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| Section | Weight | Objectives |
|---|---|---|
| Inferential Statistics & Study Design | 20% | - Sampling methods and bias - Hypothesis testing framework and interpretation - Confidence intervals for means/proportions - Observational studies vs experiments |
| Descriptive Statistics | 25% | - Measures of center: mean, median, mode - Graphical displays: histograms, boxplots, scatterplots - Measures of spread: range, IQR, variance, standard deviation - Types of data: categorical, discrete, continuous |
| Basic Numeracy & Algebra | 15% | - Exponents, roots, and basic formulas - Arithmetic operations, fractions, decimals, percentages - Linear equations, inequalities, graphing functions |
| Probability Concepts | 20% | - Discrete and continuous probability distributions - Probability rules, independent and dependent events - Normal distribution and empirical rule - Conditional probability and Venn diagrams |
| Correlation & Regression | 20% | - Interpreting slope, intercept, and R-squared - Predictions and limitations of regression - Correlation coefficient and interpretation - Simple linear regression models |
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NEW QUESTION # 155
Two events mutually exclusive # P(A and B) = ?
Answer: C
Explanation:
Mutually exclusive events cannot happen at the same time. Therefore, the probability of their intersection is zero: P(A and B) = 0. For instance, on a single coin toss, the event "heads" and the event "tails" are mutually exclusive because a single toss cannot produce both outcomes. Option B applies to independent events when calculating the probability of both occurring, but mutually exclusive events with positive probabilities are not independent because one event prevents the other from occurring. Option C is used to find P(A or B) for mutually exclusive events, not P(A and B). Option D would mean both events always occur together, which directly contradicts mutual exclusivity. The word "and" is the key operator: it asks for overlap. Since mutually exclusive events have no overlap, the result is 0. Study Guide references/topics: mutually exclusive events, intersections, joint probability, addition rule.
NEW QUESTION # 156
A die is rolled twice. Probability of rolling two sixes?
Answer: B
Explanation:
Rolling a standard six-sided die twice creates two independent events. The probability of rolling a six on the first roll is 1/6 because there is one favorable outcome, six, out of six equally likely outcomes. The probability of rolling a six on the second roll is also 1/6. Because the outcome of the first roll does not affect the outcome of the second roll, the multiplication rule for independent events applies. Thus, P(six and six) = 1/6 × 1/6 = 1
/36. Another way to verify the result is to list the sample space: two die rolls produce 6 × 6 = 36 equally likely ordered outcomes. Only one outcome, (6, 6), satisfies the condition of two sixes. Therefore, the probability is
1 favorable outcome out of 36 total outcomes. Study Guide references/topics: independent events, multiplication rule, sample space, theoretical probability.
NEW QUESTION # 157
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 # 158
A dataset: 4, 8, 12, 16, 20. Mean = ?
Answer: C
Explanation:
The mean is the arithmetic average of a data set. To compute it, add all values and divide by the number of observations. For the dataset 4, 8, 12, 16, and 20, the sum is 4 + 8 + 12 + 16 + 20 = 60. There are 5 values, so the mean is 60 ÷ 5 = 12. The mean represents the balance point of the data. This dataset is evenly spaced around 12: 4 and 20 are equally distant from 12, and 8 and 16 are equally distant from 12. That symmetry supports the computed result. Option A, 10, is too low because it does not account for the higher values 16 and 20. Option C, 14, is too high, and option D is one of the data values but not the average. The correct answer is 12. Study Guide references/topics: mean, arithmetic average, measures of center, quantitative data.
NEW QUESTION # 159
Probability of rolling 1, 2, or 3 on die = ?
Answer: A
Explanation:
A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The event "rolling 1, 2, or 3" has three favorable outcomes: 1, 2, and 3. The probability is therefore favorable outcomes divided by total outcomes: 3/6. This fraction simplifies to 1/2. Option B, 1/3, would correspond to two favorable outcomes out of six. Option C, 1/6, is the probability of rolling one specific number only. Option D, 2/3, would require four favorable outcomes out of six. Since exactly half of the die faces are 1, 2, or 3, the correct probability is one- half. This is a direct application of theoretical probability with equally likely outcomes. Study Guide references/topics: die probability, favorable outcomes, sample space, theoretical probability.
NEW QUESTION # 160
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