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

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.

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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.

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For a Poisson distribution, the mean is λ, pronounced “lambda.” The parameter λ represents the expected number of events occurring in a fixed interval, such as calls per hour, defects per batch, or arrivals per minute. A defining feature of the Poisson distribution is that both the mean and variance are equal to λ. Therefore, if X follows a Poisson distribution with λ = 4, then the expected value, or mean, is 4, and the variance is also 4. Option B and option C are only correct in special cases where λ happens to equal 0 or 1, not generally. Option D is conceptually related because the Poisson variance equals λ, but the question asks for the mean, and the parameter name is λ. The Poisson distribution is used for discrete counts of events over a fixed interval under a constant average rate. Study Guide references/topics: Poisson distribution, expected value, variance, λ parameter.

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For a Poisson distribution, the parameter λ represents the mean number of events occurring in a fixed interval. Therefore, if λ = 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 λ, 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, λ parameter, expected value, count data.

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The z-score formula is z = (x − μ)/σ, where x is the observed value, μ is the population mean, and σ is the population standard deviation. A z-score standardizes a raw value by measuring how many standard deviations it lies above or below the mean. The numerator, x − μ, gives the deviation from the mean. Dividing by σ converts that deviation into standard deviation units. If z is positive, the value is above the mean. If z is negative, the value is below the mean. If z = 0, the value equals the mean. Option B incorrectly adds the value and mean, which does not measure deviation. Option C ignores the mean, and option D ignores the observed value entirely. The z-score is central for normal distribution calculations because it allows values from different scales to be compared using the standard normal distribution. Study Guide references/topics: z-scores, standard normal distribution, mean, standard deviation.

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A z-score measures how many standard deviations a data value is from the mean. The formula is z = (x − μ)/σ, where x is the observed value, μ is the mean, and σ is the standard deviation. Since standard deviation is positive, the sign of the z-score depends on the numerator, x − μ. If x is less than the mean, then x − μ is negative, producing a negative z-score. Therefore, a negative z-score indicates that the observed value is below the mean. A positive z-score indicates a value above the mean, while a z-score of 0 indicates a value exactly equal to the mean. Option D is incorrect because the sign of the z-score gives direct positional information relative to the mean. Study Guide references/topics: z-scores, standard normal distribution, mean, standard deviation.

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The coefficient of determination, R², measures the proportion of variability in the response variable Y that is explained by the regression model using explanatory variable X. If R² = 0.64, then 64% of the variation in Y is explained by its linear relationship with X. The remaining 36% is unexplained by the model and may reflect other variables, random variation, measurement error, or nonlinear patterns. Option B is incorrect because R² is not the same as correlation. In simple linear regression, correlation r would be either +0.8 or −0.8, depending on the slope direction, because r² = 0.64. Option C is incorrect because the slope measures change in Y per one-unit change in X, not explained variation. Option D is incorrect because R² is not a probability of making a correct prediction. It is a proportion of variation accounted for by the model. Study Guide references/topics: regression, coefficient of determination, explained variation, linear modeling.

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A z-score expresses how many standard deviations a value lies above or below the mean. The formula is z = (x − μ)/σ. To solve for the raw value x, rearrange the formula: x = μ + zσ. Here, the mean μ is 100, the standard deviation σ is 15, and z = 2. Substitute these values: x = 100 + 2(15) = 100 + 30 = 130. Therefore, the raw value is 130. This means the value is two standard deviations above the mean. Option B, 115, would correspond to z = 1. Option C, 125, is not exactly two standard deviations above 100 when σ = 15. Option D, 110, is less than one standard deviation above the mean. The correct value must add two full standard deviations to the mean. Study Guide references/topics: z-scores, standardization, normal distribution, raw score conversion.

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For a normal distribution, the empirical rule, also called the 68-95-99.7 rule, describes how data are distributed around the mean. Approximately 68% of observations fall within 1 standard deviation of the mean, approximately 95% fall within 2 standard deviations, and approximately 99.7% fall within 3 standard deviations. The question asks specifically for the percentage within 2 standard deviations, so the correct value is 95%. This interval is written as μ − 2σ to μ + 2σ, where μ is the mean and σ is the standard deviation. Option B corresponds to one standard deviation, not two. Option C is an imprecise version of the three-standard-deviation rule, and option D has no standard empirical-rule interpretation in this context. This rule is frequently used to estimate ranges in bell-shaped distributions. Study Guide references/topics: normal distribution, empirical rule, standard deviation, distribution spread.

Mutually exclusive events cannot occur at the same time. If event A occurs, event B cannot occur, and if event B occurs, event A cannot occur. Therefore, the probability of both events occurring together is zero: P(A and B) = 0. For example, when rolling one die, the event “roll a 2” and the event “roll a 5” are mutually exclusive because one roll cannot produce both outcomes simultaneously. Option B applies to independent events, not mutually exclusive events. In fact, mutually exclusive events with positive probabilities are not independent because the occurrence of one eliminates the possibility of the other. Option C is part of the addition rule for mutually exclusive events when calculating P(A or B), not P(A and B). Option D would mean both events always occur together, which contradicts mutual exclusivity. Study Guide references/topics: mutually exclusive events, joint probability, addition rule, probability foundations.

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