Pass the WGU Courses and Certificates Applied-Probability-and-Statistics Questions and answers with CertsForce

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.

============

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.

A standard six-sided die has three even outcomes: 2, 4, and 6. Thus, the probability that one die lands on an even number is 3/6 = 1/2. When two dice are rolled, the outcomes are independent because the result of one die does not affect the other. Therefore, the probability that both dice are even is 1/2 × 1/2 = 1/4. Equivalently, there are 36 ordered outcomes for two dice. The even-even outcomes are formed from 3 even choices on the first die and 3 even choices on the second die, giving 3 × 3 = 9 favorable outcomes. Then 9/36 simplifies to 1/4. Option B gives the probability of one die being even, not both. Options C and D do not match the count or multiplication rule. Study Guide references/topics: independent events, dice probability, multiplication rule, sample space.

============

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.

============

For a Poisson random variable, the probability of exactly k events is P(X = k) = e^−λ λ^k / k!, where λ is the mean number of events per interval. Here, λ = 4 and the question asks for exactly 2 events, so k = 2. Substitute into the formula: P(X = 2) = e^−4 × 4² / 2!. Since 4² = 16 and 2! = 2, this becomes e^−4 × 8. Numerically, e^−4 is about 0.0183, and 0.0183 × 8 ≈ 0.1465. Thus, the probability is approximately 0.1465. Options B, C, and D are unsupported by the Poisson probability formula. The Poisson model is appropriate for counts of events occurring independently over a fixed interval at a constant average rate. Study Guide references/topics: Poisson distribution, λ parameter, exact event probability, discrete distributions.

============

A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The even outcomes are 2, 4, and 6. That gives 3 favorable outcomes out of 6 total outcomes. Therefore, the probability of rolling an even number is 3/6, which simplifies to 1/2. Option A, 1/3, would correspond to 2 favorable outcomes out of 6, which is not correct here. Option C, 1/6, is the probability of rolling one specific number, such as only a 2. Option D, 2/3, would require 4 favorable outcomes out of 6. Since exactly half of the die faces are even and half are odd, the probability of rolling an even number is one-half. Study Guide references/topics: theoretical probability, equally likely outcomes, sample space, favorable outcomes.

============

When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To find the probability that the sum is 7, count all ordered pairs that produce 7. These are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 favorable outcomes. Therefore, the probability is 6/36, which simplifies to 1/6. Since the answer choices include both 6/36 and 1/6, the listed answer from the PDF uses the unsimplified equivalent form, 6/36. Option C, 5/36, omits one valid ordered pair. Option D, 1/12, is too small because it would represent only 3 favorable outcomes out of 36. This problem reinforces that dice outcomes are ordered when two dice are rolled, so (1,6) and (6,1) are counted separately. Study Guide references/topics: sample spaces, dice probability, favorable outcomes, theoretical probability.

============

A Type I error occurs when the null hypothesis, H₀, is true but the test decision rejects it. In plain statistical language, this is a false positive: the analysis concludes that there is evidence of an effect, difference, or relationship when in reality the null condition is true. The probability of making a Type I error is denoted by α, the significance level, commonly set at 0.05. A Type II error is different; it occurs when the null hypothesis is false but the test fails to reject it. That is a false negative. Option C is incorrect because rejecting a true null is not a correct decision. Option D describes not rejecting or accepting the null, not the stated event. The critical phrase is “rejecting H₀ when true,” which directly defines a Type I error. Study Guide references/topics: hypothesis testing, null hypothesis, Type I error, significance level.

============

The correlation coefficient r measures the direction and strength of a linear relationship between two quantitative variables. In the scatterplot, the points are widely dispersed, so the relationship is weak rather than strong. The fitted trend line slopes slightly downward, indicating a negative association: as golf handicap increases, math SAT score tends to decrease slightly. Because the pattern is weak and negative, r should be close to 0 but less than 0. The best match is −0.10. Option A, −0.63, would represent a moderately strong negative linear relationship, which would require the points to cluster more tightly around a downward-sloping line. Option C, 0.10, has the right weak magnitude but the wrong direction because it is positive. Option D, 0.63, is both too strong and positive. The visual evidence supports only a very slight negative linear association. References/topics from the Study Guide: scatterplots, correlation coefficient, positive and negative association, strength of linear relationship.

============

A box plot displays the five-number summary of a quantitative data set: minimum, first quartile, median, third quartile, and maximum. The median is represented by the vertical line inside the box. In the given box plot, that internal vertical line is positioned at 80 on the score axis. Therefore, the median test score for Class A is 80 points. The value 70 corresponds to the left edge of the box, which represents the first quartile, not the median. The value near 88 represents the right edge of the box, which is the third quartile. The whiskers extend to the approximate minimum and maximum scores, but those values do not determine the median. The median divides the ordered data into two equal halves, meaning about 50% of the students scored at or below 80 and about 50% scored at or above 80. References/topics from the Study Guide: box plots, five-number summary, median, quartiles.

============