Free Practice Questions for WGU Applied-Probability-and-Statistics Exam

This question requires comparing conditional percentages from a two-way table. For residents, 150 out of 200 are in favor of the proposed transportation system. The resident support rate is therefore 150/200 = 0.75, or 75%. For business owners, 30 out of 100 are in favor. The business owner support rate is 30/100 = 0.30, or 30%. The difference is 75% − 30% = 45 percentage points, which is a large gap. Therefore, a resident is much more likely than a business owner to favor the new transportation system. Option A understates the magnitude of the difference by calling it slight. Options B and C reverse the direction of the comparison because residents have the higher support rate, not the lower one. The appropriate statistical tool here is conditional percentage because support is being compared within each respondent group. References/topics from the Study Guide: contingency tables, conditional percentages, two-variable categorical data, comparative proportions.

The sample space of a probability experiment is the complete set of all possible outcomes. In this experiment, the spinner has four equal sections labeled A, B, C, and D. Since one spin can land on exactly one of those four labeled sections, the possible outcomes are A, B, C, and D. Therefore, the sample space is {A, B, C, D}. Option A incorrectly adds E, which is not listed as a section on the spinner. Option B groups labels into combined outcomes, but the experiment is a single spin, not a two-part or paired event. Option C uses numerical labels, but the spinner sections are identified by letters, not numbers. Equal section size affects the probability of each outcome, making each outcome have probability 1/4, but it does not change the sample space itself. References/topics from the Study Guide: probability experiments, outcomes, sample space, equally likely events.

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The median is the center value of an ordered dataset. The values 3, 5, 7, 9, and 11 are already arranged from least to greatest. Because there are five observations, the median is the third value, with two observations below it and two observations above it. The third value is 7, so the median is 7. The median is a measure of central tendency that describes the middle position of the data rather than the arithmetic average. In this dataset, the mean is also 7 because the values are evenly spaced around 7, but the median is determined by position, not by summing and dividing. Option B is the second value, option C is the fourth value, and option D is not a data value. The correct answer is the value that splits the ordered list into two equal halves. Study Guide references/topics: median, ordered data, measures of center, descriptive statistics.

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The central limit theorem states that, for sufficiently large sample sizes, the sampling distribution of the sample mean is approximately normal, regardless of the shape of the original population distribution, provided observations are independent and drawn appropriately. This is why option A is correct. The theorem does not require the population itself to be normal; if the population is normal, the sample mean is normally distributed for any sample size, but the central limit theorem is especially powerful because it applies broadly for large n. Option C is false because sample variance is an estimate of population variance, not automatically equal to it. Option D is false because standard error generally decreases as sample size increases but does not become exactly zero unless sample size is infinite or variability is absent. The central limit theorem supports confidence intervals and hypothesis tests for means. Study Guide references/topics: central limit theorem, sampling distribution, sample mean, normal approximation.

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The random variable X counts the number of heads obtained in 4 coin flips. Since it is a count, it can take only specific whole-number values: 0, 1, 2, 3, or 4. It cannot take fractional values such as 2.5 heads. A random variable with countable possible outcomes is classified as discrete. This situation also fits a binomial framework because there is a fixed number of independent trials, each trial has two outcomes, and the probability of heads remains constant for a fair coin. A continuous random variable, by contrast, can take any value over an interval, such as time, weight, or height. The number of heads is not measured on a continuum; it is counted. Therefore, the correct classification is discrete. Study Guide references/topics: discrete random variables, binomial setting, coin-flip outcomes, probability distributions.

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A box plot displays the five-number summary, including the median, which is shown by the vertical line inside each box. In the exhibit, the median for Class A is positioned at approximately 65, while the median for Class B is positioned at approximately 80. The difference is therefore 80 − 65 = 15 points. Since Class A’s median is lower, the correct interpretation is that the median test score of Class A is 15 points less than that of Class B. This comparison uses a categorical grouping variable, class, and a quantitative response variable, test score. The box plot is appropriate because it allows the distributions of two groups to be compared visually using medians, quartiles, ranges, and potential spread. Options A, B, and D do not match the median locations shown by the center lines in the boxes. References/topics from the Study Guide: box plots, median, five-number summary, comparing distributions.

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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.

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The symbol α represents the significance level of a hypothesis test and is the probability of making a Type I error. A Type I error occurs when the null hypothesis is true but the test rejects it. Therefore, α = 0.05 means there is a 5% risk of rejecting a true null hypothesis under the testing procedure. This is often described as the false positive rate. Option B describes a Type II error, which involves failing to reject a false null hypothesis. Option C is incorrect because a 5% significance level corresponds to a 95% confidence level in related confidence-interval contexts, not 5% confidence. Option D is unrelated; α is not a sample-size percentage. The interpretation must connect α directly to the probability of incorrectly rejecting H₀ when H₀ is actually true. Study Guide references/topics: hypothesis testing, significance level, Type I error, null hypothesis.

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A standard deck contains 52 cards. There are 4 kings and 4 queens, one of each rank in each suit. The event is drawing a king or a queen. Since a single card cannot be both a king and a queen, these two events are mutually exclusive. Therefore, the number of favorable outcomes is 4 + 4 = 8. The probability is favorable outcomes divided by total outcomes: 8/52. This fraction can be simplified to 2/13, but the answer choice provides the unsimplified correct form. Option B, 4/52, counts only kings or only queens, not both. Option C, 1/13, is equivalent to 4/52 and again represents one rank only. Option D, 1/26, is too small and does not match the count of favorable cards. The word “or” signals that both qualifying ranks should be included. Study Guide references/topics: card probability, mutually exclusive events, addition rule, sample space.

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Empirical probability is probability determined from observed data rather than from a purely theoretical model. It is calculated by dividing the number of times an event actually occurs by the total number of trials or observations. For example, if a basketball player makes 72 free throws in 100 attempts, the empirical probability of a made free throw is 72/100 = 0.72. This differs from theoretical probability, which is derived from known equally likely outcomes, such as rolling a 3 on a fair six-sided die. It also differs from a guess or subjective estimate because empirical probability requires actual recorded evidence. The term “uniform” describes a distribution in which outcomes or intervals have equal probability, not the basis for empirical probability. Since the question asks for the definition of empirical probability, the correct selection is observed data. Study Guide references/topics: empirical probability, relative frequency, observed outcomes, probability interpretation.

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