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| Section | Objectives |
|---|---|
| Probability Theory | - Probability Distributions
|
| Descriptive Statistics | - Two Variable Data Analysis
|
| Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Regression and Modeling | - Linear Relationships
|
>> Applied-Probability-and-Statistics Exam Sample <<
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NEW QUESTION # 22
Probability of drawing red card from deck = ?
Answer: B
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 # 23
Conditional probability P(A|B) = P(A and B)/P(B) #
Answer: D
Explanation:
The formula for conditional probability is P(A|B) = P(A and B)/P(B), provided P(B) > 0. The notation P(A|B) is read as "the probability of A given B." The denominator P(B) restricts the sample space to cases in which B occurs, and the numerator P(A and B) counts the portion of that restricted group in which A also occurs.
Therefore, the statement is true. Option B is incorrect because the formula is the standard definition of conditional probability. Option C is incorrect because if A and B are mutually exclusive, then P(A and B) = 0, which creates a special case rather than the general definition. Option D is invalid because conditional probability is undefined when P(B) = 0. The condition must have positive probability. Study Guide references
/topics: conditional probability, joint probability, event intersection, probability rules.
NEW QUESTION # 24
Boxplot shows:
Answer: B
Explanation:
A boxplot displays the distribution of a quantitative variable using the five-number summary and potential outliers. The main features are the median, first quartile, third quartile, lower whisker, upper whisker, and any marked outliers. The box extends from Q1 to Q3 and represents the interquartile range, which contains the middle 50% of the data. The line inside the box marks the median. Whiskers show the spread of non-outlier values, depending on the graphing convention. Option B is incorrect because a boxplot does not show only the mean; many boxplots do not display the mean at all. Option C is incorrect because variance is a numerical measure of spread, not a standard direct feature of a boxplot. Option D is incorrect because a boxplot is not a probability model. It is a descriptive display for summarizing center, spread, and unusual observations. Study Guide references/topics: boxplots, median, quartiles, interquartile range, outliers.
NEW QUESTION # 25
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: C
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 # 26
Probability of rolling sum = 12 = ?
Answer: B
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 # 27
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