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| Section | Objectives |
|---|---|
| Topic 1: Probability Theory | - Probability Distributions
|
| Topic 2: Descriptive Statistics | - Two Variable Data Analysis
|
| Topic 3: Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Topic 4: Regression and Modeling | - Linear Relationships
|
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NEW QUESTION # 81
Central limit theorem applies when:
Answer: B
Explanation:
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.
NEW QUESTION # 82
Probability of rolling an even number on a six-sided die?
Answer: B
Explanation:
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.
NEW QUESTION # 83
Sum of probabilities in sample space = ?
Answer: D
Explanation:
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.
NEW QUESTION # 84
Best chart for categorical data?
Answer: D
Explanation:
A bar chart is the best display for categorical data because it shows the frequency or relative frequency of each category using separate bars. Categorical variables classify observations into groups, such as favorite color, type of transportation, political affiliation, or product category. A bar chart allows the viewer to compare category counts directly by bar height. A histogram is not appropriate because it is used for quantitative data grouped into numerical intervals. A scatterplot is used to examine the relationship between two quantitative variables, such as height and weight. A boxplot summarizes a quantitative distribution using quartiles, median, spread, and outliers. Since the question asks specifically about categorical data, the display must preserve category labels rather than numerical intervals or continuous measurement. Therefore, the bar chart is the correct choice. Study Guide references/topics: categorical data, bar charts, frequency displays, descriptive statistics.
NEW QUESTION # 85
A commuter has a .4 probability of taking the bus to work, a .4 probability of driving a car, and a .2 probability of cycling. The probability of being late is .15 when taking the bus, .05 when driving a car, and .1 when cycling.
What is the probability of taking the bus and being on time, or driving a car and being on time?
Answer: B
Explanation:
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.
NEW QUESTION # 86
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