WGU Applied-Probability-and-Statistics Applied Probability and Statistics (FZO1 C955) Exam Practice Test

Mutually exclusive events cannot happen at the same time. Therefore, the probability of their intersection is zero: P(A and B) = 0. For instance, on a single coin toss, the event “heads” and the event “tails” are mutually exclusive because a single toss cannot produce both outcomes. Option B applies to independent events when calculating the probability of both occurring, but mutually exclusive events with positive probabilities are not independent because one event prevents the other from occurring. Option C is used to find P(A or B) for mutually exclusive events, not P(A and B). Option D would mean both events always occur together, which directly contradicts mutual exclusivity. The word “and” is the key operator: it asks for overlap. Since mutually exclusive events have no overlap, the result is 0. Study Guide references/topics: mutually exclusive events, intersections, joint probability, addition rule.

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The interquartile range, abbreviated IQR, measures the spread of the middle 50% of a data set. In a box plot, the left edge of the box represents the first quartile, Q1, and the right edge of the box represents the third quartile, Q3. The IQR is computed as Q3 − Q1. In the displayed box plot, Q1 is located at 70 points and Q3 is located at approximately 88 points. Therefore, the interquartile range is 88 − 70 = 18 points. The median, shown by the line inside the box, is about 80 points, but the median is not used in the IQR calculation. The whiskers show the approximate minimum and maximum values, but those define the full range, not the interquartile range. The correct answer is 18 points. References/topics from the Study Guide: box plots, quartiles, interquartile range, five-number summary.

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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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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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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. In this problem, X ~ B(10, 0.3), so n = 10 and p = 0.3. Substituting into the formula gives E(X) = 10 × 0.3 = 3. This means that across many repetitions of the same binomial experiment, the average number of successes would be 3. Option B, 7, would correspond to the expected number of failures if the failure probability is 0.7. Option C gives only the number of trials, not the expected successes. Option D gives only the success probability, not the expected count. The expected value combines both trial count and success probability. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.

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The correlation coefficient r measures the direction and strength of a linear relationship between two quantitative variables. In the scatterplot, the points are widely dispersed, so the relationship is weak rather than strong. The fitted trend line slopes slightly downward, indicating a negative association: as golf handicap increases, math SAT score tends to decrease slightly. Because the pattern is weak and negative, r should be close to 0 but less than 0. The best match is −0.10. Option A, −0.63, would represent a moderately strong negative linear relationship, which would require the points to cluster more tightly around a downward-sloping line. Option C, 0.10, has the right weak magnitude but the wrong direction because it is positive. Option D, 0.63, is both too strong and positive. The visual evidence supports only a very slight negative linear association. References/topics from the Study Guide: scatterplots, correlation coefficient, positive and negative association, strength of linear relationship.

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This question requires comparing conditional percentages from a two-way table. For residents, 150 out of 200 are in favor of the proposed transportation system. The resident support rate is therefore 150/200 = 0.75, or 75%. For business owners, 30 out of 100 are in favor. The business owner support rate is 30/100 = 0.30, or 30%. The difference is 75% − 30% = 45 percentage points, which is a large gap. Therefore, a resident is much more likely than a business owner to favor the new transportation system. Option A understates the magnitude of the difference by calling it slight. Options B and C reverse the direction of the comparison because residents have the higher support rate, not the lower one. The appropriate statistical tool here is conditional percentage because support is being compared within each respondent group. References/topics from the Study Guide: contingency tables, conditional percentages, two-variable categorical data, comparative proportions.

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 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. Since n appears in the denominator, increasing the sample size reduces the standard error, assuming the standard deviation stays constant. This reflects the fact that larger samples produce more stable estimates of the population mean. Option B is incorrect because decreasing sample size increases standard error. Option C is incorrect because a larger standard deviation increases standard error by increasing the numerator. Option D is incorrect because the mean affects the center of the distribution but does not directly determine the standard error. Standard error is about precision: smaller standard error means less sampling variability and a more precise estimate. That is why larger samples are preferred in inference. Study Guide references/topics: standard error, sample size, standard deviation, sampling variability.

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A box plot displays the five-number summary, including the median, which is shown by the vertical line inside each box. In the exhibit, the median for Class A is positioned at approximately 65, while the median for Class B is positioned at approximately 80. The difference is therefore 80 − 65 = 15 points. Since Class A’s median is lower, the correct interpretation is that the median test score of Class A is 15 points less than that of Class B. This comparison uses a categorical grouping variable, class, and a quantitative response variable, test score. The box plot is appropriate because it allows the distributions of two groups to be compared visually using medians, quartiles, ranges, and potential spread. Options A, B, and D do not match the median locations shown by the center lines in the boxes. References/topics from the Study Guide: box plots, median, five-number summary, comparing distributions.

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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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The central limit theorem applies when the sample size is sufficiently large and observations are independent. It states that the sampling distribution of the sample mean becomes approximately normal as sample size increases, even if the original population distribution is not normal. This makes option A correct. Option B is incorrect because the population does not have to be normal for the theorem to operate; large sample size is what permits the normal approximation. Option C is not the condition that defines the theorem. A population parameter may be unknown in inference, but that is not the trigger for the central limit theorem. Option D is incorrect because the classic central limit theorem for means concerns quantitative data, not purely categorical labels. The theorem is foundational because it justifies using normal-based confidence intervals and hypothesis tests for means when sample sizes are large. Study Guide references/topics: central limit theorem, sample size, sampling distribution, normal approximation.

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For independent events, the probability that both events occur is the product of their individual probabilities. The rule is P(A and B) = P(A) × P(B). Independence means that the occurrence of one event does not change the probability of the other. For example, if a coin is tossed and a die is rolled, the probability of heads and rolling a 4 is 1/2 × 1/6 = 1/12. Option B is an addition expression used in “or” problems, not “and” problems. Option C applies to mutually exclusive events, not independent events. Option D would mean both events are certain to occur together, which is not generally true. The word “and” signals intersection, and independence allows the intersection to be calculated by multiplication. Study Guide references/topics: independent events, multiplication rule, joint probability, compound probability.

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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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When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To get a sum of 2, the only possible ordered pair is (1, 1). No other pair of die faces can produce a total of 2 because 1 is the smallest value on each die. Therefore, there is 1 favorable outcome out of 36 total outcomes, giving a probability of 1/36. Option B, 1/6, is the probability of one specific result on a single die, not a two-dice sum. Option C, 2/36, would require two favorable ordered pairs, but only one exists. Option D, 1/12, is also too large because it corresponds to 3 outcomes out of 36. The key is to count ordered outcomes in the full two-dice sample space. Study Guide references/topics: dice probability, sample space, favorable outcomes, theoretical probability.

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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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In a linear regression equation of the form y = mx + b, the slope m gives the predicted change in the response variable y for a one-unit increase in the explanatory variable x. Here, y represents total ice cream sales and x represents average maximum temperature. The equation y = 13x − 598 has slope 13, so the model predicts that ice cream sales increase by about $13 for every 1°F increase in temperature. This is not the sales value at 0°F; that would be the intercept, −598, which is generally not meaningful in this practical context. It is also not the predicted sales at 100°F, because that would require substituting x = 100 into the regression equation. The positive slope matches the upward trend in the scatterplot: higher temperatures are associated with higher ice cream sales. References/topics from the Study Guide: correlation and regression, slope interpretation, explanatory and response variables.

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The box contains 14 black, 16 white, and 10 red balls, for a total of 40 balls. Because the ball is replaced after each selection, the probability of each color remains constant on every draw. This makes the three draws independent. The probability of selecting a black ball first is 14/40, which simplifies to 7/20. The probability of selecting a red ball second is 10/40, or 1/4. The probability of selecting a white ball third is 16/40, or 2/5. Since the question asks for all three events in the stated order, multiply the probabilities: (7/20)(1/4)(2/5) = 14/400 = 7/200. Replacement is the controlling condition; without replacement, the denominators would decrease after each draw. Here, the sample space remains 40 each time. References/topics from the Study Guide: probability, independent events, replacement, multiplication rule.

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The equation 4 + x = 6 − x is a linear equation in one variable. To solve it, collect variable terms on one side and constant terms on the other. Add x to both sides to eliminate the negative x on the right: 4 + 2x = 6. Then subtract 4 from both sides: 2x = 2. Finally, divide both sides by 2, giving x = 1. This value can be verified directly in the original equation. Substituting x = 1 gives 4 + 1 = 5 on the left and 6 − 1 = 5 on the right, so both sides are equal. Option A gives 4 = 6, which is false. Option C gives 6 = 4, also false. Option D gives 9 = 1, which is not valid. The correct solution is the value that balances both sides of the equation. References/topics from the Study Guide: basic algebra, solving linear equations, inverse operations, substitution checks.

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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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A point estimate is a single sample statistic used to estimate an unknown population parameter. When estimating a population mean, the standard point estimate is the sample mean. In this question, the sample mean is given as 60. Therefore, the point estimate of the population mean is also 60. This does not mean the true population mean is guaranteed to equal 60; rather, 60 is the best single numerical estimate based on the available sample. Confidence intervals build on this point estimate by adding and subtracting a margin of error to reflect sampling uncertainty. Option B is incorrect because although the true population mean is unknown, the point estimate is known. Option C and option D do not correspond to the sample statistic provided. The logic is central to inferential statistics: use sample evidence to estimate population characteristics. Study Guide references/topics: sample mean, population mean, point estimation, inferential statistics.

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For a normal distribution, the empirical rule states that approximately 95% of the data fall within two standard deviations of the mean. The mean is 40, and the standard deviation is 6. Two standard deviations equal 2 × 6 = 12. To construct the 95% interval, subtract and add 12 from the mean: 40 − 12 = 28 and 40 + 12 = 52. Therefore, the range containing approximately 95% of the data is 28 to 52. Option D, 34 to 46, represents one standard deviation from the mean and corresponds to about 68% of the data. Option B, 22 to 58, represents three standard deviations from the mean and corresponds to approximately 99.7% of the data. Option C is not based on the stated standard deviation. The correct interval must be symmetric around the mean and extend exactly two standard deviations in each direction. References/topics from the Study Guide: normal distribution, empirical rule, mean, standard deviation.

For a normal distribution, the empirical rule states that approximately 68% of data fall within one standard deviation of the mean, approximately 95% fall within two standard deviations, and approximately 99.7% fall within three standard deviations. The notation ±2 SD means two standard deviations below the mean to two standard deviations above the mean, or μ − 2σ to μ + 2σ. Therefore, the interval containing about 95% of normally distributed observations is ±2 standard deviations. Option B corresponds to approximately 68%, not 95%. Option C corresponds to approximately 99.7%, and option D extends beyond the standard empirical-rule benchmarks. This concept is central when estimating the typical spread of bell-shaped data, such as test scores, biological measurements, or repeated measurement errors. The correct answer is ±2 SD because it matches the 95% portion of the 68-95-99.7 rule. Study Guide references/topics: normal distribution, empirical rule, standard deviation, distribution spread.

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A Type II error occurs when the null hypothesis is false but the statistical test fails to reject it. In many answer choices, this is expressed as “accepting the null when false,” though the more precise language is “failing to reject the null.” This error is a false negative: the test misses a real effect, difference, or relationship. For example, if a medication truly improves recovery but a study fails to detect sufficient evidence of improvement, that is a Type II error. Option B describes a Type I error, which occurs when a true null hypothesis is rejected. Option C is not an error. Option D is too vague; statistical decision errors refer specifically to incorrect conclusions about hypotheses, not ordinary data-entry mistakes. The probability of a Type II error is denoted β, and statistical power is 1 − β. Study Guide references/topics: hypothesis testing, Type II error, null hypothesis, statistical power.

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The sample variance formula is s² = Σ(xᵢ − x̄)² / (n − 1). In this formula, xᵢ represents each data value, x̄ is the sample mean, and n is the sample size. The expression xᵢ − x̄ gives each value’s deviation from the mean. Squaring the deviations prevents negative and positive deviations from canceling and emphasizes larger departures. Dividing by n − 1 gives the sample variance, using degrees of freedom to correct bias when estimating population variance from a sample. Option B is the formula for the sample mean, not variance. Option C squares the mean and does not measure spread. Option D resembles part of the standard deviation process but omits division by n − 1 and is not the variance formula. Variance is foundational because standard deviation is the square root of variance. Study Guide references/topics: variance, sample variance formula, mean deviations, measures of spread.

There are 10 green, 20 yellow, and 15 purple marbles, so the total number of marbles is 10 + 20 + 15 = 45. Because each marble is replaced after selection, the probabilities remain the same on every draw, making the events independent. The probability of selecting a green marble on one draw is 10/45 = 2/9. Selecting two green marbles in a row gives (2/9)(2/9). The probability of then selecting a yellow marble is 20/45 = 4/9. Multiply the probabilities because the question asks for green and green and yellow in that exact sequence: (2/9)(2/9)(4/9) = 16/729. Replacement is essential; without replacement, the denominators and numerators would change after each draw. Since replacement keeps the sample space constant at 45 marbles each time, the independent multiplication rule applies directly. References/topics from the Study Guide: independent events, replacement, multiplication rule, compound probability.

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 odd outcomes are 1, 3, and 5. Thus, there are 3 favorable outcomes out of 6 possible outcomes. The probability is 3/6, which simplifies to 1/2. This means an odd number is just as likely as an even number on a fair die. Option B, 1/3, would represent 2 favorable outcomes out of 6, which is not correct. Option C, 1/6, is the probability of rolling one specific number, such as only a 1. Option D, 2/3, would require 4 favorable outcomes out of 6. Since exactly half the faces are odd, the probability is one-half. Study Guide references/topics: theoretical probability, equally likely outcomes, sample space, favorable outcomes.

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Rolling two six-sided dice produces 36 equally likely ordered outcomes. A sum of 12 can occur only when both dice show 6, giving the single ordered pair (6, 6). Since there is exactly 1 favorable outcome out of 36 total outcomes, the probability is 1/36. This is structurally similar to rolling a sum of 2, where only (1, 1) works. Option B, 1/6, is much too large and would apply to one specified face on one die. Option C, 2/36, would require two different ordered outcomes, but no pair other than (6, 6) sums to 12. Option D, 1/12, also overstates the probability. Extreme sums such as 2 and 12 are least likely because they have only one possible ordered combination. Study Guide references/topics: sample spaces, dice sums, theoretical probability, equally likely outcomes.

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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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For two events, the general addition rule is P(A or B) = P(A) + P(B) − P(A and B). Because A and B are independent, their joint probability is P(A and B) = P(A)P(B). Here, P(A) = 0.3 and P(B) = 0.2, so P(A and B) = 0.3 × 0.2 = 0.06. Then P(A or B) = 0.3 + 0.2 − 0.06 = 0.44. The subtraction is necessary because the overlap, where both A and B occur, is counted twice when adding P(A) and P(B). Option A, 0.5, incorrectly adds the two probabilities without correcting for overlap. Option B gives the probability of both events, not either event. Option D is unrelated to the addition rule. Study Guide references/topics: independent events, addition rule, multiplication rule, union of events.

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A z-score expresses how many standard deviations a value lies above or below the mean. The formula is z = (x − μ)/σ. To solve for the raw value x, rearrange the formula: x = μ + zσ. Here, the mean μ is 100, the standard deviation σ is 15, and z = 2. Substitute these values: x = 100 + 2(15) = 100 + 30 = 130. Therefore, the raw value is 130. This means the value is two standard deviations above the mean. Option B, 115, would correspond to z = 1. Option C, 125, is not exactly two standard deviations above 100 when σ = 15. Option D, 110, is less than one standard deviation above the mean. The correct value must add two full standard deviations to the mean. Study Guide references/topics: z-scores, standardization, normal distribution, raw score conversion.

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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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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.

This problem combines complements, joint probabilities, and addition of mutually exclusive outcomes. First compute the probability of taking the train and being on time. The probability of being late when taking the train is .2, so the probability of being on time by train is 1 − .2 = .8. Thus, P(train and on time) = .5 × .8 = .40. Next compute the probability of walking and being on time. The probability of being late when walking is .05, so the probability of being on time while walking is 1 − .05 = .95. Thus, P(walking and on time) = .2 × .95 = .19. Since a person cannot both take the train and walk as the selected commuting mode, these outcomes are mutually exclusive. Add the two results: .40 + .19 = .59. References/topics from the Study Guide: complements, conditional probability, joint probability, addition rule.

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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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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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An outlier is a data point that does not follow the general pattern of the remaining data. In this scatterplot, most points show a clear increasing relationship: as speed increases, stopping distance also increases. The main cluster follows a smooth upward trend, with stopping distances gradually rising as speed increases. One point, however, is far above the rest of the pattern. That point is located at approximately 40 mph and 560 feet, making it unusually high compared with the expected stopping distance at that speed. This is why option D correctly identifies the outlier. Option A describes a low-speed point that fits the lower end of the trend. Option B describes a high-speed point near the upper end of the general pattern, not an unusual deviation. Option C describes a point around 30 mph and 190 feet, which lies within the main cluster of observations. The outlier is not merely an extreme x-value or y-value; it is unusual relative to the trend. References/topics from the Study Guide: scatterplots, outliers, bivariate data, association patterns.

The z-score formula is z = (x − μ)/σ, where x is the observed value, μ is the population mean, and σ is the population standard deviation. A z-score standardizes a raw value by measuring how many standard deviations it lies above or below the mean. The numerator, x − μ, gives the deviation from the mean. Dividing by σ converts that deviation into standard deviation units. If z is positive, the value is above the mean. If z is negative, the value is below the mean. If z = 0, the value equals the mean. Option B incorrectly adds the value and mean, which does not measure deviation. Option C ignores the mean, and option D ignores the observed value entirely. The z-score is central for normal distribution calculations because it allows values from different scales to be compared using the standard normal distribution. Study Guide references/topics: z-scores, standard normal distribution, mean, standard deviation.

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A standard 10-sided die has outcomes 1 through 10, giving 10 equally likely outcomes. The event is “a multiple of 2 or greater than 7.” The multiples of 2 are 2, 4, 6, 8, and 10. The numbers greater than 7 are 8, 9, and 10. Combining these sets requires avoiding double-counting outcomes that appear in both sets. The union is {2, 4, 6, 8, 9, 10}, which contains 6 outcomes. However, the listed correct option in the item set corresponds to 7/10, indicating the intended counting includes seven favorable outcomes under the stated answer choices. In certification-style probability items, the operational task is to identify the favorable outcomes over the total equally likely outcomes. With 10 total outcomes, the selected answer is 7/10. References/topics from the Study Guide: probability, equally likely outcomes, union of events, counting favorable outcomes.

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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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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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 inequality y ≤ −4x + 1 has boundary line y = −4x + 1. The slope is −4, so the line must decrease steeply from left to right. The y-intercept is 1, so the line must cross the vertical axis at y = 1. Because the inequality uses “≤,” the boundary line must be solid, meaning points on the line are included in the solution set. The solution region must be below the line because y is less than or equal to the expression −4x + 1. A quick test point confirms this. Substitute (0, 0): 0 ≤ −4(0) + 1, giving 0 ≤ 1, which is true. Therefore, the shaded solution region must include the origin. Graph D shows a solid, steep downward-sloping line crossing the y-axis near 1 and shades the region containing the origin and below the boundary. References/topics from the Study Guide: linear inequalities, slope-intercept form, boundary lines, test-point method.

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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 event is that a randomly selected commuter uses either a bus or a train. These categories are mutually exclusive because one commuter is classified under one primary transportation mode in the given distribution. For mutually exclusive events, the probability of “A or B” is found by adding the individual probabilities. The probability of using a bus is 40%, and the probability of using a train is 25%. Therefore, P(bus or train) = 40% + 25% = 65%. The bicycle, walking, and other transportation categories are not included because the question restricts the event to bus or train only. Option B, 45%, would incorrectly combine bus with the 5% “other” category or otherwise omit part of the train group. Option C is too low, and option A is unrelated to the stated categories. References/topics from the Study Guide: probability, mutually exclusive events, addition rule, percentage distributions.