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WGU Applied-Probability-and-Statistics Exam Syllabus Topics:
| Section | Objectives |
|---|
| Topic 1: Statistical Inference | - Estimation and Confidence Intervals
- 1. Confidence interval interpretation
- 2. Point estimates
- Hypothesis Testing (Introductory Level)
- 1. Null vs alternative hypothesis
- 2. p-values interpretation
|
| Topic 2: Descriptive Statistics | - 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)
- Two Variable Data Analysis
- 1. Outliers and relationships
- 2. Scatter plots interpretation
- 3. Correlation
|
| Topic 3: 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 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) Applied-Probability-and-Statistics Prüfungsfragen mit Lösungen (Q16-Q21):
16. Frage
The pie chart shows the contribution of different renewable energy sources to the total energy production.

What is true about the given data?
- A. Biomass and geothermal together contribute more than wind energy to the total production.
- B. Hydropower and biomass together contribute more than solar energy to the total production.
- C. Solar power contributes the most to the total production.
- D. Geothermal contributes the least to the total production.
Antwort: C
Begründung:
A pie chart displays how a whole is divided into categorical parts, usually as percentages totaling 100%. The chart shows solar at 35%, hydropower at 30%, wind at 20%, geothermal at 10%, and biomass at 5%. The largest percentage is solar, so solar power contributes the most to total energy production. Option A is false because hydropower plus biomass equals 30% + 5% = 35%, which is equal to solar, not more than solar.
Option B is false because geothermal is 10%, while biomass is only 5%; therefore, biomass contributes the least. Option C is false because biomass plus geothermal equals 5% + 10% = 15%, which is less than wind at
20%. The correct interpretation requires comparing category percentages accurately rather than relying on visual size alone. References/topics from the Study Guide: pie charts, categorical data, percentages, interpreting graphical displays.
17. Frage
A teacher plots the test scores of a class using the box plot.

What is the interquartile range?
- A. 22 points
- B. 10 points
- C. 8 points
- D. 18 points
Antwort: D
Begründung:
The interquartile range, abbreviated IQR, measures the spread of the middle 50% of a data set. In a box plot, the left edge of the box represents the first quartile, Q1, and the right edge of the box represents the third quartile, Q3. The IQR is computed as Q3 # Q1. In the displayed box plot, Q1 is located at 70 points and Q3 is located at approximately 88 points. Therefore, the interquartile range is 88 # 70 = 18 points. The median, shown by the line inside the box, is about 80 points, but the median is not used in the IQR calculation. The whiskers show the approximate minimum and maximum values, but those define the full range, not the interquartile range. The correct answer is 18 points. References/topics from the Study Guide: box plots, quartiles, interquartile range, five-number summary.
18. Frage
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?
Antwort: B
Begründung:
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.
19. Frage
Mean of Poisson # = 5 = ?
Antwort: B
Begründung:
For a Poisson distribution, the parameter # represents the mean number of events occurring in a fixed interval.
Therefore, if # = 5, the mean is 5. This means that over many repeated intervals of the same size, the long-run average number of events per interval would be 5. The Poisson distribution is used for count data, such as calls per hour, accidents per week, or defects per batch, when events occur independently and at a constant average rate. A special property of the Poisson distribution is that its variance also equals #, so in this case the variance would also be 5. However, the question asks for the mean, so the direct answer is 5. Options B, C, and D are not supported by the stated parameter. Study Guide references/topics: Poisson distribution, # parameter, expected value, count data.
20. Frage
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?
Antwort: B
Begründung:
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.
21. Frage
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