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

SectionObjectives
Topic 1: Probability Theory- Fundamental Probability Concepts
  • 1. Conditional probability
    • 2. Basic probability rules
      • 3. Independent vs dependent events
        - Probability Distributions
        • 1. Binomial distribution basics
          • 2. Normal distribution
            Topic 2: 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 3: Regression and Modeling- Linear Relationships
                    • 1. Simple linear regression
                      • 2. Slope and intercept interpretation
                        Topic 4: 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. Outliers and relationships
                                • 2. Scatter plots interpretation
                                  • 3. Correlation

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                                    Applied-Probability-and-Statistics real questions - Testking real exam - Applied Probability and Statistics (FZO1 C955) VCE

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                                    WGU Applied Probability and Statistics (FZO1 C955) Sample Questions (Q90-Q95):

                                    NEW QUESTION # 90
                                    A commuter has a .4 probability of taking the bus to work, a .4 probability of driving a car, and a .2 probability of cycling. The probability of being late is .15 when taking the bus, .05 when driving a car, and .1 when cycling.
                                    What is the probability of taking the bus and being on time, or driving a car and being on time?

                                    Answer: A

                                    Explanation:
                                    This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes.
                                    "Taking the bus and being on time" means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 # .15 = .85. Therefore, P(bus and on time)
                                    = .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 # .05
                                    = .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes.
                                    References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.


                                    NEW QUESTION # 91
                                    Probability it does NOT rain = 1 # 0.25 = ?

                                    Answer: D

                                    Explanation:
                                    This question uses the complement rule of probability. If the probability that it rains is 0.25, then the probability that it does not rain is the complement of that event. The complement rule states that P(not A) = 1
                                    # P(A). Here, A represents the event "it rains," so P(not rain) = 1 # 0.25 = 0.75. This means there is a 75% probability that rain does not occur. Option B repeats the probability of rain rather than its complement.
                                    Option C would imply rain and no rain are equally likely, which is not supported by the given probability.
                                    Option D would mean no rain is certain, but a 0.25 probability of rain means rain is still possible. The key technical point is that an event and its complement must sum to 1. Study Guide references/topics: probability, complement rule, event probability, sample-space total.


                                    NEW QUESTION # 92
                                    Standard error decreases when:

                                    Answer: C

                                    Explanation:
                                    The standard error of the sample mean is calculated as SE = s/#n, where s is the sample standard deviation and n is the sample size. Since n appears in the denominator, increasing the sample size reduces the standard error, assuming the standard deviation stays constant. This reflects the fact that larger samples produce more stable estimates of the population mean. Option B is incorrect because decreasing sample size increases standard error. Option C is incorrect because a larger standard deviation increases standard error by increasing the numerator. Option D is incorrect because the mean affects the center of the distribution but does not directly determine the standard error. Standard error is about precision: smaller standard error means less sampling variability and a more precise estimate. That is why larger samples are preferred in inference. Study Guide references/topics: standard error, sample size, standard deviation, sampling variability.


                                    NEW QUESTION # 93
                                    Probability of rolling 1, 2, or 3 on die = ?

                                    Answer: D

                                    Explanation:
                                    A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The event "rolling 1, 2, or 3" has three favorable outcomes: 1, 2, and 3. The probability is therefore favorable outcomes divided by total outcomes: 3/6. This fraction simplifies to 1/2. Option B, 1/3, would correspond to two favorable outcomes out of six. Option C, 1/6, is the probability of rolling one specific number only. Option D, 2/3, would require four favorable outcomes out of six. Since exactly half of the die faces are 1, 2, or 3, the correct probability is one- half. This is a direct application of theoretical probability with equally likely outcomes. Study Guide references/topics: die probability, favorable outcomes, sample space, theoretical probability.


                                    NEW QUESTION # 94
                                    Sum of probabilities in sample space = ?

                                    Answer: D

                                    Explanation:
                                    The probabilities of all outcomes in a complete sample space must sum to 1. A sample space contains every possible outcome of a probability experiment, and one of those outcomes must occur. For example, when rolling a fair six-sided die, the outcomes are 1, 2, 3, 4, 5, and 6. Each has probability 1/6, and the sum is 1/6 +
                                    1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. A total probability of 0 would mean no outcome can occur, which is impossible for a valid experiment. A total greater than 1 violates probability rules because probabilities cannot exceed certainty. "Cannot exceed 2" is too broad and mathematically invalid, since the exact total must equal
                                    1. This principle is foundational for checking probability distributions and validating whether assigned probabilities are coherent. Study Guide references/topics: sample space, probability axioms, total probability, theoretical probability.


                                    NEW QUESTION # 95
                                    ......

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