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

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, the slope represents the predicted change in the response variable Y for each one-unit increase in the explanatory variable X. In slope-intercept form, ŷ = b₀ + b₁x, the slope is b₁. For example, if a regression equation predicts cost as ŷ = 25 + 4x, the slope 4 means the predicted cost increases by 4 units for each additional unit of x. The intercept, b₀, is the predicted value of Y when X = 0, so option B describes a different component. R² measures the proportion of variation in Y explained by the regression model, not the rate of change. Correlation measures strength and direction of linear association, but it is not the same as the slope because it is unitless and standardized. The slope is the operational rate of change in the model. Study Guide references/topics: linear regression, slope interpretation, response variable, explanatory variable.

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The dot plot shows a distribution centered near 3 days, with substantial frequencies at 2, 3, and 4 days and smaller frequencies at the extremes of 1 and 5 days. This shape is reasonably balanced around the center. When a distribution is approximately symmetric, the mean and median tend to be close because the lower and higher values balance each other. The median identifies the middle observation after the data are ordered, and the mean represents the balance point of the distribution. Since the tallest stack is near 3 days and the surrounding stacks on 2 and 4 days are comparable, the center of the data is approximately 3 days by both measures. There is no strong skew to pull the mean far above or below the median. Therefore, the best statement is that the mean is almost the same as the median. Options B and C would require clear skew or extreme imbalance, which the dot plot does not show. References/topics from the Study Guide: dot plots, mean, median, distribution shape, symmetry.

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The sample mean is the standard point estimate for the population mean. A point estimate is a single statistic calculated from sample data and used to estimate an unknown population parameter. If the parameter of interest is the population mean μ, the corresponding sample statistic is x̄. Therefore, the statement is true. This does not mean the sample mean is guaranteed to equal the population mean exactly; sampling variability can cause sample means to differ from the true population mean. However, the sample mean is unbiased under random sampling and becomes more stable as sample size increases. Confidence intervals expand this idea by placing a margin of error around the sample mean. Option B is incorrect because the sample mean is precisely the conventional point estimator for μ. Options C and D impose invalid restrictions. Study Guide references/topics: sample mean, population mean, point estimate, sampling variability.

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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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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 median is the measure of central tendency least affected by outliers because it depends only on the ordered position of values, not on the numerical magnitude of extreme observations. To find the median, the data are arranged from least to greatest and the middle value is identified, or the average of the two middle values is taken when the sample size is even. If an extremely large or extremely small value is added to a data set, the mean can shift substantially because every value contributes to the arithmetic average. The median, however, usually remains stable unless the extreme value changes the middle position. The mode is also not directly affected by magnitude, but it may be absent or non-unique and is often less informative for numerical data. Range is not a measure of central tendency and is highly sensitive to outliers because it uses the minimum and maximum. Study Guide references/topics: descriptive statistics, measures of center, median, resistance to outliers.

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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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Ethnic group is a categorical variable because it classifies students into named groups rather than measuring a numerical quantity. Categorical data are commonly displayed with bar graphs or pie charts. A bar graph shows the frequency or relative frequency of each category using separate bars, making it easy to compare group sizes. A pie chart shows how the whole student population is divided among categories, making it useful for displaying proportions or percentages. Therefore, the correct pair is bar graph and pie chart. A box plot is used for quantitative data and summarizes values using quartiles and a median, so it is not appropriate for ethnic categories. A line graph is typically used to show change over time or ordered numerical trends. A histogram is used for distributions of quantitative data grouped into intervals, not named categories. Since the goal is to display category membership, both selected graphs must support categorical frequency or percentage displays. References/topics from the Study Guide: categorical data displays, bar graphs, pie charts, relative frequency.

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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.