBD. The total possible outcomes are the permutations of choosing and arranging 4 out of 7, which is 7P4 = 7 × 6 × 5 × 4. The favorable outcomes fix A first, then choose and arrange 3 from the remaining 6, which is 6P3 = 6 × 5 × 4. Thus, the probability is 6P3 / 7P4. A common mistake is using combinations, treating order as irrelevant; always ask, 'Would switching two letters' positions create a new outcome?' Here, yes, like swapping first and second changes the password, so permutations are needed.
A supermarket manager compared weekly data over a year and found that weeks with higher advertising spending (quantitative) tended to have higher total sales revenue (quantitative), a positive association. The manager did not randomly assign ad spending levels; spending varied based on season and promotions (observational study). Which statement is the most reasonable conclusion?
Explanation: We're examining correlation versus causation in business data. Correlation means variables are associated, like higher ad spending with higher sales. To claim causation, random assignment or control is needed to rule out confounders. This observational study with varying ad levels doesn't allow concluding ad spending causes sales increases—seasonal factors might influence both. Answer A is justified by the design limiting us to association. It's common to assume long-term data eliminates confounders, but without control, causation isn't proven. Check: 'Were participants randomly assigned to treatments?' for causation evaluation.
A bag is said to contain 40% blue marbles, so P(blue)=0.40 on each draw with replacement. You draw 25 times and get 6 blue marbles. You simulate 100 runs of 25 draws under P(blue)=0.40. In the simulations, 22 out of 100 runs produced 6 or fewer blues. Which conclusion is most reasonable?
Explanation: The idea is to verify if observed data, such as 6 blue marbles in 25 draws, match a model like P(blue)=0.40. Random chance can lead to odd outcomes sometimes, but only very rare ones typically prompt serious doubt. Simulation recreates the draws repeatedly under the model to measure how often we see results as extreme, here 6 or fewer blues. Since 22 out of 100 simulations showed that, it's fairly common (22%), so it doesn't provide strong evidence against the model. This means the result is plausible and doesn't raise much doubt. A common mix-up is expecting randomness to alternate colors evenly, but small samples naturally fluctuate. For broader use, define extremes (e.g., 6 or less), run simulations, and check if the frequency indicates rarity or not.
Two movies—Movie X and Movie Y—are tied in a vote for what to watch. The group wants a fair tie-breaker, meaning each movie has probability 21 of being chosen. Which procedure is fair?
Explanation: The skill here is using probability to make fair decisions. Fairness means that each movie has an equal probability of being selected, specifically 1/2 for Movie X and Movie Y. In the correct method, a fair coin is flipped once, with heads for Movie X and tails for Movie Y. This assigns equal chances because the coin has two equally likely sides, so each movie has a probability of 1/2. One incorrect method is rolling a die where 1–4 means Movie X and 5–6 means Movie Y, introducing bias by giving Movie X a probability of 2/3 and Movie Y only 1/3. Remember, fairness is about equal chances in each selection, not about the same movie being chosen equally often over multiple selections. To apply this strategy elsewhere, list all possible outcomes and check if each option is equally likely.
A lab culture doubles in size every hour. The number of cells (in thousands) is recorded below.
Time (hours): 0, 1, 2, 3, 4, 5 Cells (thousands): 3, 6, 12, 24, 48, 96
Which type of function best models the relationship between time and the number of cells?
Explanation: Modeling bivariate data involves identifying constant first differences (linear), constant ratios (exponential), or curved patterns (quadratic). The standout cue is the constant ratios in cell counts: 6/3=2, 12/6=2, 24/12=2, 48/24=2, 96/48=2, exactly doubling each hour. This multiplicative pattern directly fits the doubling growth of a lab culture. Exponential is reasonable for population growth scenarios with constant proportional increases. Often, people mistake this for linear due to doubling differences (3,6,12,24,48), but consistent ratios distinguish it. Always compute ratios after differences to detect exponential trends.
A phone store offers two warranty plans for a $600 phone. Assume the only possible repair event within a year is a single screen break.
Strategy A (Buy warranty): Pay 90upfront.Outcomes:withprobability0.15thescreenbreaksandyoupay0 for repair; with probability 0.85 the screen does not break and you pay $0 for repair.
Strategy B (No warranty): Pay 0upfront.Outcomes:withprobability0.15thescreenbreaksandyoupay300 for repair; with probability 0.85 the screen does not break and you pay $0 for repair.
From the consumer’s perspective, over many repetitions, which option would result in a lower average total cost (in dollars)?
Explanation: This problem requires comparing strategies using expected value to find the lower average total cost. For Strategy A (buying warranty), you always pay 90upfront,sotheexpectedtotalcostissimply90 regardless of whether the screen breaks. For Strategy B (no warranty), the expected cost is calculated as: (0.15 × 300)+(0.85×0) = 45.Comparingtheseexpectedvalues,StrategyBhastheloweraveragetotalcostat45 versus 90forStrategyA.Overmanyrepetitions,choosingnottobuythewarrantysavesanaverageof45 per phone. A tempting distractor is Strategy A because it guarantees you never pay the 300repaircost,butthisignoresthatyou′repaying90 every time regardless. The transfer strategy is to calculate the average cost by multiplying each outcome by its probability, not focus on avoiding the worst-case scenario.
According to the sample space S={1,2,3,4,5,6,7,8} for selecting one card labeled 1–8 from a bag, event A is the set of even outcomes and event B is the set of outcomes greater than 5. Which expression represents A∩B (A and B)?
Explanation: This question involves identifying the intersection of two events in a sample space. An event is a subset of the sample space, which here is S = {1,2,3,4,5,6,7,8}, representing possible outcomes when selecting a card. The intersection A ∩ B means the outcomes that satisfy both event A (even numbers) and event B (numbers greater than 5), or in plain language, 'A and B.' The correct set {6,8} matches this because 6 is even and greater than 5, and 8 is even and greater than 5, both from S. A tempting distractor like {2,4,6,8} fails because it represents only A, confusing intersection with the union of events. To approach these problems, translate the words 'and' to intersection, then list outcomes from S that satisfy both conditions exactly. Always ensure the outcomes match the listed elements in S precisely, without adding or assuming extras.
A city planner fit the model y^=200+15x to predict daily subway riders (y, in thousands) from the number of downtown events (x). The residuals for x = 0 through 6 events were:
x: 0, 1, 2, 3, 4, 5, 6 residual: 2, -1, 1, -2, 0, 2, -2
Which statement best describes how well the model fits the data?
Explanation: To evaluate model fit, residuals show if errors are random or patterned, guiding model choice. Residual is actual - predicted (y - ŷ); positive indicates underestimation, negative overestimation. Random scatter around 0 looks like irregular ups and downs with no trends. Patterns suggest missed elements, like curvature or heteroscedasticity. Here, residuals (2, -1, 1, -2, 0, 2, -2) fluctuate randomly without clear trends, supporting a good linear fit for subway riders. Misconceptions include thinking sign changes always mean patterns, but true randomness can have alternations; small sizes don't excuse patterns elsewhere. Focus on absence of patterns, not just residual magnitude, for confirming model fit.
A city’s public library system wants to estimate the proportion of all registered library cardholders in the city who prefer e-books over printed books. The library randomly selects 120 cardholders from the full list of registered cardholders and finds that 48 of them prefer e-books. What population parameter is being estimated?
Explanation: In statistics, we use samples to make inferences about populations. Here, the population is all registered library cardholders in the city - this is the entire group we want to know about. The sample is the 120 randomly selected cardholders who were actually surveyed. The parameter of interest is the true proportion of ALL cardholders who prefer e-books, which is unknown and what we're trying to estimate. The sample statistic (48/120) helps us estimate this population parameter. Random sampling matters because it reduces bias and allows us to generalize from our sample to the population. We can reasonably infer that the population proportion is likely near 48/120, but we cannot know the exact value with certainty.
A coin is claimed to be fair: P(heads)=0.5. In 20 flips, you observe 18 heads. You decide to evaluate the claim by simulation: repeat 20-flip experiments many times using a random number generator with P(heads)=0.5 and count how often you get 18 or more heads. Would this result cause you to question the model? Why? (Here “as extreme or more extreme” means ≥18 heads out of 20.)
Explanation: This question examines whether getting 18 heads in 20 coin flips is consistent with a fair coin model where P(heads) = 0.5. With a fair coin, we'd expect about 10 heads, but random results vary—the question is whether 18 heads is unusually extreme. To evaluate this, we'd simulate many sets of 20 flips using P(heads) = 0.5 and count how often we get 18 or more heads. For a fair coin, getting 18+ heads out of 20 would be extremely rare (probability ≈ 0.0002), so observing this outcome would strongly suggest the coin isn't fair. A key misconception is believing that random outcomes should alternate regularly—true randomness can produce streaks. Another error is thinking past results affect future flips (gambler's fallacy). The evaluation strategy remains: simulate under the model, check how often results as extreme occur, and if very rarely, doubt the model.
A card is drawn from a standard 52-card deck first (event A: the card is a heart). Without replacing it, a second card is drawn then (event B: the second card is a heart). You are told P(A)=5213 and, given the first card was a heart, P(B∣A)=5112. What is the probability that both A and B occur?
Explanation: This question tests the Multiplication Rule for probability. The "and" means both cards must be hearts: first card is a heart, then second card is also a heart. The first event is drawing a heart with probability P(A) = 13/52. The second event is drawing another heart given the first was a heart, with conditional probability P(B|A) = 12/51 (only 12 hearts remain out of 51 cards). To find P(A and B), we multiply: P(A) × P(B|A) = 13/52 × 12/51 = 156/2652 = 1/17. A common mistake is treating the draws as independent and using 13/52 for both, but drawing without replacement creates dependence. Always check whether events affect each other.
Six students—P, Q, R, S, T, and U—need a fair way to decide who cleans up after a lab. Fair means each student has probability 61 of being chosen. Which method gives each student an equal chance?
Explanation: This problem tests using probability to make fair decisions. Fairness means each student has an equal probability of being chosen—here, 1/6 for each of the six students. Rolling a standard die with one student assigned to each number (choice A) gives each student exactly 1/6 probability since each die face is equally likely to appear. This method assigns equal chances because the die's symmetry ensures each outcome has the same likelihood. The spinner method (choice D) creates bias by giving P and Q each 2/6 = 1/3 probability while others get only 1/6, and choosing by handwriting neatness (choice C) introduces subjective bias. Remember that fairness concerns equal probability, not equal results over multiple trials. To check fairness, list all possible outcomes and verify each person appears exactly once with equal likelihood.
Two delivery services recorded the number of packages delivered per driver in a day.
Summary statistics:
Which statement best compares the typical number of packages and consistency using appropriate measures?
Explanation: To compare distributions accurately, we must match our measures to the data's shape. Service A is roughly symmetric (mean = median = 52), suggesting mean and standard deviation work well. However, Service B is right-skewed (mean 52.5 > median 50), with high outliers pulling the mean up. When one distribution is skewed, we use median and IQR for both to ensure fair comparison. For center: Service A's median (52) is higher than Service B's (50), so Service A typically delivers more packages. For spread: Service A's IQR (5) is slightly smaller than Service B's (6), making Service A slightly more consistent. The skewness in Service B inflates its mean and standard deviation. A misconception is using range alone, which ignores the distribution's shape. The strategy is: check symmetry → use median/IQR with any skewness → interpret appropriately.