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

This is an observational study because the scientists collect existing information about individuals’ smoking habits and lung cancer diagnoses without assigning treatments or controlling exposure. In an experimental study, researchers impose a condition, treatment, or intervention and then measure the resulting effect. Here, the researchers are not requiring some individuals to smoke and others not to smoke; they are simply recording naturally occurring characteristics and outcomes. The purpose is to examine association, or correlation, between two variables: smoking status and lung cancer diagnosis. Both variables are categorical, so the analysis would likely involve proportions, conditional percentages, or contingency tables. Option A is incorrect because analyzing correlations does not make a study experimental. Option B is incorrect because “testing effects” would require manipulation or assignment. Option C identifies a true feature, a large sample size, but sample size does not determine study type. References/topics from the Study Guide: observational studies, experimental studies, association, categorical variables.

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The probability of selecting a student who does not play soccer is found by identifying the complement of the soccer group. There are 25 students in the class, and 10 students play soccer. Therefore, the number of students who do not play soccer is 25 − 10 = 15. The probability is then the number of favorable outcomes divided by the total number of possible outcomes: 15/25 = 0.60. The information about basketball and neither sport provides context, but the direct complement method is sufficient because the question asks only who does not play soccer. Option B, .40, represents the probability of selecting a student who does play soccer, since 10/25 = .40. Option A, .30, is associated with neither sport if interpreted approximately from 7/25, but it is not the requested event. Option C, .50, does not match the count structure. References/topics from the Study Guide: probability, complements, favorable outcomes, relative frequency.

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The mean is the arithmetic average of a quantitative data set. To calculate it, add all observed values and divide by the number of observations. The jelly bean counts are 50, 60, 80, 80, 100, 100, 100, 120, 120, and 140. Their sum is 950. There are 10 jars, so the mean is 950 ÷ 10 = 95. This means the average number of jelly beans per jar is 95. Option A, 100, may seem plausible because 100 appears frequently and is near the center of the data, but frequency alone determines the mode, not the mean. Option C, 80, is one of the repeated values but not the average. Option B, 120, is too high because several observations fall well below 120. The mean uses every value in the data set, including the low values 50 and 60 and the high value 140. References/topics from the Study Guide: mean, measures of center, quantitative data, arithmetic average.

A Type II error occurs when the null hypothesis is false but the test fails to reject it. Many introductory answer choices phrase this as “accept H₀ when false,” though the more technically careful wording is “fail to reject H₀ when false.” This is a false negative: the real effect or difference exists, but the test does not detect sufficient evidence. The probability of a Type II error is denoted β. Option B describes a Type I error, which is rejecting a true null hypothesis. Option C is incorrect because 1 − α is not β; α is the Type I error probability, while 1 − β is statistical power. Option D is too vague and does not identify a formal hypothesis-testing error. The defining condition for Type II error is missing a real effect by failing to reject a false null hypothesis. Study Guide references/topics: Type II error, beta, hypothesis testing, statistical power.

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A Type II error occurs when the null hypothesis is false, but the test fails to reject it. Many introductory materials describe this as “accepting H₀ when false,” although the more precise statistical wording is “failing to reject H₀.” This is a false negative: a real effect, difference, or relationship exists, but the test does not detect enough evidence to conclude it. For example, if a new teaching method truly improves scores but the test fails to find significance, the result is a Type II error. Option B describes a Type I error, which is rejecting a true null hypothesis. Option C is a correct test decision, not an error. Option D is invalid because Type II error is a defined concept. The probability of Type II error is β, and statistical power is 1 − β. Study Guide references/topics: hypothesis testing, Type II error, null hypothesis, statistical power.

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The data describe frequencies for a single variable: weekly exercise duration. Each duration, 1 hour through 4 hours, has an associated count of participants. A bar chart is the most appropriate display because it uses separate bars to show how many observations fall into each category or discrete value. Here, the bar above 1 hour would have height 3, the bar above 2 hours would have height 6, the bar above 3 hours would have height 4, and the bar above 4 hours would have height 2. A line graph is inappropriate because the data are not measuring change over time. A pie chart could show proportions, but the question specifically asks for frequency of exercise hours, making a bar chart clearer and more direct. A scatterplot requires two quantitative variables and is used to study association, not a one-variable frequency distribution. References/topics from the Study Guide: frequency distributions, bar charts, single-variable data displays, quantitative/discrete data.

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To multiply decimals, first multiply the numbers as whole numbers and then place the decimal point according to the total number of decimal places in the factors. The values are 0.6 and 0.25. Ignoring decimals temporarily, 6 × 25 = 150. The factor 0.6 has one decimal place, and 0.25 has two decimal places, for a total of three decimal places. Therefore, 150 becomes 0.150, which is equivalent to 0.15. A second way to verify the computation is to convert the decimals into fractions: 0.6 = 6/10 and 0.25 = 25/100. Multiplying gives (6 × 25)/(10 × 100) = 150/1000 = 0.15. This result is reasonable because one-fourth of 0.6 is 0.15. The other answer choices are too large for one-fourth of 0.6 or do not result from the correct decimal multiplication. References/topics from the Study Guide: decimal multiplication, fractions, equivalent forms, numerical operations.

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The accurate display is the bar graph labeled “Number of Feet Traveled” with the vertical axis showing the number of feet and the horizontal axis identifying the student travelers. For discrete observations associated with individual students, a bar graph is appropriate because each bar represents a separate category or individual and its height represents the measured value. Option A is inaccurate because its title states “Number of Miles Traveled,” while the vertical axis is labeled in feet; this creates a unit mismatch. Option C uses a pie chart, which is designed to show parts of a whole, not individual measured distances unless the goal is proportional composition. Option D uses a line graph, which is typically reserved for ordered change, especially over time; connecting student traveler values implies continuity or trend that is not justified. The correct graph must use consistent units, clear labels, and an appropriate display type. References/topics from the Study Guide: data displays, bar graphs, categorical identifiers, quantitative measurements, graph accuracy.

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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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For a normal distribution, the empirical rule states that approximately 95% of observations lie within two standard deviations of the mean. The mean lifespan is 11.5 years, and the standard deviation is 1.5 years. Two standard deviations equal 2 × 1.5 = 3.0 years. Therefore, the interval containing approximately 95% of the data is found by subtracting and adding 3.0 years to the mean: 11.5 − 3.0 = 8.5 and 11.5 + 3.0 = 14.5. The resulting interval is 8.5 years to 14.5 years. Option C, 10.0 to 13.0, represents one standard deviation from the mean and would correspond to about 68% of the data. Options A and D are not centered correctly at the proper two-standard-deviation distance. The correct interval must be symmetric around the mean and use two standard deviations. References/topics from the Study Guide: normal distributions, empirical rule, standard deviation, measures of spread.

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