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WGU Applied-Probability-and-Statistics Exam Syllabus Topics:

SectionObjectives
Probability Theory- Probability Distributions
  • 1. Normal distribution
    • 2. Binomial distribution basics
      - Fundamental Probability Concepts
      • 1. Conditional probability
        • 2. Independent vs dependent events
          • 3. Basic probability rules
            Regression and Modeling- Linear Relationships
            • 1. Slope and intercept interpretation
              • 2. Simple linear regression
                Descriptive Statistics- Single Variable Data Analysis
                • 1. Measures of central tendency (mean, median, mode)
                  • 2. Data visualization (histograms, box plots)
                    • 3. Measures of dispersion (variance, standard deviation, range)
                      - Two Variable Data Analysis
                      • 1. Outliers and relationships
                        • 2. Correlation
                          • 3. Scatter plots interpretation
                            Statistical Inference- Estimation and Confidence Intervals
                            • 1. Point estimates
                              • 2. Confidence interval interpretation
                                - Hypothesis Testing (Introductory Level)
                                • 1. p-values interpretation
                                  • 2. Null vs alternative hypothesis

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q155-Q160):

                                    NEW QUESTION # 155
                                    P(A) = 0.3, P(B) = 0.4, A and B independent. P(A and B) = ?

                                    Answer: D

                                    Explanation:
                                    For independent events, the occurrence of one event does not change the probability of the other event. The multiplication rule states that if A and B are independent, then P(A and B) = P(A) × P(B). Here, P(A) = 0.3 and P(B) = 0.4. Therefore, P(A and B) = 0.3 × 0.4 = 0.12. This means there is a 12% probability that both events occur. Option B, 0.7, incorrectly adds the probabilities and would only be part of an "or" calculation, not an "and" calculation. Option D repeats P(A), and option C does not follow from the multiplication rule.
                                    The phrase "A and B independent" is the decisive condition because it permits direct multiplication without adjusting for conditional probability. Study Guide references/topics: independent events, multiplication rule, compound probability, joint probability.


                                    NEW QUESTION # 156
                                    Binomial expected value = ?

                                    Answer: C

                                    Explanation:
                                    For a binomial random variable, the expected value is E(X) = np, where n is the number of trials and p is the probability of success on each trial. The binomial model applies when there is a fixed number of independent trials, each trial has two outcomes, and the probability of success remains constant. The expected value represents the long-run average number of successes across repeated binomial experiments. For example, if a test has 20 independent questions and the probability of success on each is 0.75, the expected number correct is 20 × 0.75 = 15. Option B, n/p, and option C, p/n, do not represent the binomial mean. Option D, n + p, incorrectly adds the number of trials and success probability, which are not comparable units. The formula np captures both how many opportunities there are and how likely success is on each opportunity. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.


                                    NEW QUESTION # 157
                                    A teacher plots the test scores of a class using the box plot.

                                    What is the interquartile range?

                                    Answer: C

                                    Explanation:
                                    The interquartile range, abbreviated IQR, measures the spread of the middle 50% of a data set. In a box plot, the left edge of the box represents the first quartile, Q1, and the right edge of the box represents the third quartile, Q3. The IQR is computed as Q3 # Q1. In the displayed box plot, Q1 is located at 70 points and Q3 is located at approximately 88 points. Therefore, the interquartile range is 88 # 70 = 18 points. The median, shown by the line inside the box, is about 80 points, but the median is not used in the IQR calculation. The whiskers show the approximate minimum and maximum values, but those define the full range, not the interquartile range. The correct answer is 18 points. References/topics from the Study Guide: box plots, quartiles, interquartile range, five-number summary.


                                    NEW QUESTION # 158
                                    A tech company surveys its employees to assess remote work preferences. The survey reveals that 55% of employees prefer working remotely, while the remaining employees prefer working in the office. Of the employees who prefer remote work, 75% report high productivity. Among those who prefer working in the office, only 50% report high productivity.
                                    Which percentage of employees prefer working in the office and report high productivity?

                                    Answer: C

                                    Explanation:
                                    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.


                                    NEW QUESTION # 159
                                    In a normal distribution, 95% of data lies within:

                                    Answer: D

                                    Explanation:
                                    For a normal distribution, the empirical rule states that approximately 68% of data fall within one standard deviation of the mean, approximately 95% fall within two standard deviations, and approximately 99.7% fall within three standard deviations. The notation ±2 SD means two standard deviations below the mean to two standard deviations above the mean, or # # 2# to # + 2#. Therefore, the interval containing about 95% of normally distributed observations is ±2 standard deviations. Option B corresponds to approximately 68%, not
                                    95%. Option C corresponds to approximately 99.7%, and option D extends beyond the standard empirical- rule benchmarks. This concept is central when estimating the typical spread of bell-shaped data, such as test scores, biological measurements, or repeated measurement errors. The correct answer is ±2 SD because it matches the 95% portion of the 68-95-99.7 rule. Study Guide references/topics: normal distribution, empirical rule, standard deviation, distribution spread.


                                    NEW QUESTION # 160
                                    ......

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