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| Section | Objectives |
|---|---|
| Topic 1: Probability Theory | - Probability Distributions
|
| Topic 2: Descriptive Statistics | - Single Variable Data Analysis
|
| Topic 3: Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Topic 4: Regression and Modeling | - Linear Relationships
|
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NEW QUESTION # 155
A standard deck of 52 playing cards, 13 with hearts, 13 with diamonds, 13 with clovers, and 13 with spades, is shuffled, and one card is drawn at random.
What is the probability of drawing a heart?
Answer: C
Explanation:
The probability of a simple event is calculated by dividing the number of favorable outcomes by the total number of equally likely outcomes. A standard deck contains 52 cards. The favorable outcomes are the hearts, and the question states that there are 13 hearts in the deck. Therefore, the probability of drawing a heart is 13
/52. This fraction simplifies by dividing both numerator and denominator by 13, giving 1/4. Option A, 1/13, would incorrectly treat one card as favorable out of 13 rather than 13 favorable hearts out of 52 total cards.
Option B, 1/2, would imply that half the deck consists of hearts, which is false. Option D, 1/3, does not correspond to the four equally sized suits in a standard deck. Since the four suits are equally represented, each suit has probability one out of four. References/topics from the Study Guide: theoretical probability, equally likely outcomes, fractions, sample spaces.
NEW QUESTION # 156
Mean = 100, SD = 15, z = 2. Value = ?
Answer: C
Explanation:
A z-score expresses how many standard deviations a value lies above or below the mean. The formula is z = (x # #)/#. To solve for the raw value x, rearrange the formula: x = # + z#. Here, the mean # is 100, the standard deviation # is 15, and z = 2. Substitute these values: x = 100 + 2(15) = 100 + 30 = 130. Therefore, the raw value is 130. This means the value is two standard deviations above the mean. Option B, 115, would correspond to z = 1. Option C, 125, is not exactly two standard deviations above 100 when # = 15. Option D,
110, is less than one standard deviation above the mean. The correct value must add two full standard deviations to the mean. Study Guide references/topics: z-scores, standardization, normal distribution, raw score conversion.
NEW QUESTION # 157
Expected value of binomial X ~ B(10, 0.3) = ?
Answer: B
Explanation:
For a binomial random variable, the expected value is E(X) = np, where n is the number of trials and p is the probability of success on each trial. In this problem, X ~ B(10, 0.3), so n = 10 and p = 0.3. Substituting into the formula gives E(X) = 10 × 0.3 = 3. This means that across many repetitions of the same binomial experiment, the average number of successes would be 3. Option B, 7, would correspond to the expected number of failures if the failure probability is 0.7. Option C gives only the number of trials, not the expected successes. Option D gives only the success probability, not the expected count. The expected value combines both trial count and success probability. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.
NEW QUESTION # 158
A teacher plots the test scores of a class using the box plot.
What is true about the test scores?
Answer: D
Explanation:
A box plot displays the five-number summary of a quantitative data set: minimum, first quartile, median, third quartile, and maximum. The median is represented by the vertical line inside the box. In the given box plot, that internal vertical line is positioned at 80 on the score axis. Therefore, the median test score for Class A is
80 points. The value 70 corresponds to the left edge of the box, which represents the first quartile, not the median. The value near 88 represents the right edge of the box, which is the third quartile. The whiskers extend to the approximate minimum and maximum scores, but those values do not determine the median. The median divides the ordered data into two equal halves, meaning about 50% of the students scored at or below
80 and about 50% scored at or above 80. References/topics from the Study Guide: box plots, five-number summary, median, quartiles.
NEW QUESTION # 159
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: C
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 # 160
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