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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Descriptive Statistics | 25% | - Measures of spread: range, IQR, variance, standard deviation - Measures of center: mean, median, mode - Types of data: categorical, discrete, continuous - Graphical displays: histograms, boxplots, scatterplots |
| Topic 2: Basic Numeracy & Algebra | 15% | - Linear equations, inequalities, graphing functions - Exponents, roots, and basic formulas - Arithmetic operations, fractions, decimals, percentages |
| Topic 3: Probability Concepts | 20% | - Conditional probability and Venn diagrams - Probability rules, independent and dependent events - Normal distribution and empirical rule - Discrete and continuous probability distributions |
| Topic 4: Correlation & Regression | 20% | - Simple linear regression models - Predictions and limitations of regression - Interpreting slope, intercept, and R-squared - Correlation coefficient and interpretation |
| Topic 5: Inferential Statistics & Study Design | 20% | - Sampling methods and bias - Hypothesis testing framework and interpretation - Observational studies vs experiments - Confidence intervals for means/proportions |
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NEW QUESTION # 144
Determine z in the equation:
1/(6z) = 2/9
Answer: B
Explanation:
The equation 1/(6z) = 2/9 is a proportion because it sets two fractions equal to each other. To solve it, use cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the denominator of the first fraction by the numerator of the second fraction: 1 × 9 = 2 × 6z. This gives 9 = 12z. To isolate z, divide both sides by 12: z = 9/12. This fraction can also be simplified to 3/4, but the answer choices list the unsimplified equivalent form 9/12. The other options do not satisfy the original equation. For example, if z = 2/54, then 6z is very small and the left side becomes much larger than 2/9. If z =
12/9 or 18/2, the denominator becomes too large, making the left side too small. References/topics from the Study Guide: proportions, solving equations, cross multiplication, equivalent fractions.
NEW QUESTION # 145
Probability of drawing red card from deck = ?
Answer: A
Explanation:
A standard deck has 52 cards divided into four suits: hearts, diamonds, clubs, and spades. Hearts and diamonds are red, while clubs and spades are black. Each suit has 13 cards, so the number of red cards is 13 +
13 = 26. The probability of drawing a red card is therefore the number of favorable outcomes divided by the total number of outcomes: 26/52. This simplifies to 1/2. Option B, 1/4, would describe the probability of drawing one specific suit, such as hearts only. Option C, 1/13, would describe one specific rank in the deck, such as a particular value across suits. Option D does not correspond to the card structure of a standard deck.
Since exactly half the deck is red, the correct probability is 26/52 = 1/2. Study Guide references/topics: card probability, favorable outcomes, sample space, theoretical probability.
NEW QUESTION # 146
Expected value of X = 1×0.2 + 2×0.5 + 3×0.3 = ?
Answer: A
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 # 147
Best chart for categorical data?
Answer: A
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 # 148
t-test used when:
Answer: C
Explanation:
A t-test is used when making inferences about a population mean and the population standard deviation, #, is unknown. In practical settings, # is rarely available, so the sample standard deviation, s, is used as an estimate.
That substitution introduces additional uncertainty, which is why the t-distribution is used instead of the standard normal distribution. The t-distribution has heavier tails, especially for small samples, reflecting the extra variability caused by estimating # from the sample. Option B describes a z-test setting, where the population standard deviation is known. Option C is incorrect because categorical population data are usually analyzed with proportions, chi-square tests, or related categorical procedures, not mean-based t-tests. Option D is invalid because t-tests have a clear inferential role. The controlling condition is unknown #, with inference focused on means. Study Guide references/topics: t-tests, unknown population standard deviation, sample standard deviation, inferential statistics.
NEW QUESTION # 149
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