Applied-Probability-and-Statistics Exam Dumps - WGU Courses and Certificates Questions and Answers

A t-distribution is used for inference about a population mean when the population standard deviation σ is unknown, especially with a small sample. In that situation, the sample standard deviation s must estimate σ, creating extra uncertainty. The t-distribution accounts for this added uncertainty with heavier tails than the standard normal distribution. As sample size increases, the t-distribution becomes closer to the normal distribution. Option B describes a z-procedure setting, where σ is known and the normal distribution may be used. Option C is incorrect because categorical data are usually analyzed with proportions or chi-square procedures, not t-distributions for means. Option D is incomplete: t-procedures typically require quantitative continuous or approximately continuous data, but the decisive condition in the answer set is unknown σ with a small sample. Study Guide references/topics: t-distribution, unknown standard deviation, small samples, inference for means.

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Expected value is the long-run average value of a random variable. For a discrete random variable, it is calculated by multiplying each possible value by its probability and then adding those products. Here, the expression is already structured as an expected value calculation: 1×0.2 + 2×0.5 + 3×0.3. Compute each product: 1×0.2 = 0.2, 2×0.5 = 1.0, and 3×0.3 = 0.9. Add them: 0.2 + 1.0 + 0.9 = 2.1. Therefore, the expected value is 2.1. This does not mean the random variable must equal 2.1 in a single trial; it means that over many repetitions, the average outcome would approach 2.1. Option B is close but omits part of the weighted contribution. Options C and D do not match the weighted-average computation. Study Guide references/topics: expected value, discrete random variables, weighted average, probability distributions.

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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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This is a binomial probability problem because there is a fixed number of independent trials, each trial has two outcomes, and the probability of success is constant. Let success be flipping heads. There are 3 coin flips, and the probability of heads on each flip is 1/2. The event “exactly two heads” can occur in three ordered ways: HHT, HTH, and THH. Each sequence has probability (1/2)(1/2)(1/2) = 1/8. Since there are three favorable sequences, the total probability is 3 × 1/8 = 3/8. Equivalently, use the binomial formula: C(3,2)(1/2)^2(1/2)^1 = 3 × 1/8 = 3/8. Option B would be too large, and option D counts only one favorable sequence rather than all three. Study Guide references/topics: binomial probability, combinations, independent trials, coin-flip sample spaces.

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The equation y = 40x + 75 is written in slope-intercept form, y = mx + b. In this form, m represents the rate of change and b represents the initial value. The constant term 75 means the graph must cross the y-axis at 75, because the studio charges $75 before any hourly time is added. The coefficient 40 means that for every additional hour of studio time, the total cost increases by $40. Therefore, the correct graph must begin at y = 75 when x = 0 and rise at a steady linear rate of 40 dollars per hour. For example, at x = 6 hours, the model gives y = 40(6) + 75 = 315, so the line should pass near the point (6, 315). Among the displayed graphs, Graph 4 matches both the y-intercept and the slope. References/topics from the Study Guide: basic algebra, linear equations, slope-intercept form, interpreting rate of change.

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The standard error of the mean measures the expected sampling variability of the sample mean. It is calculated by dividing the sample standard deviation by the square root of the sample size: SE = s / √n. In this question, the standard deviation is 5 and the sample size is 25. Therefore, SE = 5 / √25 = 5 / 5 = 1. The mean of 50 identifies the center of the sample distribution but is not directly used in the standard error calculation. Option B, 5, is the standard deviation, not the standard error. Option C, 25, is the sample size. Option D, 0.2, would result from incorrectly dividing 5 by 25 rather than by the square root of 25. The distinction between standard deviation and standard error is important: standard deviation describes spread among individual observations, while standard error describes spread among sample means. Study Guide references/topics: standard error, standard deviation, sample size, sampling distributions.

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In a standard deck of playing cards, every card is either red or black. Red cards consist of hearts and diamonds, and black cards consist of clubs and spades. There are 26 red cards and 26 black cards, for a total of 52 cards. The event “red or black” includes all possible cards in the deck. Therefore, the probability is 52 favorable outcomes out of 52 total outcomes, which equals 1. A probability of 1 means the event is certain. Option B, 1/2, would be the probability of drawing a red card alone or a black card alone, not red or black. Option C would mean the event is impossible, which is false. Option D would correspond to one suit out of four, such as hearts only. The word “or” combines both color categories, covering the entire sample space. Study Guide references/topics: sample space, certainty, addition rule, card probability.

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Probability values must lie between 0 and 1, inclusive. A probability of 0 means an event is impossible, while a probability of 1 means an event is certain. Values between 0 and 1 represent events with varying degrees of likelihood. A negative probability has no valid interpretation in standard probability theory because probability measures relative likelihood or long-run frequency, and neither can be less than zero. Option B is allowed because impossible events have probability 0. Option C is allowed because certain events have probability 1. Option D is allowed because ordinary uncertain events have probabilities strictly between 0 and 1. For example, the probability of rolling a 3 on a fair die is 1/6, which lies between 0 and 1. The only invalid option listed is negative probability. Study Guide references/topics: probability axioms, probability range, impossible events, certain events.

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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.

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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