Pass the WGU Courses and Certificates Applied-Probability-and-Statistics Questions and answers with ExamsMirror

In simple linear regression, R² is the square of the correlation coefficient r. Therefore, to recover the possible correlation value from R², take the square root: r = ±√R². With R² = 0.81, the square root is √0.81 = 0.9, so r may be +0.9 or −0.9. The sign cannot be determined from R² alone because squaring removes direction. A positive r would indicate an upward linear association, while a negative r would indicate a downward linear association. Option D gives only +0.9, which would be correct only if the slope or scatterplot confirmed a positive relationship. Option B incorrectly treats R² as though it were r. Option C incorrectly keeps 0.81 after applying the square-root relationship. The technically complete answer is ±0.9. Study Guide references/topics: coefficient of determination, correlation coefficient, regression, explained variation.

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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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This study involves two categorical variables: customer age group and preferred shopping method. Age group is categorical because customers are classified into groups such as teens, adults, and seniors. Shopping method is also categorical because customers are classified as preferring either online or in-store shopping. When two categorical variables are analyzed together, conditional percentages are the most appropriate numerical summary because they describe how the distribution of one categorical variable changes within levels of another categorical variable. For example, the analyst could compute the percentage of teens who prefer online shopping, the percentage of adults who prefer online shopping, and the percentage of seniors who prefer online shopping. A five-number summary is used for quantitative data, not categorical data. Mean difference requires numerical measurements. A correlation coefficient measures the strength and direction of a linear relationship between two quantitative variables, so it is not appropriate here. References/topics from the Study Guide: two-variable categorical data, conditional percentages, categorical association, numerical summaries.

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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.

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 symbol α represents the significance level of a hypothesis test and is the probability of making a Type I error. A Type I error occurs when the null hypothesis is true but the test rejects it. Therefore, α = 0.05 means there is a 5% risk of rejecting a true null hypothesis under the testing procedure. This is often described as the false positive rate. Option B describes a Type II error, which involves failing to reject a false null hypothesis. Option C is incorrect because a 5% significance level corresponds to a 95% confidence level in related confidence-interval contexts, not 5% confidence. Option D is unrelated; α is not a sample-size percentage. The interpretation must connect α directly to the probability of incorrectly rejecting H₀ when H₀ is actually true. Study Guide references/topics: hypothesis testing, significance level, Type I error, null hypothesis.

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The Poisson distribution is used to model the number of events occurring in a fixed interval of time, space, area, or volume when the events occur independently at a constant average rate. It is especially common for rare event counts, such as calls arriving at a help desk per hour, accidents at an intersection per month, defects per production batch, or website errors per day. The defining parameter of the Poisson distribution is λ, the average number of events per interval. A binomial distribution instead models the number of successes in a fixed number of independent trials with a constant probability of success. A normal distribution models continuous, bell-shaped measurements. A uniform distribution assigns equal probability across outcomes or intervals. Because the prompt specifically says “rare event count per interval,” the technical match is the Poisson distribution. Study Guide references/topics: Poisson distribution, event counts, rate parameter λ, discrete probability distributions.

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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.

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.

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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