WGU Applied-Probability-and-Statistics Practice Test Questions Answers

The question asks for a joint percentage: employees who both prefer working in the office and report high productivity. First identify the percentage who prefer office work. Since 55% prefer remote work, the remaining 45% prefer working in the office. The problem then states that among office-preferring employees, 50% report high productivity. This is a conditional percentage: P(high productivity | office preference) = 50%. To find the overall percentage of all employees who satisfy both conditions, multiply the base percentage by the conditional percentage: 45% × 50% = 0.45 × 0.50 = 0.225 = 22.5%. Option D gives only the conditional rate within the office group, not the percentage of all employees. Option C gives only the office-preference percentage, not productivity within that group. Option B does not match the correct joint probability calculation. The correct selection is therefore 22.5%. References/topics from the Study Guide: probability, conditional percentages, joint probability, two-way categorical reasoning.

============

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.

============

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.

============

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.

============

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.

============