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

A boxplot summarizes the distribution of a quantitative variable using key positional statistics. It displays the median, first quartile, third quartile, and typically the minimum and maximum non-outlier values. It may also mark outliers separately. The box itself spans from Q1 to Q3, representing the interquartile range, or middle 50% of the data. The line inside the box marks the median. Whiskers extend to values within the non-outlier range, depending on the graphing convention. A boxplot does not primarily show frequency counts; histograms, dot plots, and bar charts are better for frequencies. A scatterplot displays the relationship between two quantitative variables. Probability is not directly displayed by a standard boxplot, although distributional interpretation can be supported by it. The correct answer is therefore median, quartiles, and outliers. Study Guide references/topics: boxplots, five-number summary, quartiles, outliers.

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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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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 box plot displays the five-number summary of a quantitative data set: minimum, first quartile, median, third quartile, and maximum. The median is represented by the vertical line inside the box. In the given box plot, that internal vertical line is positioned at 80 on the score axis. Therefore, the median test score for Class A is 80 points. The value 70 corresponds to the left edge of the box, which represents the first quartile, not the median. The value near 88 represents the right edge of the box, which is the third quartile. The whiskers extend to the approximate minimum and maximum scores, but those values do not determine the median. The median divides the ordered data into two equal halves, meaning about 50% of the students scored at or below 80 and about 50% scored at or above 80. References/topics from the Study Guide: box plots, five-number summary, median, quartiles.

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The inequality y < x − 1/2 has boundary line y = x − 1/2. This line has slope 1 and y-intercept −1/2, so it crosses the y-axis slightly below the origin and rises one unit for every one unit moved to the right. Because the inequality is strict, using “ < ” rather than “≤,” the boundary line must be dashed, showing that points on the line are not included in the solution set. The solution region must be shaded below the line because y is less than the expression x − 1/2. A useful verification method is the test point (0, 0). Substituting gives 0 < 0 − 1/2, or 0 < −1/2, which is false. Therefore, the region containing the origin should not be shaded. The fourth graph has the dashed boundary with intercept −1/2 and shades the region below the line while excluding the origin. References/topics from the Study Guide: graphing linear inequalities, slope-intercept form, dashed boundaries, test-point method.

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A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The event “rolling 1, 2, or 3” has three favorable outcomes: 1, 2, and 3. The probability is therefore favorable outcomes divided by total outcomes: 3/6. This fraction simplifies to 1/2. Option B, 1/3, would correspond to two favorable outcomes out of six. Option C, 1/6, is the probability of rolling one specific number only. Option D, 2/3, would require four favorable outcomes out of six. Since exactly half of the die faces are 1, 2, or 3, the correct probability is one-half. This is a direct application of theoretical probability with equally likely outcomes. Study Guide references/topics: die probability, favorable outcomes, sample space, theoretical probability.