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| Section | Objectives |
|---|---|
| Probability Theory | - Fundamental Probability Concepts
|
| Statistical Inference | - Estimation and Confidence Intervals
|
| Descriptive Statistics | - Two Variable Data Analysis
|
| Regression and Modeling | - Linear Relationships
|
>> Answers Applied-Probability-and-Statistics Real Questions <<
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NEW QUESTION # 144
A dataset: 4, 8, 12, 16, 20. Mean = ?
Answer: A
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 # 145
A city council surveys residents and business owners about a proposed new public transportation system. The results of the survey are shown in the following 2 × 2 contingency table.
Which statement is true?
Answer: C
Explanation:
This question requires comparing conditional percentages from a two-way table. For residents, 150 out of 200 are in favor of the proposed transportation system. The resident support rate is therefore 150/200 = 0.75, or
75%. For business owners, 30 out of 100 are in favor. The business owner support rate is 30/100 = 0.30, or
30%. The difference is 75% # 30% = 45 percentage points, which is a large gap. Therefore, a resident is much more likely than a business owner to favor the new transportation system. Option A understates the magnitude of the difference by calling it slight. Options B and C reverse the direction of the comparison because residents have the higher support rate, not the lower one. The appropriate statistical tool here is conditional percentage because support is being compared within each respondent group. References/topics from the Study Guide: contingency tables, conditional percentages, two-variable categorical data, comparative proportions.
NEW QUESTION # 146
Probability of rolling sum = 12 = ?
Answer: C
Explanation:
Rolling two six-sided dice produces 36 equally likely ordered outcomes. A sum of 12 can occur only when both dice show 6, giving the single ordered pair (6, 6). Since there is exactly 1 favorable outcome out of 36 total outcomes, the probability is 1/36. This is structurally similar to rolling a sum of 2, where only (1, 1) works. Option B, 1/6, is much too large and would apply to one specified face on one die. Option C, 2/36, would require two different ordered outcomes, but no pair other than (6, 6) sums to 12. Option D, 1/12, also overstates the probability. Extreme sums such as 2 and 12 are least likely because they have only one possible ordered combination. Study Guide references/topics: sample spaces, dice sums, theoretical probability, equally likely outcomes.
NEW QUESTION # 147
Type II error # is:
Answer: C
Explanation:
A Type II error occurs when the null hypothesis is false but the test fails to reject it. Many introductory answer choices phrase this as "accept H# when false," though the more technically careful wording is "fail to reject H# when false." This is a false negative: the real effect or difference exists, but the test does not detect sufficient evidence. The probability of a Type II error is denoted #. Option B describes a Type I error, which is rejecting a true null hypothesis. Option C is incorrect because 1 # # is not #; # is the Type I error probability, while 1 # # is statistical power. Option D is too vague and does not identify a formal hypothesis- testing error. The defining condition for Type II error is missing a real effect by failing to reject a false null hypothesis. Study Guide references/topics: Type II error, beta, hypothesis testing, statistical power.
NEW QUESTION # 148
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?
Answer: C
Explanation:
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.
NEW QUESTION # 149
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