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Real4Prep offers verified, authentic WGU Applied-Probability-and-Statistics Real Questions and answers, which are essential for passing the Applied Probability and Statistics (FZO1 C955) (Applied-Probability-and-Statistics). These questions and answers have been designed by Sitecore experts and can be easily downloaded on a PC, MacBook, or smartphone for comfortable and convenient learning.
| Section | Objectives |
|---|---|
| Regression and Modeling | - Linear Relationships
|
| Descriptive Statistics | - Single Variable Data Analysis
|
| Statistical Inference | - Estimation and Confidence Intervals
|
| Probability Theory | - Fundamental Probability Concepts
|
>> Reliable Applied-Probability-and-Statistics Study Guide <<
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NEW QUESTION # 15
Variance of Poisson # = 5 = ?
Answer: C
Explanation:
A defining property of the Poisson distribution is that its variance equals its mean, and both are equal to #.
Since the question states # = 5, the variance is also 5. This property distinguishes the Poisson distribution from many other probability distributions. The mean represents the expected number of events per interval, while the variance describes the spread of the event count around that mean. In a Poisson model with # = 5, event counts tend to vary around 5, and the numerical variance is 5. Option B, 4, option C, 0, and option D, 1, do not follow from the Poisson variance rule. The result is not obtained by squaring # or taking its square root; it is simply equal to #. Study Guide references/topics: Poisson distribution, variance, mean, # parameter.
NEW QUESTION # 16
Sample mean = 40, SD = 8, n = 16. Standard error = ?
Answer: D
Explanation:
The standard error of the sample mean is calculated as SE = s/#n, where s is the sample standard deviation and n is the sample size. Here, the standard deviation is 8 and the sample size is 16. The square root of 16 is 4, so SE = 8/4 = 2. The sample mean of 40 is not used directly in the standard error formula; it identifies the center of the sample, not the variability of the sample mean across repeated samples. Option B, 8, is the standard deviation, not the standard error. Option D, 4, is the square root of the sample size. Option C, 0.5, would result from an incorrect calculation. Standard error decreases as sample size increases because larger samples provide more precise estimates. Study Guide references/topics: standard error, standard deviation, sample size, sampling distribution.
NEW QUESTION # 17
Conditional probability formula:
Answer: D
Explanation:
Conditional probability measures the probability that event A occurs given that event B has already occurred.
The formula is P(A|B) = P(A and B)/P(B), assuming P(B) is greater than zero. The denominator P(B) restricts the sample space to cases where B occurs, and the numerator P(A and B) counts cases where both A and B occur within that restricted space. Option B is incorrect because it divides P(A) by P(B) without requiring overlap between A and B. Option C reverses the conditioning and would relate to P(B|A) only if the numerator were P(A and B). Option D applies to independent events when finding P(A and B), not to conditional probability in general. The vertical bar in P(A|B) is read as "given," and it signals that the denominator should be the probability of the given event. Study Guide references/topics: conditional probability, joint probability, event intersection, probability rules.
NEW QUESTION # 18
A person has a .5 probability of taking a train to work, a .3 probability of carpooling, and a .2 probability of walking. The probability of being late is .2 when taking the train, .1 when carpooling, and .05 when walking.
What is the probability of taking the train and being on time, or walking and being on time?
Answer: D
Explanation:
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.
NEW QUESTION # 19
A t-distribution is used when:
Answer: C
Explanation:
A t-distribution is used for inference about a population mean when the population standard deviation # is unknown, especially with a small sample. In that situation, the sample standard deviation s must estimate #, 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 # 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 # with a small sample. Study Guide references/topics: t-distribution, unknown standard deviation, small samples, inference for means.
NEW QUESTION # 20
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