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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Basic Numeracy & Algebra | 15% | - Linear equations, inequalities, graphing functions - Exponents, roots, and basic formulas - Arithmetic operations, fractions, decimals, percentages |
| Topic 2: Probability Concepts | 20% | - Probability rules, independent and dependent events - Conditional probability and Venn diagrams - Normal distribution and empirical rule - Discrete and continuous probability distributions |
| Topic 3: Inferential Statistics & Study Design | 20% | - Observational studies vs experiments - Confidence intervals for means/proportions - Hypothesis testing framework and interpretation - Sampling methods and bias |
| Topic 4: Correlation & Regression | 20% | - Interpreting slope, intercept, and R-squared - Predictions and limitations of regression - Simple linear regression models - Correlation coefficient and interpretation |
| Topic 5: Descriptive Statistics | 25% | - Types of data: categorical, discrete, continuous - Graphical displays: histograms, boxplots, scatterplots - Measures of spread: range, IQR, variance, standard deviation - Measures of center: mean, median, mode |
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NEW QUESTION # 127
Probability of rolling sum = 2 on 2 dice = ?
Answer: C
Explanation:
When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To get a sum of 2, the only possible ordered pair is (1, 1). No other pair of die faces can produce a total of 2 because 1 is the smallest value on each die. Therefore, there is 1 favorable outcome out of 36 total outcomes, giving a probability of 1/36. Option B, 1/6, is the probability of one specific result on a single die, not a two-dice sum.
Option C, 2/36, would require two favorable ordered pairs, but only one exists. Option D, 1/12, is also too large because it corresponds to 3 outcomes out of 36. The key is to count ordered outcomes in the full two- dice sample space. Study Guide references/topics: dice probability, sample space, favorable outcomes, theoretical probability.
NEW QUESTION # 128
Expected value of X = 1×0.2 + 2×0.5 + 3×0.3 = ?
Answer: C
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 # 129
What is the mode of this dataset: 3, 5, 7, 7, 10?
Answer: C
Explanation:
The mode is the value that occurs most frequently in a data set. In the set 3, 5, 7, 7, 10, the value 3 appears once, 5 appears once, 10 appears once, and 7 appears twice. Since 7 has the highest frequency, it is the mode.
This question tests recognition of a measure of center that differs from the mean and median. The mean would require summing all values and dividing by the number of observations, while the median would identify the middle value after ordering the data. The data are already ordered, and the middle value is also 7, but the reason the correct answer is 7 here is specifically because it appears more often than any other value. Options A, B, and D each name values that occur only once, so none can be the mode. A dataset may have one mode, multiple modes, or no mode, but this dataset is unimodal. References/topics from the Study Guide: descriptive statistics, measures of center, frequency, mode.
NEW QUESTION # 130
Chi-square test used for:
Answer: B
Explanation:
A chi-square test is commonly used to analyze categorical data, especially to determine whether two categorical variables are associated. In a chi-square test of independence, data are arranged in a contingency table, and observed cell counts are compared with expected cell counts under the assumption that the variables are independent. If the observed counts differ substantially from the expected counts, there is evidence of association between the categorical variables. For example, a chi-square test could examine whether voting preference is associated with age group or whether product preference differs by region.
Option B describes comparing means, which is typically handled by a t-test or related procedure. Option C refers to regression, which models relationships between variables, often involving quantitative outcomes.
Option D refers to variance, which is tested using different procedures depending on context. The key phrase is "categorical variables": chi-square methods work with counts in categories. Study Guide references/topics:
categorical data, contingency tables, chi-square test, association.
NEW QUESTION # 131
Dataset: 3, 5, 7, 9, 11. Median = ?
Answer: C
Explanation:
The median is the center value of an ordered dataset. The values 3, 5, 7, 9, and 11 are already arranged from least to greatest. Because there are five observations, the median is the third value, with two observations below it and two observations above it. The third value is 7, so the median is 7. The median is a measure of central tendency that describes the middle position of the data rather than the arithmetic average. In this dataset, the mean is also 7 because the values are evenly spaced around 7, but the median is determined by position, not by summing and dividing. Option B is the second value, option C is the fourth value, and option D is not a data value. The correct answer is the value that splits the ordered list into two equal halves. Study Guide references/topics: median, ordered data, measures of center, descriptive statistics.
NEW QUESTION # 132
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