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| Section | Objectives |
|---|---|
| Descriptive Statistics | - Data visualization
|
| Regression and Correlation | - Relationship analysis
|
| Probability Distributions | - Continuous distributions
|
| Statistical Inference | - Estimation
|
| Probability | - Fundamental probability concepts
|
>> Applied-Probability-and-Statistics Fragen Antworten <<
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47. Frage
Probability of drawing king or queen from deck = ?
Antwort: A
Begründung:
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.
48. Frage
A commuter has a .4 probability of taking the bus to work, a .4 probability of driving a car, and a .2 probability of cycling. The probability of being late is .15 when taking the bus, .05 when driving a car, and .1 when cycling.
What is the probability of taking the bus and being on time, or driving a car and being on time?
Antwort: A
Begründung:
This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes.
"Taking the bus and being on time" means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 # .15 = .85. Therefore, P(bus and on time)
= .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 # .05
= .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes.
References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.
49. Frage
Random variable X = number of heads in 4 coin flips #
Antwort: A
Begründung:
The random variable X counts the number of heads obtained in 4 coin flips. Since it is a count, it can take only specific whole-number values: 0, 1, 2, 3, or 4. It cannot take fractional values such as 2.5 heads. A random variable with countable possible outcomes is classified as discrete. This situation also fits a binomial framework because there is a fixed number of independent trials, each trial has two outcomes, and the probability of heads remains constant for a fair coin. A continuous random variable, by contrast, can take any value over an interval, such as time, weight, or height. The number of heads is not measured on a continuum; it is counted. Therefore, the correct classification is discrete. Study Guide references/topics: discrete random variables, binomial setting, coin-flip outcomes, probability distributions.
50. Frage
One-sample t-test compares:
Antwort: C
Begründung:
A one-sample t-test is used to compare a sample mean to a hypothesized or known population mean when the population standard deviation is unknown and the data are approximately normal or the sample size is sufficiently large. The test statistic evaluates how far the sample mean is from the hypothesized mean in standard error units. Option A is therefore correct. A two-sample t-test compares means from two independent groups, so option B describes a different test. Tests of variances use procedures such as chi-square or F-based methods depending on context, so option C is not appropriate. Tests of proportions use z procedures for categorical success/failure data, not a one-sample t-test for means. The t-test is part of inferential statistics because it uses sample evidence to make a decision about a population parameter. Study Guide references
/topics: one-sample t-test, sample mean, population mean, hypothesis testing.
51. Frage
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?
Antwort: B
Begründung:
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.
52. Frage
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