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

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

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q35-Q40):

                                    NEW QUESTION # 35
                                    Probability of rolling sum = 2 on 2 dice = ?

                                    Answer: A

                                    Explanation:
                                    When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To get a sum of 2, the only possible ordered pair is (1, 1). No other pair of die faces can produce a total of 2 because 1 is the smallest value on each die. Therefore, there is 1 favorable outcome out of 36 total outcomes, giving a probability of 1/36. Option B, 1/6, is the probability of one specific result on a single die, not a two-dice sum.
                                    Option C, 2/36, would require two favorable ordered pairs, but only one exists. Option D, 1/12, is also too large because it corresponds to 3 outcomes out of 36. The key is to count ordered outcomes in the full two- dice sample space. Study Guide references/topics: dice probability, sample space, favorable outcomes, theoretical probability.


                                    NEW QUESTION # 36
                                    Two mutually exclusive events: P(A and B) = ?

                                    Answer: C

                                    Explanation:
                                    Mutually exclusive events cannot occur at the same time. If event A occurs, event B cannot occur, and if event B occurs, event A cannot occur. Therefore, the probability of both events occurring together is zero: P (A and B) = 0. For example, when rolling one die, the event "roll a 2" and the event "roll a 5" are mutually exclusive because one roll cannot produce both outcomes simultaneously. Option B applies to independent events, not mutually exclusive events. In fact, mutually exclusive events with positive probabilities are not independent because the occurrence of one eliminates the possibility of the other. Option C is part of the addition rule for mutually exclusive events when calculating P(A or B), not P(A and B). Option D would mean both events always occur together, which contradicts mutual exclusivity. Study Guide references/topics:
                                    mutually exclusive events, joint probability, addition rule, probability foundations.


                                    NEW QUESTION # 37
                                    There are 20 men in a room whose heights are reflected in inches as 68, 70, 73, 67, 71, 68, 76, 70, 73, 66, 68,
                                    69, 71, 74, 73, 72, 65, 60, 70, and 78 inches respectively.
                                    What is the median of these data?

                                    Answer: A

                                    Explanation:
                                    The median is the middle value of an ordered data set. Because there are 20 observations, the number of data values is even, so the median is the average of the 10th and 11th values after the data are arranged from least to greatest. Ordering the heights gives 60, 65, 66, 67, 68, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73, 73, 73, 74, 76,
                                    78. The 10th value is 70, and the 11th value is also 70. The median is therefore (70 + 70) ÷ 2 = 70. This value divides the ordered heights into two equal halves: ten values are at or below 70, and ten values are at or above
                                    70. The median is resistant to extreme values, so the low value 60 and the high value 78 do not distort it the way they could affect the mean. References/topics from the Study Guide: median, ordered data, measures of center, quantitative data.


                                    NEW QUESTION # 38
                                    There are 10 green, 20 yellow, and 15 purple marbles in a jar.
                                    What is the probability of selecting 2 green marbles in a row and then 1 yellow marble, replacing the marble each time?

                                    Answer: D

                                    Explanation:
                                    There are 10 green, 20 yellow, and 15 purple marbles, so the total number of marbles is 10 + 20 + 15 = 45.
                                    Because each marble is replaced after selection, the probabilities remain the same on every draw, making the events independent. The probability of selecting a green marble on one draw is 10/45 = 2/9. Selecting two green marbles in a row gives (2/9)(2/9). The probability of then selecting a yellow marble is 20/45 = 4/9.
                                    Multiply the probabilities because the question asks for green and green and yellow in that exact sequence: (2
                                    /9)(2/9)(4/9) = 16/729. Replacement is essential; without replacement, the denominators and numerators would change after each draw. Since replacement keeps the sample space constant at 45 marbles each time, the independent multiplication rule applies directly. References/topics from the Study Guide: independent events, replacement, multiplication rule, compound probability.


                                    NEW QUESTION # 39
                                    Uniform distribution 0-5: P(X < 3) = ?

                                    Answer: C

                                    Explanation:
                                    For a continuous uniform distribution on the interval from 0 to 5, probability is proportional to interval length.
                                    The total interval length is 5 # 0 = 5. The event X < 3 corresponds to the interval from 0 to 3, which has length 3. Therefore, P(X < 3) = 3/5 = 0.6. This works because the uniform distribution assigns constant density across the entire interval, so any subinterval's probability equals its length divided by the total length.
                                    Option B would correspond to half the interval, such as X < 2.5. Option C would correspond to length 2 out of
                                    5. Option D incorrectly treats the cutoff value 3 as if it were a percentage rather than an interval boundary.
                                    The correct answer is 0.6, meaning 60% of the uniform distribution lies below 3. Study Guide references
                                    /topics: uniform distribution, continuous probability, interval length, probability density.


                                    NEW QUESTION # 40
                                    ......

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