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

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

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This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes. “Taking the bus and being on time” means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 − .15 = .85. Therefore, P(bus and on time) = .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 − .05 = .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes. References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.

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For independent events, the occurrence of one event does not change the probability of the other event. The multiplication rule states that if A and B are independent, then P(A and B) = P(A) × P(B). Here, P(A) = 0.3 and P(B) = 0.4. Therefore, P(A and B) = 0.3 × 0.4 = 0.12. This means there is a 12% probability that both events occur. Option B, 0.7, incorrectly adds the probabilities and would only be part of an “or” calculation, not an “and” calculation. Option D repeats P(A), and option C does not follow from the multiplication rule. The phrase “A and B independent” is the decisive condition because it permits direct multiplication without adjusting for conditional probability. Study Guide references/topics: independent events, multiplication rule, compound probability, joint probability.

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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.

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

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The 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 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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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.