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

SectionObjectives
Topic 1: Descriptive Statistics- Data summarization
  • 1. Measures of variability (range, variance, standard deviation)
    • 2. Measures of central tendency (mean, median, mode)
      - Data visualization
      • 1. Histograms and frequency distributions
        • 2. Box plots and interpretation
          Topic 2: Probability- Probability rules
          • 1. Bayes’ theorem (introductory level)
            • 2. Addition and multiplication rules
              - Fundamental probability concepts
              • 1. Events and sample spaces
                • 2. Conditional probability and independence
                  Topic 3: Statistical Inference- Hypothesis testing
                  • 1. t-tests and z-tests (basic application)
                    • 2. Null and alternative hypotheses
                      - Estimation
                      • 1. Confidence intervals for means and proportions
                        Topic 4: Regression and Correlation- Relationship analysis
                        • 1. Correlation coefficient interpretation
                          • 2. Simple linear regression basics
                            Topic 5: Probability Distributions- Continuous distributions
                            • 1. Standard normal and z-scores
                              • 2. Normal distribution
                                - Discrete distributions
                                • 1. Poisson distribution (introductory use cases)
                                  • 2. Binomial distribution

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q70-Q75):

                                    NEW QUESTION # 70
                                    A person has a .5 probability of taking a train to work, a .3 probability of carpooling, and a .2 probability of walking. The probability of being late is .2 when taking the train, .1 when carpooling, and .05 when walking.
                                    What is the probability of taking the train and being on time, or walking and being on time?

                                    Answer: C

                                    Explanation:
                                    This problem combines complements, joint probabilities, and addition of mutually exclusive outcomes. First compute the probability of taking the train and being on time. The probability of being late when taking the train is .2, so the probability of being on time by train is 1 # .2 = .8. Thus, P(train and on time) = .5 × .8 = .40.
                                    Next compute the probability of walking and being on time. The probability of being late when walking is .
                                    05, so the probability of being on time while walking is 1 # .05 = .95. Thus, P(walking and on time) = .2 × .95
                                    = .19. Since a person cannot both take the train and walk as the selected commuting mode, these outcomes are mutually exclusive. Add the two results: .40 + .19 = .59. References/topics from the Study Guide:
                                    complements, conditional probability, joint probability, addition rule.


                                    NEW QUESTION # 71
                                    In a class of 40 students, the following data were collected on their favorite sport:
                                    * Soccer: 10 students
                                    * Basketball: 15 students
                                    * Tennis: 5 students
                                    * Swimming: 10 students
                                    What is the relative frequency of students whose favorite sport is not basketball?

                                    Answer: B

                                    Explanation:
                                    Relative frequency is the proportion of observations in a category compared with the total number of observations. The class has 40 students. Basketball accounts for 15 students, so the number of students whose favorite sport is not basketball is 40 # 15 = 25. The relative frequency is therefore 25/40. Dividing gives
                                    0.625. This means 62.5% of the class selected a sport other than basketball. The same result can be found by adding the non-basketball categories directly: soccer 10, tennis 5, and swimming 10, giving 10 + 5 + 10 = 25 students. Option A, 0.25, corresponds to 10 out of 40 and would represent soccer or swimming alone. Option B, 0.50, corresponds to 20 out of 40. Option D, 0.75, would imply 30 students, which is too high. References
                                    /topics from the Study Guide: relative frequency, categorical data, frequency tables, proportions.


                                    NEW QUESTION # 72
                                    Conditional probability P(A|B) = P(A and B)/P(B) #

                                    Answer: D

                                    Explanation:
                                    The formula for conditional probability is P(A|B) = P(A and B)/P(B), provided P(B) > 0. The notation P(A|B) is read as "the probability of A given B." The denominator P(B) restricts the sample space to cases in which B occurs, and the numerator P(A and B) counts the portion of that restricted group in which A also occurs.
                                    Therefore, the statement is true. Option B is incorrect because the formula is the standard definition of conditional probability. Option C is incorrect because if A and B are mutually exclusive, then P(A and B) = 0, which creates a special case rather than the general definition. Option D is invalid because conditional probability is undefined when P(B) = 0. The condition must have positive probability. Study Guide references
                                    /topics: conditional probability, joint probability, event intersection, probability rules.


                                    NEW QUESTION # 73
                                    Standard error decreases when:

                                    Answer: A

                                    Explanation:
                                    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.


                                    NEW QUESTION # 74
                                    A school surveyed student participation in extracurricular activities. The survey found:
                                    * The drama club had 50% of students participate.
                                    * The science club had 30% of students participate.
                                    * Both clubs had 15% of students participate.
                                    What is the probability that a student participated in the drama club, given that they attended the science club?

                                    Answer: B

                                    Explanation:
                                    This question asks for a conditional probability. The phrase "given that they attended the science club" means the science club group becomes the restricted reference group. The required probability is P(drama club | science club), which is calculated as P(drama and science) divided by P(science). The problem states that 15% of students participated in both clubs, so P(drama and science) = 0.15. It also states that 30% participated in the science club, so P(science) = 0.30. Therefore, P(drama | science) = 0.15 ÷ 0.30 = 0.50. This means that among students who attended the science club, half also participated in drama club. Option A gives the joint probability, not the conditional probability. Option B gives only the science club participation rate. Option C does not match the conditional probability calculation. References/topics from the Study Guide: conditional probability, joint probability, probability ratios, two-event probability.


                                    NEW QUESTION # 75
                                    ......

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