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| Section | Objectives |
|---|---|
| Descriptive Statistics | - Data summarization
|
| Probability | - Probability rules
|
| Regression and Correlation | - Relationship analysis
|
| Statistical Inference | - Hypothesis testing
|
| Probability Distributions | - Continuous distributions
|
>> New Applied-Probability-and-Statistics Test Pdf <<
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NEW QUESTION # 25
What is the product of 0.6 and 0.25?
Answer: D
Explanation:
To multiply decimals, first multiply the numbers as whole numbers and then place the decimal point according to the total number of decimal places in the factors. The values are 0.6 and 0.25. Ignoring decimals temporarily, 6 × 25 = 150. The factor 0.6 has one decimal place, and 0.25 has two decimal places, for a total of three decimal places. Therefore, 150 becomes 0.150, which is equivalent to 0.15. A second way to verify the computation is to convert the decimals into fractions: 0.6 = 6/10 and 0.25 = 25/100. Multiplying gives (6 ×
25)/(10 × 100) = 150/1000 = 0.15. This result is reasonable because one-fourth of 0.6 is 0.15. The other answer choices are too large for one-fourth of 0.6 or do not result from the correct decimal multiplication.
References/topics from the Study Guide: decimal multiplication, fractions, equivalent forms, numerical operations.
NEW QUESTION # 26
Probability of rolling an even number on a six-sided die?
Answer: C
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 # 27
A dataset: 2, 4, 6, 8, 10. Variance = ?
Answer: B
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 # 28
There are 10 green, 20 yellow, and 15 purple marbles in a jar.
What is the probability of selecting 2 green marbles in a row and then 1 yellow marble, replacing the marble each time?
Answer: B
Explanation:
There are 10 green, 20 yellow, and 15 purple marbles, so the total number of marbles is 10 + 20 + 15 = 45.
Because each marble is replaced after selection, the probabilities remain the same on every draw, making the events independent. The probability of selecting a green marble on one draw is 10/45 = 2/9. Selecting two green marbles in a row gives (2/9)(2/9). The probability of then selecting a yellow marble is 20/45 = 4/9.
Multiply the probabilities because the question asks for green and green and yellow in that exact sequence: (2
/9)(2/9)(4/9) = 16/729. Replacement is essential; without replacement, the denominators and numerators would change after each draw. Since replacement keeps the sample space constant at 45 marbles each time, the independent multiplication rule applies directly. References/topics from the Study Guide: independent events, replacement, multiplication rule, compound probability.
NEW QUESTION # 29
t-test used when:
Answer: B
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 # 30
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