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| Section | Objectives |
|---|---|
| Topic 1: Probability Theory | - Probability Distributions
|
| Topic 2: Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Topic 3: Regression and Modeling | - Linear Relationships
|
| Topic 4: Descriptive Statistics | - Two Variable Data Analysis
|
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NEW QUESTION # 105
Empirical probability is based on:
Answer: B
Explanation:
Empirical probability is based on observed data collected from experiments, surveys, simulations, or repeated trials. It is calculated as the relative frequency of an event: number of times the event occurs divided by the total number of trials or observations. For example, if a machine produces 12 defective items in a sample of
300, the empirical probability of a defect is 12/300 = 0.04. This differs from classical or theoretical probability, which is based on equally likely outcomes and mathematical structure, such as a fair die having probability 1/6 for each face. It also differs from subjective probability, which is based on personal judgment or expert belief rather than observed frequency. The word "empirical" signals evidence obtained through observation, so observed frequency is the correct basis. Study Guide references/topics: empirical probability, relative frequency, observed data, probability interpretation.
NEW QUESTION # 106
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: B
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 # 107
Probability of drawing king or queen from deck = ?
Answer: B
Explanation:
A standard deck contains 52 cards. There are 4 kings and 4 queens, one of each rank in each suit. The event is drawing a king or a queen. Since a single card cannot be both a king and a queen, these two events are mutually exclusive. Therefore, the number of favorable outcomes is 4 + 4 = 8. The probability is favorable outcomes divided by total outcomes: 8/52. This fraction can be simplified to 2/13, but the answer choice provides the unsimplified correct form. Option B, 4/52, counts only kings or only queens, not both. Option C,
1/13, is equivalent to 4/52 and again represents one rank only. Option D, 1/26, is too small and does not match the count of favorable cards. The word "or" signals that both qualifying ranks should be included. Study Guide references/topics: card probability, mutually exclusive events, addition rule, sample space.
NEW QUESTION # 108
A dataset: 2, 4, 6, 8, 10. Variance = ?
Answer: D
Explanation:
Variance measures the average squared distance of data values from the mean. For the dataset 2, 4, 6, 8, 10, the mean is (2 + 4 + 6 + 8 + 10) ÷ 5 = 30 ÷ 5 = 6. The deviations from the mean are #4, #2, 0, 2, and 4.
Squaring these deviations gives 16, 4, 0, 4, and 16. The sum of squared deviations is 40. Using the sample variance formula, divide by n # 1, where n = 5. Thus, sample variance = 40 ÷ 4 = 10. Option A is correct.
Option C, 6.25, does not result from the standard sample variance computation for these values. The important distinction is whether a problem is using sample variance or population variance; the provided answer set and calculation use the sample variance formula. Study Guide references/topics: variance, mean, squared deviations, sample variance.
NEW QUESTION # 109
A golf course is attempting to correlate golfing handicap with math SAT scores among local high school golfers. Ignoring potential confounding variables such as socioeconomic status, the golf course creates the following scatterplot.
What is the estimated value of r, the correlation coefficient, between these variables?
Answer: C
Explanation:
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.
NEW QUESTION # 110
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