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| Section | Objectives |
|---|---|
| Probability | - Fundamental probability concepts
|
| Descriptive Statistics | - Data summarization
|
| Regression and Correlation | - Relationship analysis
|
| Statistical Inference | - Estimation
|
| Probability Distributions | - Discrete distributions
|
>> Applied-Probability-and-Statistics Valid Exam Answers <<
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NEW QUESTION # 131
What percent of data lies within 2 standard deviations of the mean in a normal distribution?
Answer: A
Explanation:
For a normal distribution, the empirical rule, also called the 68-95-99.7 rule, describes how data are distributed around the mean. Approximately 68% of observations fall within 1 standard deviation of the mean, approximately 95% fall within 2 standard deviations, and approximately 99.7% fall within 3 standard deviations. The question asks specifically for the percentage within 2 standard deviations, so the correct value is 95%. This interval is written as # # 2# to # + 2#, where # is the mean and # is the standard deviation.
Option B corresponds to one standard deviation, not two. Option C is an imprecise version of the three- standard-deviation rule, and option D has no standard empirical-rule interpretation in this context. This rule is frequently used to estimate ranges in bell-shaped distributions. Study Guide references/topics: normal distribution, empirical rule, standard deviation, distribution spread.
NEW QUESTION # 132
In a city's public transportation system, the distribution of commuters by mode of transport is as follows: 40% use buses, 25% use trains, 20% use bicycles, 10% walk, and 5% use other means of transportation.
What is the probability that a randomly selected commuter uses either a bus or a train?
Answer: A
Explanation:
The event is that a randomly selected commuter uses either a bus or a train. These categories are mutually exclusive because one commuter is classified under one primary transportation mode in the given distribution.
For mutually exclusive events, the probability of "A or B" is found by adding the individual probabilities. The probability of using a bus is 40%, and the probability of using a train is 25%. Therefore, P(bus or train) = 40%
+ 25% = 65%. The bicycle, walking, and other transportation categories are not included because the question restricts the event to bus or train only. Option B, 45%, would incorrectly combine bus with the 5% "other" category or otherwise omit part of the train group. Option C is too low, and option A is unrelated to the stated categories. References/topics from the Study Guide: probability, mutually exclusive events, addition rule, percentage distributions.
NEW QUESTION # 133
A person has a .5 probability of taking a train to work, a .3 probability of carpooling, and a .2 probability of walking. The probability of being late is .2 when taking the train, .1 when carpooling, and .05 when walking.
What is the probability of taking the train and being on time, or walking and being on time?
Answer: B
Explanation:
This problem combines complements, joint probabilities, and addition of mutually exclusive outcomes. First compute the probability of taking the train and being on time. The probability of being late when taking the train is .2, so the probability of being on time by train is 1 # .2 = .8. Thus, P(train and on time) = .5 × .8 = .40.
Next compute the probability of walking and being on time. The probability of being late when walking is .
05, so the probability of being on time while walking is 1 # .05 = .95. Thus, P(walking and on time) = .2 × .95
= .19. Since a person cannot both take the train and walk as the selected commuting mode, these outcomes are mutually exclusive. Add the two results: .40 + .19 = .59. References/topics from the Study Guide:
complements, conditional probability, joint probability, addition rule.
NEW QUESTION # 134
Probability of heads on 3 coin flips exactly twice?
Answer: B
Explanation:
This is a binomial probability problem because there is a fixed number of independent trials, each trial has two outcomes, and the probability of success is constant. Let success be flipping heads. There are 3 coin flips, and the probability of heads on each flip is 1/2. The event "exactly two heads" can occur in three ordered ways: HHT, HTH, and THH. Each sequence has probability (1/2)(1/2)(1/2) = 1/8. Since there are three favorable sequences, the total probability is 3 × 1/8 = 3/8. Equivalently, use the binomial formula: C(3,2)(1/2)