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| Section | Objectives |
|---|---|
| Probability Theory | - Fundamental Probability Concepts
|
| Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Regression and Modeling | - Linear Relationships
|
| Descriptive Statistics | - Single Variable Data Analysis
|
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NEW QUESTION # 17
In a game, a coin is tossed, and a spinner with 8 equal spaces numbered 1 through 8 is spun.
What is the probability of getting heads on the coin and a number less than 3 on the spinner?
Answer: B
Explanation:
This experiment has two independent stages: tossing a coin and spinning a numbered spinner. The probability of getting heads on a fair coin is 1/2. The spinner has 8 equally likely outcomes numbered 1 through 8. A number less than 3 means either 1 or 2, so there are 2 favorable spinner outcomes out of 8 total outcomes.
Thus, P(number less than 3) = 2/8 = 1/4. Since the coin result does not affect the spinner result, the events are independent. For independent events connected by "and," multiply the probabilities: 1/2 × 1/4 = 1/8.
Equivalently, the combined sample space has 2 coin outcomes times 8 spinner outcomes, or 16 total equally likely outcomes. The favorable outcomes are heads with 1 and heads with 2, giving 2/16 = 1/8. References
/topics from the Study Guide: independent events, sample spaces, multiplication rule, theoretical probability.
NEW QUESTION # 18
Probability of rolling an even number on a six-sided die?
Answer: A
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 # 19
P(A) = 0.3, P(B) = 0.4, A and B independent. P(A and B) = ?
Answer: B
Explanation:
For independent events, the occurrence of one event does not change the probability of the other event. The multiplication rule states that if A and B are independent, then P(A and B) = P(A) × P(B). Here, P(A) = 0.3 and P(B) = 0.4. Therefore, P(A and B) = 0.3 × 0.4 = 0.12. This means there is a 12% probability that both events occur. Option B, 0.7, incorrectly adds the probabilities and would only be part of an "or" calculation, not an "and" calculation. Option D repeats P(A), and option C does not follow from the multiplication rule.
The phrase "A and B independent" is the decisive condition because it permits direct multiplication without adjusting for conditional probability. Study Guide references/topics: independent events, multiplication rule, compound probability, joint probability.
NEW QUESTION # 20
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 # 21
Expected value of X = 1×0.2 + 2×0.5 + 3×0.3 = ?
Answer: D
Explanation:
Expected value is the long-run average value of a random variable. For a discrete random variable, it is calculated by multiplying each possible value by its probability and then adding those products. Here, the expression is already structured as an expected value calculation: 1×0.2 + 2×0.5 + 3×0.3. Compute each product: 1×0.2 = 0.2, 2×0.5 = 1.0, and 3×0.3 = 0.9. Add them: 0.2 + 1.0 + 0.9 = 2.1. Therefore, the expected value is 2.1. This does not mean the random variable must equal 2.1 in a single trial; it means that over many repetitions, the average outcome would approach 2.1. Option B is close but omits part of the weighted contribution. Options C and D do not match the weighted-average computation. Study Guide references
/topics: expected value, discrete random variables, weighted average, probability distributions.
NEW QUESTION # 22
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