Applied-Probability-and-Statistics Applied Probability and Statistics (FZO1 C955) Questions and Answers

A defining property of the Poisson distribution is that its variance equals its mean, and both are equal to λ. Since the question states λ = 5, the variance is also 5. This property distinguishes the Poisson distribution from many other probability distributions. The mean represents the expected number of events per interval, while the variance describes the spread of the event count around that mean. In a Poisson model with λ = 5, event counts tend to vary around 5, and the numerical variance is 5. Option B, 4, option C, 0, and option D, 1, do not follow from the Poisson variance rule. The result is not obtained by squaring λ or taking its square root; it is simply equal to λ. Study Guide references/topics: Poisson distribution, variance, mean, λ parameter.

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A boxplot displays the distribution of a quantitative variable using the five-number summary and potential outliers. The main features are the median, first quartile, third quartile, lower whisker, upper whisker, and any marked outliers. The box extends from Q1 to Q3 and represents the interquartile range, which contains the middle 50% of the data. The line inside the box marks the median. Whiskers show the spread of non-outlier values, depending on the graphing convention. Option B is incorrect because a boxplot does not show only the mean; many boxplots do not display the mean at all. Option C is incorrect because variance is a numerical measure of spread, not a standard direct feature of a boxplot. Option D is incorrect because a boxplot is not a probability model. It is a descriptive display for summarizing center, spread, and unusual observations. Study Guide references/topics: boxplots, median, quartiles, interquartile range, outliers.

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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 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.

A bar chart is the best display for categorical data because it shows the frequency or relative frequency of each category using separate bars. Categorical variables classify observations into groups, such as favorite color, type of transportation, political affiliation, or product category. A bar chart allows the viewer to compare category counts directly by bar height. A histogram is not appropriate because it is used for quantitative data grouped into numerical intervals. A scatterplot is used to examine the relationship between two quantitative variables, such as height and weight. A boxplot summarizes a quantitative distribution using quartiles, median, spread, and outliers. Since the question asks specifically about categorical data, the display must preserve category labels rather than numerical intervals or continuous measurement. Therefore, the bar chart is the correct choice. Study Guide references/topics: categorical data, bar charts, frequency displays, descriptive statistics.

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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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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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This experiment has two independent stages: tossing a coin and spinning a numbered spinner. The probability of getting heads on a fair coin is 1/2. The spinner has 8 equally likely outcomes numbered 1 through 8. A number less than 3 means either 1 or 2, so there are 2 favorable spinner outcomes out of 8 total outcomes. Thus, P(number less than 3) = 2/8 = 1/4. Since the coin result does not affect the spinner result, the events are independent. For independent events connected by “and,” multiply the probabilities: 1/2 × 1/4 = 1/8. Equivalently, the combined sample space has 2 coin outcomes times 8 spinner outcomes, or 16 total equally likely outcomes. The favorable outcomes are heads with 1 and heads with 2, giving 2/16 = 1/8. References/topics from the Study Guide: independent events, sample spaces, multiplication rule, theoretical probability.

The correlation coefficient r describes the direction and strength of a linear relationship between two quantitative variables. Its values range from −1 to +1. A negative value indicates that as one variable increases, the other tends to decrease. Since r = −0.5, the direction is negative. Its magnitude, 0.5, is not close enough to zero to indicate no relationship, but it is also not close enough to −1 to represent an extremely strong relationship. Therefore, the best description is a moderate negative linear relationship. Option B is incorrect because the sign is negative, not positive. Option C is incorrect for both direction and strength. Option D would be appropriate only when r is near 0, indicating little or no linear association. A scatterplot with r = −0.5 would generally show a downward trend, but with noticeable scatter around the trend. Study Guide references/topics: correlation coefficient, negative association, scatterplots, linear relationship strength.

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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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For a continuous uniform distribution from 0 to 10, probability is proportional to interval length because every subinterval of equal length has equal probability density. The total interval length is 10 − 0 = 10. The event X < 4 corresponds to the interval from 0 to 4, which has length 4. Therefore, P(X < 4) = 4/10 = 0.4. This means 40% of the distribution lies below 4. Option B, 0.6, would correspond to P(X > 4), since the interval from 4 to 10 has length 6. Option C would represent half the interval, such as P(X < 5). Option D does not match the interval proportion. The essential uniform-distribution principle is area equals probability, and with constant density, area reduces to interval length divided by total length. Study Guide references/topics: uniform distribution, continuous probability, interval length, probability density.

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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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To multiply decimals, first multiply the numbers as whole numbers and then place the decimal point according to the total number of decimal places in the factors. The values are 0.6 and 0.25. Ignoring decimals temporarily, 6 × 25 = 150. The factor 0.6 has one decimal place, and 0.25 has two decimal places, for a total of three decimal places. Therefore, 150 becomes 0.150, which is equivalent to 0.15. A second way to verify the computation is to convert the decimals into fractions: 0.6 = 6/10 and 0.25 = 25/100. Multiplying gives (6 × 25)/(10 × 100) = 150/1000 = 0.15. This result is reasonable because one-fourth of 0.6 is 0.15. The other answer choices are too large for one-fourth of 0.6 or do not result from the correct decimal multiplication. References/topics from the Study Guide: decimal multiplication, fractions, equivalent forms, numerical operations.

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The scatterplot shows a clear downward pattern: as x increases, y tends to decrease. This establishes a negative correlation. The strength of a correlation is judged by how closely the points follow a general linear pattern. Here, although the points are not perfectly aligned, they form a visible descending band from upper left to lower right. That pattern is much stronger than a weak association, where the points would appear widely scattered with only a slight trend. Therefore, the best description is a strong negative correlation. Options C and D are incorrect because they describe positive correlation, which would require y to increase as x increases. Option A correctly identifies the negative direction but understates the strength of the pattern. In correlation terminology, a strong negative value would have r substantially below 0 and closer to −1 than to 0. References/topics from the Study Guide: scatterplots, negative correlation, correlation strength, bivariate quantitative data.

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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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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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The probability of a simple event is calculated by dividing the number of favorable outcomes by the total number of equally likely outcomes. A standard deck contains 52 cards. The favorable outcomes are the hearts, and the question states that there are 13 hearts in the deck. Therefore, the probability of drawing a heart is 13/52. This fraction simplifies by dividing both numerator and denominator by 13, giving 1/4. Option A, 1/13, would incorrectly treat one card as favorable out of 13 rather than 13 favorable hearts out of 52 total cards. Option B, 1/2, would imply that half the deck consists of hearts, which is false. Option D, 1/3, does not correspond to the four equally sized suits in a standard deck. Since the four suits are equally represented, each suit has probability one out of four. References/topics from the Study Guide: theoretical probability, equally likely outcomes, fractions, sample spaces.

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In a linear regression equation, usually written as ŷ = b₀ + b₁x, the intercept b₀ represents the predicted value of Y when X equals 0. It is the point where the regression line crosses the vertical axis. For example, if a model is ŷ = 12 + 3x, the intercept 12 means the predicted response is 12 when x = 0. The slope, b₁, is different; it represents the predicted change in Y for each one-unit increase in X. Correlation measures the strength and direction of a linear association, not the intercept. The mean is a measure of center for a variable and is not the same as a regression intercept. In applied settings, the intercept is meaningful only when X = 0 is within a realistic or relevant range of the data. Study Guide references/topics: linear regression, intercept interpretation, slope-intercept form, response prediction.

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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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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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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 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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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 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 standard error of the mean measures the expected sampling variability of the sample mean. It is calculated by dividing the sample standard deviation by the square root of the sample size: SE = s / √n. In this question, the standard deviation is 5 and the sample size is 25. Therefore, SE = 5 / √25 = 5 / 5 = 1. The mean of 50 identifies the center of the sample distribution but is not directly used in the standard error calculation. Option B, 5, is the standard deviation, not the standard error. Option C, 25, is the sample size. Option D, 0.2, would result from incorrectly dividing 5 by 25 rather than by the square root of 25. The distinction between standard deviation and standard error is important: standard deviation describes spread among individual observations, while standard error describes spread among sample means. Study Guide references/topics: standard error, standard deviation, sample size, sampling distributions.

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The binomial distribution models the number of successes in a fixed number of independent trials when each trial has only two possible outcomes, usually labeled success and failure, and the probability of success remains constant from trial to trial. These four conditions define the binomial setting: fixed number of trials, independent trials, two outcomes per trial, and constant success probability. Examples include the number of heads in 10 coin flips, the number of defective items in a sample when the defect probability is constant, or the number of students who pass an exam out of a fixed group. The normal distribution describes continuous bell-shaped data, not counts of successes. The Poisson distribution models counts of events occurring over a fixed interval when events occur at a constant average rate. The uniform distribution assigns equal probability across outcomes or intervals. Because the question explicitly states “number of successes” and “fixed independent trials,” the correct model is binomial. Study Guide references/topics: binomial distribution, independent trials, success probability, discrete random variables.

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The probabilities of all possible outcomes in a sample space must sum to 1. In this experiment, the bag contains only red, blue, and green balls, so these three outcomes exhaust the sample space. The probability of drawing a red ball is 0.3, and the probability of drawing a blue ball is 0.6. Their combined probability is 0.3 + 0.6 = 0.9. The remaining probability must belong to drawing a green ball. Therefore, P(green) = 1 − 0.9 = 0.1. This can also be viewed as a complement problem: green is the complement of drawing either red or blue when those are the only other colors available. Option B repeats the red probability, option C would make the total probability exceed 1, and option D is the combined probability of red or blue rather than green. References/topics from the Study Guide: probability sum rule, complements, sample space, mutually exclusive outcomes.

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A one-sample t-test is used to compare a sample mean to a hypothesized or known population mean when the population standard deviation is unknown and the data are approximately normal or the sample size is sufficiently large. The test statistic evaluates how far the sample mean is from the hypothesized mean in standard error units. Option A is therefore correct. A two-sample t-test compares means from two independent groups, so option B describes a different test. Tests of variances use procedures such as chi-square or F-based methods depending on context, so option C is not appropriate. Tests of proportions use z procedures for categorical success/failure data, not a one-sample t-test for means. The t-test is part of inferential statistics because it uses sample evidence to make a decision about a population parameter. Study Guide references/topics: one-sample t-test, sample mean, population mean, hypothesis testing.

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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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The coefficient of determination, R², measures the proportion of variability in the response variable Y that is explained by the regression model using explanatory variable X. If R² = 0.64, then 64% of the variation in Y is explained by its linear relationship with X. The remaining 36% is unexplained by the model and may reflect other variables, random variation, measurement error, or nonlinear patterns. Option B is incorrect because R² is not the same as correlation. In simple linear regression, correlation r would be either +0.8 or −0.8, depending on the slope direction, because r² = 0.64. Option C is incorrect because the slope measures change in Y per one-unit change in X, not explained variation. Option D is incorrect because R² is not a probability of making a correct prediction. It is a proportion of variation accounted for by the model. Study Guide references/topics: regression, coefficient of determination, explained variation, linear modeling.

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A z-score standardizes a raw value by expressing its distance from the mean in standard deviation units. The formula is z = (x − μ)/σ, where x is the observed value, μ is the mean, and σ is the standard deviation. The numerator x − μ measures how far the value is from the mean. Dividing by σ converts that distance into standard deviation units. A positive z-score means the value is above the mean, a negative z-score means the value is below the mean, and z = 0 means the value equals the mean. Option B divides the raw value by the mean and does not measure standardized distance. Option C reverses the ratio incorrectly. Option D gives only the raw deviation and fails to standardize by the standard deviation. Z-scores are central in normal distribution calculations and comparisons across different scales. Study Guide references/topics: z-score, standardization, mean, standard deviation.

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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Rolling a standard six-sided die twice creates two independent events. The probability of rolling a six on the first roll is 1/6 because there is one favorable outcome, six, out of six equally likely outcomes. The probability of rolling a six on the second roll is also 1/6. Because the outcome of the first roll does not affect the outcome of the second roll, the multiplication rule for independent events applies. Thus, P(six and six) = 1/6 × 1/6 = 1/36. Another way to verify the result is to list the sample space: two die rolls produce 6 × 6 = 36 equally likely ordered outcomes. Only one outcome, (6, 6), satisfies the condition of two sixes. Therefore, the probability is 1 favorable outcome out of 36 total outcomes. Study Guide references/topics: independent events, multiplication rule, sample space, theoretical probability.

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The scatterplot shows time on the horizontal axis and plant length on the vertical axis. As time increases, the plotted plant lengths also increase. This upward pattern indicates a positive relationship. The points are not randomly scattered; they follow a clear rising trend from approximately 3 inches at the beginning to more than 9 inches near the end of the observation period. Because the points cluster closely around an increasing pattern, the relationship is strong rather than weak. A negative relationship would require plant length to decrease as time increases, which is not shown. A weak positive relationship would show only a slight upward tendency with substantial scatter, but this graph shows a consistent increase across the weeks. Therefore, the best description is a strong positive relationship between time and plant height or length. References/topics from the Study Guide: scatterplots, association, positive correlation, strength of relationship.

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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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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 pie chart displays how a whole is divided into categorical parts, usually as percentages totaling 100%. The chart shows solar at 35%, hydropower at 30%, wind at 20%, geothermal at 10%, and biomass at 5%. The largest percentage is solar, so solar power contributes the most to total energy production. Option A is false because hydropower plus biomass equals 30% + 5% = 35%, which is equal to solar, not more than solar. Option B is false because geothermal is 10%, while biomass is only 5%; therefore, biomass contributes the least. Option C is false because biomass plus geothermal equals 5% + 10% = 15%, which is less than wind at 20%. The correct interpretation requires comparing category percentages accurately rather than relying on visual size alone. References/topics from the Study Guide: pie charts, categorical data, percentages, interpreting graphical displays.

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In any valid probability model, the probabilities of all outcomes in the sample space must sum to 1. The sample space is the complete set of possible outcomes for an experiment, and exactly one outcome from that set must occur. A total probability of 1 represents certainty. For example, when rolling a fair six-sided die, the six outcomes each have probability 1/6, and their sum is 6 × 1/6 = 1. A total of 0 would imply no possible outcome can occur, which is not a valid probability model. A total greater than 1 violates the maximum boundary of probability, and a total less than 1 would mean the sample space is incomplete or some probability has been omitted. This rule is one of the axioms of probability and is used to verify distributions. Study Guide references/topics: sample space, probability axioms, total probability, probability distributions.

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The median is the middle value of a data set after the values are placed in ascending order. The dataset is 3, 6, 9, which is already ordered from least to greatest. Since there are three observations, the median is the second value, because one value lies below it and one value lies above it. Therefore, the median is 6. The median is a measure of center and is especially useful when data may be skewed or contain outliers because it depends on position rather than the magnitude of every value. The mean for this dataset is also 6, since (3 + 6 + 9) ÷ 3 = 6, but the question specifically asks for the median. Options B and C are not values in the dataset, and option D is the minimum value, not the middle. Study Guide references/topics: median, ordered data, measures of central tendency, descriptive statistics.

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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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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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For independent events, the occurrence of one event does not change the probability of the other. Conditional probability P(A|B) means the probability of A occurring given that B has occurred. If A and B are independent, knowing B occurred gives no new information about A. Therefore, P(A|B) = P(A). This is one of the defining properties of independence. For example, if a coin toss and a die roll are independent, knowing the die landed on 4 does not change the probability that the coin landed heads; it remains 1/2. Option B reverses the event being measured. Option C would imply that A becomes impossible after B, which describes a different situation. Option D is an addition expression and does not represent conditional probability. The correct relationship is that the conditional probability equals the original probability when events are independent. Study Guide references/topics: independent events, conditional probability, probability rules, event relationships.

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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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