WGU Updated Applied-Probability-and-Statistics Exam Questions and Answers by alaya

Variance measures the average squared distance of data values from the mean. For the dataset 2, 4, 6, 8, 10, the mean is (2 + 4 + 6 + 8 + 10) ÷ 5 = 30 ÷ 5 = 6. The deviations from the mean are −4, −2, 0, 2, and 4. Squaring these deviations gives 16, 4, 0, 4, and 16. The sum of squared deviations is 40. Using the sample variance formula, divide by n − 1, where n = 5. Thus, sample variance = 40 ÷ 4 = 10. Option A is correct. Option C, 6.25, does not result from the standard sample variance computation for these values. The important distinction is whether a problem is using sample variance or population variance; the provided answer set and calculation use the sample variance formula. Study Guide references/topics: variance, mean, squared deviations, sample variance.

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This experiment has two independent stages: tossing a coin and spinning a numbered spinner. The probability of getting heads on a fair coin is 1/2. The spinner has 8 equally likely outcomes numbered 1 through 8. A number less than 3 means either 1 or 2, so there are 2 favorable spinner outcomes out of 8 total outcomes. Thus, P(number less than 3) = 2/8 = 1/4. Since the coin result does not affect the spinner result, the events are independent. For independent events connected by “and,” multiply the probabilities: 1/2 × 1/4 = 1/8. Equivalently, the combined sample space has 2 coin outcomes times 8 spinner outcomes, or 16 total equally likely outcomes. The favorable outcomes are heads with 1 and heads with 2, giving 2/16 = 1/8. References/topics from the Study Guide: independent events, sample spaces, multiplication rule, theoretical probability.

A chi-square test is commonly used to analyze categorical data, especially to determine whether two categorical variables are associated. In a chi-square test of independence, data are arranged in a contingency table, and observed cell counts are compared with expected cell counts under the assumption that the variables are independent. If the observed counts differ substantially from the expected counts, there is evidence of association between the categorical variables. For example, a chi-square test could examine whether voting preference is associated with age group or whether product preference differs by region. Option B describes comparing means, which is typically handled by a t-test or related procedure. Option C refers to regression, which models relationships between variables, often involving quantitative outcomes. Option D refers to variance, which is tested using different procedures depending on context. The key phrase is “categorical variables”: chi-square methods work with counts in categories. Study Guide references/topics: categorical data, contingency tables, chi-square test, association.

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This question uses the complement rule of probability. If the probability that it rains is 0.25, then the probability that it does not rain is the complement of that event. The complement rule states that P(not A) = 1 − P(A). Here, A represents the event “it rains,” so P(not rain) = 1 − 0.25 = 0.75. This means there is a 75% probability that rain does not occur. Option B repeats the probability of rain rather than its complement. Option C would imply rain and no rain are equally likely, which is not supported by the given probability. Option D would mean no rain is certain, but a 0.25 probability of rain means rain is still possible. The key technical point is that an event and its complement must sum to 1. Study Guide references/topics: probability, complement rule, event probability, sample-space total.

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