Es ist uns allen bekannt, dass IT-Branche eine neue Branche und auch eine Kette ist, die die wirtschaftliche Entwicklung fördert. So ist ihre Position nicht zu ignorieren. Die WGU Applied-Probability-and-Statistics IT-Zertifizierung ist eine Methode für den Wettbewerb. Durch die WGU Applied-Probability-and-Statistics Zertifizierung werden Sie sich in allen Aspekten verbessern. Aber es ist nicht so einfach, die Prüfung zu bestehen. So empfehle ich Ihnen unsere originale Fragen. Wenn Sie die Schulungsressourcen wählen, ist ExamFragen die erste Wahl. Seine Erfolgsquote beträgt 100%. Und Sie können die WGU Applied-Probability-and-Statistics Prüfung sicher bestehen.
| Section | Weight | Objectives |
|---|---|---|
| Probability Concepts | 20% | - Probability rules, independent and dependent events - Conditional probability and Venn diagrams - Discrete and continuous probability distributions - Normal distribution and empirical rule |
| Descriptive Statistics | 25% | - Measures of spread: range, IQR, variance, standard deviation - Measures of center: mean, median, mode - Graphical displays: histograms, boxplots, scatterplots - Types of data: categorical, discrete, continuous |
| Basic Numeracy & Algebra | 15% | - Arithmetic operations, fractions, decimals, percentages - Exponents, roots, and basic formulas - Linear equations, inequalities, graphing functions |
| Inferential Statistics & Study Design | 20% | - Sampling methods and bias - Observational studies vs experiments - Hypothesis testing framework and interpretation - Confidence intervals for means/proportions |
| Correlation & Regression | 20% | - Correlation coefficient and interpretation - Predictions and limitations of regression - Simple linear regression models - Interpreting slope, intercept, and R-squared |
>> Applied-Probability-and-Statistics Fragenkatalog <<
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105. Frage
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?
Antwort: D
Begründung:
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.
106. Frage
Review the following inequality:
y < x # 1/2
Which shaded portion in the graphs corresponds to this inequality?
Antwort: D
Begründung:
The inequality y < x # 1/2 has boundary line y = x # 1/2. This line has slope 1 and y-intercept #1/2, so it crosses the y-axis slightly below the origin and rises one unit for every one unit moved to the right. Because the inequality is strict, using " < " rather than "#," the boundary line must be dashed, showing that points on the line are not included in the solution set. The solution region must be shaded below the line because y is less than the expression x # 1/2. A useful verification method is the test point (0, 0). Substituting gives 0 < 0 #
1/2, or 0 < #1/2, which is false. Therefore, the region containing the origin should not be shaded. The fourth graph has the dashed boundary with intercept #1/2 and shades the region below the line while excluding the origin. References/topics from the Study Guide: graphing linear inequalities, slope-intercept form, dashed boundaries, test-point method.
107. Frage
Boxplot shows:
Antwort: A
Begründung:
A boxplot summarizes the distribution of a quantitative variable using key positional statistics. It displays the median, first quartile, third quartile, and typically the minimum and maximum non-outlier values. It may also mark outliers separately. The box itself spans from Q1 to Q3, representing the interquartile range, or middle
50% of the data. The line inside the box marks the median. Whiskers extend to values within the non-outlier range, depending on the graphing convention. A boxplot does not primarily show frequency counts; histograms, dot plots, and bar charts are better for frequencies. A scatterplot displays the relationship between two quantitative variables. Probability is not directly displayed by a standard boxplot, although distributional interpretation can be supported by it. The correct answer is therefore median, quartiles, and outliers. Study Guide references/topics: boxplots, five-number summary, quartiles, outliers.
108. Frage
Type II error = ?
Antwort: D
Begründung:
A Type II error occurs when the null hypothesis is false, but the test fails to reject it. Many introductory materials describe this as "accepting H# when false," although the more precise statistical wording is "failing to reject H#." This is a false negative: a real effect, difference, or relationship exists, but the test does not detect enough evidence to conclude it. For example, if a new teaching method truly improves scores but the test fails to find significance, the result is a Type II error. Option B describes a Type I error, which is rejecting a true null hypothesis. Option C is a correct test decision, not an error. Option D is invalid because Type II error is a defined concept. The probability of Type II error is #, and statistical power is 1 # #. Study Guide references/topics: hypothesis testing, Type II error, null hypothesis, statistical power.
109. Frage
Probability of at least one six in 2 dice rolls = ?
Antwort: B
Begründung:
"At least one six" means one six or two sixes across the two rolls. The most efficient method is to use the complement rule. The complement of at least one six is no six on either roll. On one die roll, the probability of not rolling a six is 5/6. Since the two die rolls are independent, the probability of no six on both rolls is (5/6) (5/6) = 25/36. Therefore, the probability of at least one six is 1 # 25/36 = 11/36. This includes all outcomes with a six on the first roll, a six on the second roll, or sixes on both rolls. Option B gives the probability of a six on one single roll only. Option C gives the probability of rolling two sixes. Option D gives the complement, no sixes. Study Guide references/topics: complement rule, independent events, dice probability, sample spaces.
110. Frage
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