Applied-Probability-and-Statistics Exam Dumps - WGU Courses and Certificates Questions and Answers

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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The probability of a simple event is calculated by dividing the number of favorable outcomes by the total number of equally likely outcomes. A standard deck contains 52 cards. The favorable outcomes are the hearts, and the question states that there are 13 hearts in the deck. Therefore, the probability of drawing a heart is 13/52. This fraction simplifies by dividing both numerator and denominator by 13, giving 1/4. Option A, 1/13, would incorrectly treat one card as favorable out of 13 rather than 13 favorable hearts out of 52 total cards. Option B, 1/2, would imply that half the deck consists of hearts, which is false. Option D, 1/3, does not correspond to the four equally sized suits in a standard deck. Since the four suits are equally represented, each suit has probability one out of four. References/topics from the Study Guide: theoretical probability, equally likely outcomes, fractions, sample spaces.

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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.

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A histogram is used to display the frequency distribution of quantitative numerical data, especially continuous data grouped into intervals or bins. For example, a histogram can show the distribution of heights, test scores, commute times, or weights. Each bar represents a numerical interval, and the height of the bar represents how many observations fall within that interval. Unlike a bar chart, the bars in a histogram usually touch because the intervals form a continuous numerical scale. Option B is incorrect because categorical data are better displayed with a bar chart or pie chart. Option C is incorrect because a scatterplot displays the relationship between two quantitative variables, not the frequency distribution of one variable. Option D is incorrect because a boxplot summarizes median, quartiles, spread, and outliers rather than showing detailed frequency counts. The correct graph for continuous numerical frequency is a histogram. Study Guide references/topics: histograms, quantitative data, frequency distributions, graphical displays.

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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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To add and subtract fractions, first rewrite all fractions using a common denominator. The denominators are 5, 3, and 4. Their least common denominator is 60. Convert each fraction: 2/5 = 24/60, 1/3 = 20/60, and 1/4 = 15/60. Now combine the numerators while keeping the common denominator: 24/60 + 20/60 − 15/60 = (24 + 20 − 15)/60 = 29/60. The fraction 29/60 is already in simplest form because 29 is prime and does not divide 60. Option A, 13/60, and option B, 11/60, result from incorrect numerator operations. Option D, 19/60, also fails to preserve the correct addition before subtraction. The essential technical step is establishing a common denominator before performing arithmetic operations. References/topics from the Study Guide: fractions, least common denominator, fraction addition, fraction subtraction.

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The standard error of the mean is calculated as SE = s/√n, where s is the sample standard deviation and n is the sample size. Because n appears in the denominator under a square root, increasing the sample size decreases the standard error when the standard deviation is held constant. This reflects a central idea in sampling: larger samples tend to produce more stable and precise estimates of the population mean. Option C is incorrect because increasing the standard deviation increases the standard error, not decreases it. Option B is incomplete and does not identify what is decreasing. Option D is not the direct driver of standard error; confidence level affects the critical value and margin of error, but the standard error itself is determined by variability and sample size. The correct relationship is inverse: as sample size increases, standard error decreases. Study Guide references/topics: standard error, sample size, sampling variability, precision of estimates.

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A box plot displays the five-number summary, including the median, which is shown by the vertical line inside each box. In the exhibit, the median for Class A is positioned at approximately 65, while the median for Class B is positioned at approximately 80. The difference is therefore 80 − 65 = 15 points. Since Class A’s median is lower, the correct interpretation is that the median test score of Class A is 15 points less than that of Class B. This comparison uses a categorical grouping variable, class, and a quantitative response variable, test score. The box plot is appropriate because it allows the distributions of two groups to be compared visually using medians, quartiles, ranges, and potential spread. Options A, B, and D do not match the median locations shown by the center lines in the boxes. References/topics from the Study Guide: box plots, median, five-number summary, comparing distributions.

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This problem compares conditional percentages across two groups. For faculty members, 40 out of 50 support the recycling initiative, so the faculty support rate is 40/50 = 0.80, or 80%. For students, 200 out of 500 support the initiative, so the student support rate is 200/500 = 0.40, or 40%. Comparing 80% and 40% shows that faculty members are twice as likely as students to support the initiative. This is not a slight difference; it is a large difference of 40 percentage points. Therefore, the correct statement is that faculty members are much more likely than students to support the recycling initiative. The table contains two categorical variables: group type and recycling opinion. Conditional percentages are the appropriate numerical tool because the goal is to compare support within each group. References/topics from the Study Guide: contingency tables, two-way categorical data, conditional percentages, comparative proportions.

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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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