Applied-Probability-and-Statistics Questions and Answers

The standard error of the sample mean is calculated as SE = s/√n, where s is the sample standard deviation and n is the sample size. Here, the standard deviation is 8 and the sample size is 16. The square root of 16 is 4, so SE = 8/4 = 2. The sample mean of 40 is not used directly in the standard error formula; it identifies the center of the sample, not the variability of the sample mean across repeated samples. Option B, 8, is the standard deviation, not the standard error. Option D, 4, is the square root of the sample size. Option C, 0.5, would result from an incorrect calculation. Standard error decreases as sample size increases because larger samples provide more precise estimates. Study Guide references/topics: standard error, standard deviation, sample size, sampling distribution.

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A histogram is used to display the frequency distribution of quantitative numerical data, especially continuous data grouped into intervals or bins. For example, a histogram can show the distribution of heights, test scores, commute times, or weights. Each bar represents a numerical interval, and the height of the bar represents how many observations fall within that interval. Unlike a bar chart, the bars in a histogram usually touch because the intervals form a continuous numerical scale. Option B is incorrect because categorical data are better displayed with a bar chart or pie chart. Option C is incorrect because a scatterplot displays the relationship between two quantitative variables, not the frequency distribution of one variable. Option D is incorrect because a boxplot summarizes median, quartiles, spread, and outliers rather than showing detailed frequency counts. The correct graph for continuous numerical frequency is a histogram. Study Guide references/topics: histograms, quantitative data, frequency distributions, graphical displays.

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A probability or percentage describes how often an event is expected to occur in the long run. Here, 3% of homes are infected with termites. As a probability, this is 0.03, meaning about 3 out of every 100 homes would be expected to have termites. This is a small probability, so the event should be interpreted as unlikely. It is not impossible, because impossible would require a probability of 0%. It is also not likely, because likely events generally have probabilities substantially greater than 50%. Finally, it is not “as likely as unlikely,” because that language corresponds to a probability near 50%, where occurrence and nonoccurrence are approximately balanced. Since 3% is close to zero but not equal to zero, the technically correct interpretation is that termite infection is unlikely for a randomly selected home in the community. References/topics from the Study Guide: probability interpretation, percentages, unlikely events, complementary probability.

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The probability of selecting a student who does not play soccer is found by identifying the complement of the soccer group. There are 25 students in the class, and 10 students play soccer. Therefore, the number of students who do not play soccer is 25 − 10 = 15. The probability is then the number of favorable outcomes divided by the total number of possible outcomes: 15/25 = 0.60. The information about basketball and neither sport provides context, but the direct complement method is sufficient because the question asks only who does not play soccer. Option B, .40, represents the probability of selecting a student who does play soccer, since 10/25 = .40. Option A, .30, is associated with neither sport if interpreted approximately from 7/25, but it is not the requested event. Option C, .50, does not match the count structure. References/topics from the Study Guide: probability, complements, favorable outcomes, relative frequency.

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The phrase “given that they are enrolled in the science course” indicates conditional probability. The reference group is not all 30 students; it is only the 12 students enrolled in science. Among those 12 science students, 8 are also enrolled in math. Therefore, the conditional probability is P(math | science) = P(math and science) ÷ P(science), or more directly, 8 ÷ 12. This simplifies to 2/3, which is approximately 0.6667, rounded to .67. Option A, .27, would come from comparing the 8 students to the entire class of 30, but that ignores the given condition. Option B, .40, corresponds to 12/30, the proportion enrolled in science. Option C, .44, is approximately 8/18, which reverses the condition and would represent the probability of science given math. The correct answer uses the science group as the denominator. References/topics from the Study Guide: conditional probability, two-way events, joint enrollment, probability ratios.

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To add and subtract fractions, first rewrite all fractions using a common denominator. The denominators are 5, 3, and 4. Their least common denominator is 60. Convert each fraction: 2/5 = 24/60, 1/3 = 20/60, and 1/4 = 15/60. Now combine the numerators while keeping the common denominator: 24/60 + 20/60 − 15/60 = (24 + 20 − 15)/60 = 29/60. The fraction 29/60 is already in simplest form because 29 is prime and does not divide 60. Option A, 13/60, and option B, 11/60, result from incorrect numerator operations. Option D, 19/60, also fails to preserve the correct addition before subtraction. The essential technical step is establishing a common denominator before performing arithmetic operations. References/topics from the Study Guide: fractions, least common denominator, fraction addition, fraction subtraction.

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The correlation coefficient r measures the strength and direction of a linear relationship between two quantitative variables. Values close to 1 indicate strong positive linear association, values close to −1 indicate strong negative linear association, and values near 0 indicate no linear relationship. Therefore, r = 0 means there is no linear association detected by the correlation coefficient. It is important to interpret this precisely: r = 0 does not guarantee there is no relationship of any kind. A nonlinear relationship may still exist, but correlation measures only linear pattern. Option B and option C are incorrect because strong relationships require r to be close to 1 or −1. Option D is incorrect because perfect correlation occurs at r = 1 or r = −1, not at 0. The correct interpretation is no linear relationship. Study Guide references/topics: correlation coefficient, linear association, scatterplots, correlation interpretation.

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The equation y = 40x + 75 is written in slope-intercept form, y = mx + b. In this form, m represents the rate of change and b represents the initial value. The constant term 75 means the graph must cross the y-axis at 75, because the studio charges $75 before any hourly time is added. The coefficient 40 means that for every additional hour of studio time, the total cost increases by $40. Therefore, the correct graph must begin at y = 75 when x = 0 and rise at a steady linear rate of 40 dollars per hour. For example, at x = 6 hours, the model gives y = 40(6) + 75 = 315, so the line should pass near the point (6, 315). Among the displayed graphs, Graph 4 matches both the y-intercept and the slope. References/topics from the Study Guide: basic algebra, linear equations, slope-intercept form, interpreting rate of change.

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The sample space of a probability experiment is the complete set of all possible outcomes. In this experiment, the spinner has four equal sections labeled A, B, C, and D. Since one spin can land on exactly one of those four labeled sections, the possible outcomes are A, B, C, and D. Therefore, the sample space is {A, B, C, D}. Option A incorrectly adds E, which is not listed as a section on the spinner. Option B groups labels into combined outcomes, but the experiment is a single spin, not a two-part or paired event. Option C uses numerical labels, but the spinner sections are identified by letters, not numbers. Equal section size affects the probability of each outcome, making each outcome have probability 1/4, but it does not change the sample space itself. References/topics from the Study Guide: probability experiments, outcomes, sample space, equally likely events.

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The Poisson distribution is used to model counts of events occurring within a fixed interval of time, space, area, or volume, especially when the events are relatively rare and occur at a constant average rate. Examples include the number of calls received per hour, accidents per month, defects per batch, or arrivals at a service counter per minute. Its parameter λ represents the average number of events per interval. Option B describes the binomial distribution, which models the number of successes in a fixed number of independent success/failure trials. Option C is incorrect because Poisson outcomes are discrete counts, not continuous measurements. Option D is also incorrect because categorical data classify observations into groups, while Poisson data count event occurrences. The phrase “rare events per interval” is the defining clue for the Poisson model. Study Guide references/topics: Poisson distribution, rare-event counts, λ parameter, discrete probability models.

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The interquartile range, or IQR, measures the spread of the middle 50% of a distribution. It is calculated as Q3 − Q1, where Q1 is the first quartile and Q3 is the third quartile. In the Georgia box plot, the left edge of the box is labeled 800, representing Q1, and the right edge of the box is labeled 2800, representing Q3. Therefore, the interquartile range is 2800 − 800 = 2000 square feet. The minimum value, 500, and maximum value, 5000, are part of the full range, not the IQR. The median, 1200, identifies the center of the distribution but is not used directly in the IQR calculation. Option D, 4500, corresponds to the full range from 500 to 5000, not the interquartile range. The correct interpretation is that the middle half of Georgia house sizes spans 2000 square feet. References/topics from the Study Guide: box plots, quartiles, interquartile range, comparing distributions.

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This study contains two variables: educational level and annual income. Educational level is categorical because it places individuals into named groups, such as high school, bachelor’s, master’s, and doctorate. Although these categories have a natural order, they are still categories rather than measured numerical quantities. Annual income is quantitative because it is measured numerically and can be analyzed using arithmetic summaries such as mean, median, range, and standard deviation. Therefore, the study is classified as categorical-to-quantitative. This classification is important because it determines appropriate statistical displays and summaries. For example, side-by-side box plots or group means could be used to compare annual income across education levels. A categorical-to-categorical study would involve two label-based variables, such as education level and employment sector. A quantitative-to-quantitative study would involve two numerical variables, such as income and years of experience. References/topics from the Study Guide: categorical variables, quantitative variables, two-variable data classification, comparative summaries.

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A z-test for a population mean is used when the population standard deviation σ is known and the sampling distribution of the test statistic can be treated as normal. The test statistic has the form z = (sample statistic − hypothesized parameter) divided by the standard error. Knowing σ allows the standard error to be computed using σ/√n rather than estimating it with the sample standard deviation. Option B describes the common setting for a t-test, not a z-test. Option C is not sufficient by itself; small samples generally require stronger normality assumptions and often favor t-procedures when σ is unknown. Option D is unrelated to the classic z-test for a mean, though z-tests can also be used for proportions under appropriate large-sample conditions. In the provided answer set, the defining condition is that the population standard deviation is known. Study Guide references/topics: z-test, population standard deviation, standard error, hypothesis testing.

The coefficient of determination, R², is obtained by squaring the correlation coefficient r in simple linear regression. Here, r = 0.6, so R² = (0.6)² = 0.36. This means that 36% of the variability in the response variable is explained by the linear relationship with the explanatory variable. The remaining 64% of variability is not explained by that linear model and may be due to other variables, random variation, measurement error, or nonlinear structure. Option B gives the correlation itself, not its square. Option C would come from squaring 0.4, not 0.6. Option D would imply a perfect explanatory relationship, which would require |r| = 1. The value of R² is always between 0 and 1 and is commonly interpreted as a proportion of explained variation. Study Guide references/topics: correlation, coefficient of determination, regression, explained variation.

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The 68-95-99.7 rule, also called the empirical rule, applies to approximately normal distributions. It states that about 68% of observations fall within 1 standard deviation of the mean, about 95% fall within 2 standard deviations, and about 99.7% fall within 3 standard deviations. The question asks specifically for the percentage within 1 standard deviation, so the correct answer is 68%. In notation, this interval is μ − σ to μ + σ, where μ is the mean and σ is the standard deviation. This rule is useful for interpreting spread in bell-shaped distributions. For example, if scores have mean 80 and standard deviation 5, then approximately 68% of scores fall between 75 and 85. Option B corresponds to 2 standard deviations, and option C corresponds approximately to 3 standard deviations. Option D is not part of the empirical rule. Study Guide references/topics: empirical rule, normal distribution, standard deviation, spread.

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This question asks for a conditional probability. The phrase “given that they attended the science club” means the science club group becomes the restricted reference group. The required probability is P(drama club | science club), which is calculated as P(drama and science) divided by P(science). The problem states that 15% of students participated in both clubs, so P(drama and science) = 0.15. It also states that 30% participated in the science club, so P(science) = 0.30. Therefore, P(drama | science) = 0.15 ÷ 0.30 = 0.50. This means that among students who attended the science club, half also participated in drama club. Option A gives the joint probability, not the conditional probability. Option B gives only the science club participation rate. Option C does not match the conditional probability calculation. References/topics from the Study Guide: conditional probability, joint probability, probability ratios, two-event probability.

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A boxplot summarizes the distribution of a quantitative variable using key positional statistics. It displays the median, first quartile, third quartile, and typically the minimum and maximum non-outlier values. It may also mark outliers separately. The box itself spans from Q1 to Q3, representing the interquartile range, or middle 50% of the data. The line inside the box marks the median. Whiskers extend to values within the non-outlier range, depending on the graphing convention. A boxplot does not primarily show frequency counts; histograms, dot plots, and bar charts are better for frequencies. A scatterplot displays the relationship between two quantitative variables. Probability is not directly displayed by a standard boxplot, although distributional interpretation can be supported by it. The correct answer is therefore median, quartiles, and outliers. Study Guide references/topics: boxplots, five-number summary, quartiles, outliers.

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A histogram and a bar chart both use bars, but they display different types of data. A histogram is used for quantitative numerical data grouped into intervals, such as ages, weights, test scores, or commute times. The bars represent continuous or ordered numeric ranges, and the bars usually touch to show that the scale is continuous. A bar chart is used for categorical data, such as favorite sport, political party, product type, or survey response category. The bars are separated because the categories are distinct labels rather than continuous intervals. Option B reverses the correct uses. Option C is incorrect because the graphs are not the same even though both contain bars. Option D is invalid because option A states the standard distinction. Correct graph selection depends on identifying whether the variable is categorical or quantitative. Study Guide references/topics: histograms, bar charts, categorical data, quantitative data displays.

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A standard deck contains 52 cards. There are 4 kings and 4 queens, one of each rank in each suit. The event is drawing a king or a queen. Since a single card cannot be both a king and a queen, these two events are mutually exclusive. Therefore, the number of favorable outcomes is 4 + 4 = 8. The probability is favorable outcomes divided by total outcomes: 8/52. This fraction can be simplified to 2/13, but the answer choice provides the unsimplified correct form. Option B, 4/52, counts only kings or only queens, not both. Option C, 1/13, is equivalent to 4/52 and again represents one rank only. Option D, 1/26, is too small and does not match the count of favorable cards. The word “or” signals that both qualifying ranks should be included. Study Guide references/topics: card probability, mutually exclusive events, addition rule, sample space.

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For a Poisson random variable, the probability of exactly k events is P(X = k) = e^−λ λ^k / k!, where λ is the mean number of events per interval. Here, λ = 4 and the question asks for exactly 2 events, so k = 2. Substitute into the formula: P(X = 2) = e^−4 × 4² / 2!. Since 4² = 16 and 2! = 2, this becomes e^−4 × 8. Numerically, e^−4 is about 0.0183, and 0.0183 × 8 ≈ 0.1465. Thus, the probability is approximately 0.1465. Options B, C, and D are unsupported by the Poisson probability formula. The Poisson model is appropriate for counts of events occurring independently over a fixed interval at a constant average rate. Study Guide references/topics: Poisson distribution, λ parameter, exact event probability, discrete distributions.

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Two coin flips produce four equally likely ordered outcomes: HH, HT, TH, and TT. Exactly one head occurs in two of these outcomes: HT and TH. Therefore, the probability is 2 favorable outcomes out of 4 total outcomes, or 2/4 = 1/2. This can also be computed using the binomial model. There are n = 2 independent trials, success probability p = 1/2, and exactly one success is required. The binomial calculation is C(2,1)(1/2)^1(1/2)^1 = 2 × 1/4 = 1/2. Option B, 1/4, counts only one of the two favorable sequences. Option C, 3/4, includes outcomes with at least one head rather than exactly one head. Option D would mean certainty, which is impossible because HH and TT do not satisfy the condition. Study Guide references/topics: sample spaces, coin-flip probability, binomial probability, independent events.

A Type II error occurs when the null hypothesis is false, but the test fails to reject it. Many introductory materials describe this as “accepting H₀ when false,” although the more precise statistical wording is “failing to reject H₀.” This is a false negative: a real effect, difference, or relationship exists, but the test does not detect enough evidence to conclude it. For example, if a new teaching method truly improves scores but the test fails to find significance, the result is a Type II error. Option B describes a Type I error, which is rejecting a true null hypothesis. Option C is a correct test decision, not an error. Option D is invalid because Type II error is a defined concept. The probability of Type II error is β, and statistical power is 1 − β. Study Guide references/topics: hypothesis testing, Type II error, null hypothesis, statistical power.

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A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The even outcomes are 2, 4, and 6. That gives 3 favorable outcomes out of 6 total outcomes. Therefore, the probability of rolling an even number is 3/6, which simplifies to 1/2. Option A, 1/3, would correspond to 2 favorable outcomes out of 6, which is not correct here. Option C, 1/6, is the probability of rolling one specific number, such as only a 2. Option D, 2/3, would require 4 favorable outcomes out of 6. Since exactly half of the die faces are even and half are odd, the probability of rolling an even number is one-half. Study Guide references/topics: theoretical probability, equally likely outcomes, sample space, favorable outcomes.

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The correlation coefficient r measures the direction and strength of a linear relationship between two quantitative variables. Values of r range from −1 to 1. A negative value indicates that as one variable increases, the other variable tends to decrease. The value −0.7 is fairly close to −1, so it represents a strong negative linear relationship. It is not perfect, because perfect negative correlation would be r = −1, but it is clearly stronger than a weak association. Option B is incorrect because the sign is negative, not positive. Option C is incorrect because no linear relationship would be represented by a correlation near 0. Option D is incorrect both in direction and strength. A scatterplot with r = −0.7 would generally show points trending downward from left to right, with some scatter around the trend. Study Guide references/topics: correlation coefficient, negative association, linear relationships, scatterplot interpretation.

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A Type I error occurs when the null hypothesis, H₀, is true but the test decision rejects it. In plain statistical language, this is a false positive: the analysis concludes that there is evidence of an effect, difference, or relationship when in reality the null condition is true. The probability of making a Type I error is denoted by α, the significance level, commonly set at 0.05. A Type II error is different; it occurs when the null hypothesis is false but the test fails to reject it. That is a false negative. Option C is incorrect because rejecting a true null is not a correct decision. Option D describes not rejecting or accepting the null, not the stated event. The critical phrase is “rejecting H₀ when true,” which directly defines a Type I error. Study Guide references/topics: hypothesis testing, null hypothesis, Type I error, significance level.

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A confidence interval for a population mean is generally constructed as sample mean ± margin of error. The margin of error equals a critical value multiplied by the standard error. Therefore, the structure is sample mean ± critical value × SE. The critical value depends on the confidence level and distribution used, such as a z critical value when the population standard deviation is known or a t critical value when it is unknown. The standard error measures the sampling variability of the sample mean, often computed as s/√n when using sample standard deviation. Option B is incorrect because variance times sample size is not a margin of error. Option C is incomplete because standard deviation alone measures spread among data values, not uncertainty of the sample mean. Option D is incorrect because the median is a measure of center and is not the typical component of a mean-based confidence interval. Study Guide references/topics: confidence intervals, standard error, critical value, margin of error.

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Standard deviation measures the spread, variability, or dispersion of data values around the mean. A small standard deviation indicates that the data values are clustered closely around the mean, while a large standard deviation indicates that the values are more widely dispersed. It is calculated from deviations from the mean, squared deviations, variance, and then the square root of variance. Standard deviation is not a measure of central tendency; measures of central tendency include the mean, median, and mode. It is also not a probability, although it is used in probability distributions such as the normal distribution. Frequency refers to how often a value or category occurs, which is summarized with tables, bar charts, histograms, or dot plots. The standard deviation is essential because it quantifies how consistent or variable a dataset is. Study Guide references/topics: standard deviation, variance, spread, descriptive statistics.

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Because the first selection is replaced, the two selections are independent. Replacement means the probability structure returns to its original state after the first respondent is selected. The probability of selecting an action respondent is 30%, or 0.30. The probability of selecting a comedy respondent is 35%, or 0.35. For independent sequential events, multiply the probabilities: 0.30 × 0.35 = 0.105. Converting 0.105 to a percentage gives 10.5%. Therefore, the probability of first selecting an action respondent and then selecting a comedy respondent is 10.5%. Option B does not result from multiplying the two relevant genre percentages. Options C and D are far too large because they resemble combined or unrelated percentages rather than the probability of a two-step ordered selection. The word “and” signals multiplication, while “replacement” confirms independence between draws. References/topics from the Study Guide: independent events, replacement, multiplication rule, probability with percentages.

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The random variable X counts the number of heads obtained in 4 coin flips. Since it is a count, it can take only specific whole-number values: 0, 1, 2, 3, or 4. It cannot take fractional values such as 2.5 heads. A random variable with countable possible outcomes is classified as discrete. This situation also fits a binomial framework because there is a fixed number of independent trials, each trial has two outcomes, and the probability of heads remains constant for a fair coin. A continuous random variable, by contrast, can take any value over an interval, such as time, weight, or height. The number of heads is not measured on a continuum; it is counted. Therefore, the correct classification is discrete. Study Guide references/topics: discrete random variables, binomial setting, coin-flip outcomes, probability distributions.

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The mode is the value that occurs most frequently in a data set. In the set 3, 5, 7, 7, 10, the value 3 appears once, 5 appears once, 10 appears once, and 7 appears twice. Since 7 has the highest frequency, it is the mode. This question tests recognition of a measure of center that differs from the mean and median. The mean would require summing all values and dividing by the number of observations, while the median would identify the middle value after ordering the data. The data are already ordered, and the middle value is also 7, but the reason the correct answer is 7 here is specifically because it appears more often than any other value. Options A, B, and D each name values that occur only once, so none can be the mode. A dataset may have one mode, multiple modes, or no mode, but this dataset is unimodal. References/topics from the Study Guide: descriptive statistics, measures of center, frequency, mode.

The equation 1/(6z) = 2/9 is a proportion because it sets two fractions equal to each other. To solve it, use cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the denominator of the first fraction by the numerator of the second fraction: 1 × 9 = 2 × 6z. This gives 9 = 12z. To isolate z, divide both sides by 12: z = 9/12. This fraction can also be simplified to 3/4, but the answer choices list the unsimplified equivalent form 9/12. The other options do not satisfy the original equation. For example, if z = 2/54, then 6z is very small and the left side becomes much larger than 2/9. If z = 12/9 or 18/2, the denominator becomes too large, making the left side too small. References/topics from the Study Guide: proportions, solving equations, cross multiplication, equivalent fractions.

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For a binomial random variable, the expected value is E(X) = np, where n is the number of trials and p is the probability of success on each trial. The binomial model applies when there is a fixed number of independent trials, each trial has two outcomes, and the probability of success remains constant. The expected value represents the long-run average number of successes across repeated binomial experiments. For example, if a test has 20 independent questions and the probability of success on each is 0.75, the expected number correct is 20 × 0.75 = 15. Option B, n/p, and option C, p/n, do not represent the binomial mean. Option D, n + p, incorrectly adds the number of trials and success probability, which are not comparable units. The formula np captures both how many opportunities there are and how likely success is on each opportunity. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.

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Empirical probability is based on observed data collected from experiments, surveys, simulations, or repeated trials. It is calculated as the relative frequency of an event: number of times the event occurs divided by the total number of trials or observations. For example, if a machine produces 12 defective items in a sample of 300, the empirical probability of a defect is 12/300 = 0.04. This differs from classical or theoretical probability, which is based on equally likely outcomes and mathematical structure, such as a fair die having probability 1/6 for each face. It also differs from subjective probability, which is based on personal judgment or expert belief rather than observed frequency. The word “empirical” signals evidence obtained through observation, so observed frequency is the correct basis. Study Guide references/topics: empirical probability, relative frequency, observed data, probability interpretation.

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The mean is the arithmetic average of a data set. To compute it, add all values and divide by the number of observations. For the dataset 4, 8, 12, 16, and 20, the sum is 4 + 8 + 12 + 16 + 20 = 60. There are 5 values, so the mean is 60 ÷ 5 = 12. The mean represents the balance point of the data. This dataset is evenly spaced around 12: 4 and 20 are equally distant from 12, and 8 and 16 are equally distant from 12. That symmetry supports the computed result. Option A, 10, is too low because it does not account for the higher values 16 and 20. Option C, 14, is too high, and option D is one of the data values but not the average. The correct answer is 12. Study Guide references/topics: mean, arithmetic average, measures of center, quantitative data.

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The accurate display is the bar graph labeled “Number of Feet Traveled” with the vertical axis showing the number of feet and the horizontal axis identifying the student travelers. For discrete observations associated with individual students, a bar graph is appropriate because each bar represents a separate category or individual and its height represents the measured value. Option A is inaccurate because its title states “Number of Miles Traveled,” while the vertical axis is labeled in feet; this creates a unit mismatch. Option C uses a pie chart, which is designed to show parts of a whole, not individual measured distances unless the goal is proportional composition. Option D uses a line graph, which is typically reserved for ordered change, especially over time; connecting student traveler values implies continuity or trend that is not justified. The correct graph must use consistent units, clear labels, and an appropriate display type. References/topics from the Study Guide: data displays, bar graphs, categorical identifiers, quantitative measurements, graph accuracy.

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A t-distribution is used for inference about a population mean when the population standard deviation σ is unknown, especially with a small sample. In that situation, the sample standard deviation s must estimate σ, creating extra uncertainty. The t-distribution accounts for this added uncertainty with heavier tails than the standard normal distribution. As sample size increases, the t-distribution becomes closer to the normal distribution. Option B describes a z-procedure setting, where σ is known and the normal distribution may be used. Option C is incorrect because categorical data are usually analyzed with proportions or chi-square procedures, not t-distributions for means. Option D is incomplete: t-procedures typically require quantitative continuous or approximately continuous data, but the decisive condition in the answer set is unknown σ with a small sample. Study Guide references/topics: t-distribution, unknown standard deviation, small samples, inference for means.

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A standard six-sided die has three even outcomes: 2, 4, and 6. Thus, the probability that one die lands on an even number is 3/6 = 1/2. When two dice are rolled, the outcomes are independent because the result of one die does not affect the other. Therefore, the probability that both dice are even is 1/2 × 1/2 = 1/4. Equivalently, there are 36 ordered outcomes for two dice. The even-even outcomes are formed from 3 even choices on the first die and 3 even choices on the second die, giving 3 × 3 = 9 favorable outcomes. Then 9/36 simplifies to 1/4. Option B gives the probability of one die being even, not both. Options C and D do not match the count or multiplication rule. Study Guide references/topics: independent events, dice probability, multiplication rule, sample space.

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Expected value is the long-run average value of a random variable. For a discrete random variable, it is calculated by multiplying each possible value by its probability and then adding those products. Here, the expression is already structured as an expected value calculation: 1×0.2 + 2×0.5 + 3×0.3. Compute each product: 1×0.2 = 0.2, 2×0.5 = 1.0, and 3×0.3 = 0.9. Add them: 0.2 + 1.0 + 0.9 = 2.1. Therefore, the expected value is 2.1. This does not mean the random variable must equal 2.1 in a single trial; it means that over many repetitions, the average outcome would approach 2.1. Option B is close but omits part of the weighted contribution. Options C and D do not match the weighted-average computation. Study Guide references/topics: expected value, discrete random variables, weighted average, probability distributions.

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Mutually exclusive events cannot occur at the same time. If event A occurs, event B cannot occur, and if event B occurs, event A cannot occur. Therefore, the probability of both events occurring together is zero: P(A and B) = 0. For example, when rolling one die, the event “roll a 2” and the event “roll a 5” are mutually exclusive because one roll cannot produce both outcomes simultaneously. Option B applies to independent events, not mutually exclusive events. In fact, mutually exclusive events with positive probabilities are not independent because the occurrence of one eliminates the possibility of the other. Option C is part of the addition rule for mutually exclusive events when calculating P(A or B), not P(A and B). Option D would mean both events always occur together, which contradicts mutual exclusivity. Study Guide references/topics: mutually exclusive events, joint probability, addition rule, probability foundations.

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The correlation coefficient r measures the direction and strength of a linear relationship between two quantitative variables. It ranges from −1 to 1. A value close to −1 indicates a strong negative linear relationship, meaning that as one variable increases, the other tends to decrease in a highly consistent linear pattern. Since r = −0.9 is very close to −1, the relationship is strong and negative. Option B is incorrect because the sign is negative, not positive. Option C identifies the direction correctly but understates the strength; −0.9 is not weak. Option D would correspond to r near 0, where there is little or no linear association. A scatterplot associated with r = −0.9 would show points clustered closely around a downward-sloping line. This does not prove causation; it describes linear association only. Study Guide references/topics: correlation coefficient, negative association, strength of relationship, scatterplots.

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Variance measures the average squared distance of data values from the mean. For the dataset 2, 4, 6, 8, 10, the mean is (2 + 4 + 6 + 8 + 10) ÷ 5 = 30 ÷ 5 = 6. The deviations from the mean are −4, −2, 0, 2, and 4. Squaring these deviations gives 16, 4, 0, 4, and 16. The sum of squared deviations is 40. Using the sample variance formula, divide by n − 1, where n = 5. Thus, sample variance = 40 ÷ 4 = 10. Option A is correct. Option C, 6.25, does not result from the standard sample variance computation for these values. The important distinction is whether a problem is using sample variance or population variance; the provided answer set and calculation use the sample variance formula. Study Guide references/topics: variance, mean, squared deviations, sample variance.

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