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

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

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q100-Q105):

                                    NEW QUESTION # 100
                                    What is the result of 2/5 + 1/3 # 1/4?

                                    Answer: B

                                    Explanation:
                                    To add and subtract fractions, first rewrite all fractions using a common denominator. The denominators are
                                    5, 3, and 4. Their least common denominator is 60. Convert each fraction: 2/5 = 24/60, 1/3 = 20/60, and 1/4 =
                                    15/60. Now combine the numerators while keeping the common denominator: 24/60 + 20/60 # 15/60 = (24 +
                                    20 # 15)/60 = 29/60. The fraction 29/60 is already in simplest form because 29 is prime and does not divide
                                    60. Option A, 13/60, and option B, 11/60, result from incorrect numerator operations. Option D, 19/60, also fails to preserve the correct addition before subtraction. The essential technical step is establishing a common denominator before performing arithmetic operations. References/topics from the Study Guide: fractions, least common denominator, fraction addition, fraction subtraction.


                                    NEW QUESTION # 101
                                    Ten teachers each had a jar of jelly beans. The jars had this number of jelly beans in each respective jar: 50,
                                    60, 80, 80, 100, 100, 100, 120, 120, and 140.
                                    What is the mean of these data?

                                    Answer: B

                                    Explanation:
                                    The mean is the arithmetic average of a quantitative data set. To calculate it, add all observed values and divide by the number of observations. The jelly bean counts are 50, 60, 80, 80, 100, 100, 100, 120, 120, and
                                    140. Their sum is 950. There are 10 jars, so the mean is 950 ÷ 10 = 95. This means the average number of jelly beans per jar is 95. Option A, 100, may seem plausible because 100 appears frequently and is near the center of the data, but frequency alone determines the mode, not the mean. Option C, 80, is one of the repeated values but not the average. Option B, 120, is too high because several observations fall well below
                                    120. The mean uses every value in the data set, including the low values 50 and 60 and the high value 140.
                                    References/topics from the Study Guide: mean, measures of center, quantitative data, arithmetic average.


                                    NEW QUESTION # 102
                                    Determine z in the equation:
                                    1/(6z) = 2/9

                                    Answer: C

                                    Explanation:
                                    The equation 1/(6z) = 2/9 is a proportion because it sets two fractions equal to each other. To solve it, use cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the denominator of the first fraction by the numerator of the second fraction: 1 × 9 = 2 × 6z. This gives 9 = 12z. To isolate z, divide both sides by 12: z = 9/12. This fraction can also be simplified to 3/4, but the answer choices list the unsimplified equivalent form 9/12. The other options do not satisfy the original equation. For example, if z = 2/54, then 6z is very small and the left side becomes much larger than 2/9. If z =
                                    12/9 or 18/2, the denominator becomes too large, making the left side too small. References/topics from the Study Guide: proportions, solving equations, cross multiplication, equivalent fractions.


                                    NEW QUESTION # 103
                                    In a normal distribution, 95% of data lies within:

                                    Answer: C

                                    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 # 104
                                    A class of 25 students includes 10 who play soccer, 8 who play basketball, and 7 who play neither sport.
                                    What is the probability of randomly selecting a student who does not play soccer?

                                    Answer: A

                                    Explanation:
                                    The probability of selecting a student who does not play soccer is found by identifying the complement of the soccer group. There are 25 students in the class, and 10 students play soccer. Therefore, the number of students who do not play soccer is 25 # 10 = 15. The probability is then the number of favorable outcomes divided by the total number of possible outcomes: 15/25 = 0.60. The information about basketball and neither sport provides context, but the direct complement method is sufficient because the question asks only who does not play soccer. Option B, .40, represents the probability of selecting a student who does play soccer, since 10/25 = .40. Option A, .30, is associated with neither sport if interpreted approximately from 7/25, but it is not the requested event. Option C, .50, does not match the count structure. References/topics from the Study Guide: probability, complements, favorable outcomes, relative frequency.


                                    NEW QUESTION # 105
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