Applied-Probability-and-Statistics Exam Dumps - Applied Probability and Statistics (FZO1 C955)

The data describe frequencies for a single variable: weekly exercise duration. Each duration, 1 hour through 4 hours, has an associated count of participants. A bar chart is the most appropriate display because it uses separate bars to show how many observations fall into each category or discrete value. Here, the bar above 1 hour would have height 3, the bar above 2 hours would have height 6, the bar above 3 hours would have height 4, and the bar above 4 hours would have height 2. A line graph is inappropriate because the data are not measuring change over time. A pie chart could show proportions, but the question specifically asks for frequency of exercise hours, making a bar chart clearer and more direct. A scatterplot requires two quantitative variables and is used to study association, not a one-variable frequency distribution. References/topics from the Study Guide: frequency distributions, bar charts, single-variable data displays, quantitative/discrete data.

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For independent events, the occurrence of one event does not change the probability of the other. Conditional probability P(A|B) means the probability of A occurring given that B has occurred. If A and B are independent, knowing B occurred gives no new information about A. Therefore, P(A|B) = P(A). This is one of the defining properties of independence. For example, if a coin toss and a die roll are independent, knowing the die landed on 4 does not change the probability that the coin landed heads; it remains 1/2. Option B reverses the event being measured. Option C would imply that A becomes impossible after B, which describes a different situation. Option D is an addition expression and does not represent conditional probability. The correct relationship is that the conditional probability equals the original probability when events are independent. Study Guide references/topics: independent events, conditional probability, probability rules, event relationships.

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The probability of selecting a student who does not play soccer is found by identifying the complement of the soccer group. There are 25 students in the class, and 10 students play soccer. Therefore, the number of students who do not play soccer is 25 − 10 = 15. The probability is then the number of favorable outcomes divided by the total number of possible outcomes: 15/25 = 0.60. The information about basketball and neither sport provides context, but the direct complement method is sufficient because the question asks only who does not play soccer. Option B, .40, represents the probability of selecting a student who does play soccer, since 10/25 = .40. Option A, .30, is associated with neither sport if interpreted approximately from 7/25, but it is not the requested event. Option C, .50, does not match the count structure. References/topics from the Study Guide: probability, complements, favorable outcomes, relative frequency.

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The central limit theorem applies when the sample size is sufficiently large and observations are independent. It states that the sampling distribution of the sample mean becomes approximately normal as sample size increases, even if the original population distribution is not normal. This makes option A correct. Option B is incorrect because the population does not have to be normal for the theorem to operate; large sample size is what permits the normal approximation. Option C is not the condition that defines the theorem. A population parameter may be unknown in inference, but that is not the trigger for the central limit theorem. Option D is incorrect because the classic central limit theorem for means concerns quantitative data, not purely categorical labels. The theorem is foundational because it justifies using normal-based confidence intervals and hypothesis tests for means when sample sizes are large. Study Guide references/topics: central limit theorem, sample size, sampling distribution, normal approximation.

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This study involves two categorical variables: customer age group and preferred shopping method. Age group is categorical because customers are classified into groups such as teens, adults, and seniors. Shopping method is also categorical because customers are classified as preferring either online or in-store shopping. When two categorical variables are analyzed together, conditional percentages are the most appropriate numerical summary because they describe how the distribution of one categorical variable changes within levels of another categorical variable. For example, the analyst could compute the percentage of teens who prefer online shopping, the percentage of adults who prefer online shopping, and the percentage of seniors who prefer online shopping. A five-number summary is used for quantitative data, not categorical data. Mean difference requires numerical measurements. A correlation coefficient measures the strength and direction of a linear relationship between two quantitative variables, so it is not appropriate here. References/topics from the Study Guide: two-variable categorical data, conditional percentages, categorical association, numerical summaries.

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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 sample space of a probability experiment is the complete set of all possible outcomes. In this experiment, the spinner has four equal sections labeled A, B, C, and D. Since one spin can land on exactly one of those four labeled sections, the possible outcomes are A, B, C, and D. Therefore, the sample space is {A, B, C, D}. Option A incorrectly adds E, which is not listed as a section on the spinner. Option B groups labels into combined outcomes, but the experiment is a single spin, not a two-part or paired event. Option C uses numerical labels, but the spinner sections are identified by letters, not numbers. Equal section size affects the probability of each outcome, making each outcome have probability 1/4, but it does not change the sample space itself. References/topics from the Study Guide: probability experiments, outcomes, sample space, equally likely events.

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To evaluate the expression, substitute y = −4 everywhere y appears. The numerator is 20 − y, so replacing y with −4 gives 20 − (−4). Subtracting a negative is equivalent to addition, so the numerator becomes 24. The denominator is 5 + y, so substituting −4 gives 5 + (−4) = 1. The expression therefore becomes 24/1, which equals 24. The correct answer is D. The most common error in this problem is mishandling the negative sign in the numerator. If a student incorrectly computes 20 − (−4) as 16, the resulting value would not match the correct algebraic structure. Another important check is that the denominator is not zero; here it equals 1, so the expression is defined. References/topics from the Study Guide: substitution, evaluating algebraic expressions, integers, order of operations.