WGU Applied-Probability-and-Statistics - Applied Probability and Statistics (FZO1 C955)

For a normal distribution, the empirical rule states that approximately 95% of the data fall within two standard deviations of the mean. The mean is 40, and the standard deviation is 6. Two standard deviations equal 2 × 6 = 12. To construct the 95% interval, subtract and add 12 from the mean: 40 − 12 = 28 and 40 + 12 = 52. Therefore, the range containing approximately 95% of the data is 28 to 52. Option D, 34 to 46, represents one standard deviation from the mean and corresponds to about 68% of the data. Option B, 22 to 58, represents three standard deviations from the mean and corresponds to approximately 99.7% of the data. Option C is not based on the stated standard deviation. The correct interval must be symmetric around the mean and extend exactly two standard deviations in each direction. References/topics from the Study Guide: normal distribution, empirical rule, mean, standard deviation.

A one-sample t-test is used to compare a sample mean to a hypothesized or known population mean when the population standard deviation is unknown and the data are approximately normal or the sample size is sufficiently large. The test statistic evaluates how far the sample mean is from the hypothesized mean in standard error units. Option A is therefore correct. A two-sample t-test compares means from two independent groups, so option B describes a different test. Tests of variances use procedures such as chi-square or F-based methods depending on context, so option C is not appropriate. Tests of proportions use z procedures for categorical success/failure data, not a one-sample t-test for means. The t-test is part of inferential statistics because it uses sample evidence to make a decision about a population parameter. Study Guide references/topics: one-sample t-test, sample mean, population mean, hypothesis testing.

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A t-test is used when making inferences about a population mean and the population standard deviation, σ, is unknown. In practical settings, σ is rarely available, so the sample standard deviation, s, is used as an estimate. That substitution introduces additional uncertainty, which is why the t-distribution is used instead of the standard normal distribution. The t-distribution has heavier tails, especially for small samples, reflecting the extra variability caused by estimating σ from the sample. Option B describes a z-test setting, where the population standard deviation is known. Option C is incorrect because categorical population data are usually analyzed with proportions, chi-square tests, or related categorical procedures, not mean-based t-tests. Option D is invalid because t-tests have a clear inferential role. The controlling condition is unknown σ, with inference focused on means. Study Guide references/topics: t-tests, unknown population standard deviation, sample standard deviation, inferential statistics.

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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.

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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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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.

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When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To get a sum of 2, the only possible ordered pair is (1, 1). No other pair of die faces can produce a total of 2 because 1 is the smallest value on each die. Therefore, there is 1 favorable outcome out of 36 total outcomes, giving a probability of 1/36. Option B, 1/6, is the probability of one specific result on a single die, not a two-dice sum. Option C, 2/36, would require two favorable ordered pairs, but only one exists. Option D, 1/12, is also too large because it corresponds to 3 outcomes out of 36. The key is to count ordered outcomes in the full two-dice sample space. Study Guide references/topics: dice probability, sample space, favorable outcomes, 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 inequality y ≤ −4x + 1 has boundary line y = −4x + 1. The slope is −4, so the line must decrease steeply from left to right. The y-intercept is 1, so the line must cross the vertical axis at y = 1. Because the inequality uses “≤,” the boundary line must be solid, meaning points on the line are included in the solution set. The solution region must be below the line because y is less than or equal to the expression −4x + 1. A quick test point confirms this. Substitute (0, 0): 0 ≤ −4(0) + 1, giving 0 ≤ 1, which is true. Therefore, the shaded solution region must include the origin. Graph D shows a solid, steep downward-sloping line crossing the y-axis near 1 and shades the region containing the origin and below the boundary. References/topics from the Study Guide: linear inequalities, slope-intercept form, boundary lines, test-point method.

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A 95% confidence interval provides a plausible range of values for an unknown population mean based on sample data. The interval (45, 55) means the method used to construct the interval captures the true population mean in approximately 95% of repeated samples under the same conditions. In practical exam language, the correct interpretation is that the population mean is estimated to lie within the interval with 95% confidence. Option B states the midpoint of the interval, 50, but the question asks for the correct confidence interval interpretation, not just the center. Option C is incorrect because the population mean is not always the upper endpoint. Option D is imprecise because once an interval is computed, it is fixed; the confidence level refers to the long-run success rate of the procedure, not a changing probability attached to that exact fixed interval. Study Guide references/topics: confidence intervals, population mean, point estimate, margin of error.

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