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

SectionObjectives
Topic 1: Regression and Modeling- Linear Relationships
  • 1. Slope and intercept interpretation
    • 2. Simple linear regression
      Topic 2: Statistical Inference- Hypothesis Testing (Introductory Level)
      • 1. p-values interpretation
        • 2. Null vs alternative hypothesis
          - Estimation and Confidence Intervals
          • 1. Point estimates
            • 2. Confidence interval interpretation
              Topic 3: Probability Theory- Probability Distributions
              • 1. Binomial distribution basics
                • 2. Normal distribution
                  - Fundamental Probability Concepts
                  • 1. Conditional probability
                    • 2. Basic probability rules
                      • 3. Independent vs dependent events
                        Topic 4: Descriptive Statistics- Two Variable Data Analysis
                        • 1. Correlation
                          • 2. Outliers and relationships
                            • 3. Scatter plots interpretation
                              - Single Variable Data Analysis
                              • 1. Measures of central tendency (mean, median, mode)
                                • 2. Measures of dispersion (variance, standard deviation, range)
                                  • 3. Data visualization (histograms, box plots)

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q144-Q149):

                                    NEW QUESTION # 144
                                    The average lifespan of a dalmatian is 11.5 years, with a standard deviation of 1.5 years.
                                    What are the two values between which 95% of the data falls for the average lifespan of a dalmatian, assuming a normal distribution?

                                    Answer: A

                                    Explanation:
                                    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.


                                    NEW QUESTION # 145
                                    Independent events: P(A|B) = ?

                                    Answer: D

                                    Explanation:
                                    For independent events, the occurrence of one event does not change the probability of the other. Conditional probability P(A|B) means the probability of A occurring given that B has occurred. If A and B are independent, knowing B occurred gives no new information about A. Therefore, P(A|B) = P(A). This is one of the defining properties of independence. For example, if a coin toss and a die roll are independent, knowing the die landed on 4 does not change the probability that the coin landed heads; it remains 1/2. Option B reverses the event being measured. Option C would imply that A becomes impossible after B, which describes a different situation. Option D is an addition expression and does not represent conditional probability. The correct relationship is that the conditional probability equals the original probability when events are independent. Study Guide references/topics: independent events, conditional probability, probability rules, event relationships.


                                    NEW QUESTION # 146
                                    A single ball is drawn from an opaque bag that contains red, blue, and green balls. The probability of drawing a red ball is .3, and the probability of drawing a blue ball is .6.
                                    What is the probability of drawing a green ball?

                                    Answer: B

                                    Explanation:
                                    The probabilities of all possible outcomes in a sample space must sum to 1. In this experiment, the bag contains only red, blue, and green balls, so these three outcomes exhaust the sample space. The probability of drawing a red ball is 0.3, and the probability of drawing a blue ball is 0.6. Their combined probability is 0.3 +
                                    0.6 = 0.9. The remaining probability must belong to drawing a green ball. Therefore, P(green) = 1 # 0.9 = 0.1.
                                    This can also be viewed as a complement problem: green is the complement of drawing either red or blue when those are the only other colors available. Option B repeats the red probability, option C would make the total probability exceed 1, and option D is the combined probability of red or blue rather than green.
                                    References/topics from the Study Guide: probability sum rule, complements, sample space, mutually exclusive outcomes.


                                    NEW QUESTION # 147
                                    Probability of exactly 1 head in 2 coin flips = ?

                                    Answer: B

                                    Explanation:
                                    Two coin flips produce four equally likely ordered outcomes: HH, HT, TH, and TT. Exactly one head occurs in two of these outcomes: HT and TH. Therefore, the probability is 2 favorable outcomes out of 4 total outcomes, or 2/4 = 1/2. This can also be computed using the binomial model. There are n = 2 independent trials, success probability p = 1/2, and exactly one success is required. The binomial calculation is C(2,1)(1/2)