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| Section | Objectives |
|---|---|
| Regression and Correlation | - Relationship analysis
|
| Descriptive Statistics | - Data visualization
|
| Probability Distributions | - Continuous distributions
|
| Probability | - Probability rules
|
| Statistical Inference | - Estimation
|
>> Applied-Probability-and-Statistics Deutsch <<
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115. Frage
Ten teachers each had a jar of jelly beans. The jars had this number of jelly beans in each respective jar: 50,
60, 80, 80, 100, 100, 100, 120, 120, and 140.
What is the mean of these data?
Antwort: A
Begründung:
The mean is the arithmetic average of a quantitative data set. To calculate it, add all observed values and divide by the number of observations. The jelly bean counts are 50, 60, 80, 80, 100, 100, 100, 120, 120, and
140. Their sum is 950. There are 10 jars, so the mean is 950 ÷ 10 = 95. This means the average number of jelly beans per jar is 95. Option A, 100, may seem plausible because 100 appears frequently and is near the center of the data, but frequency alone determines the mode, not the mean. Option C, 80, is one of the repeated values but not the average. Option B, 120, is too high because several observations fall well below
120. The mean uses every value in the data set, including the low values 50 and 60 and the high value 140.
References/topics from the Study Guide: mean, measures of center, quantitative data, arithmetic average.
116. Frage
Probability of rolling 1, 2, or 3 on die = ?
Antwort: A
Begründung:
A standard six-sided die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The event "rolling 1, 2, or 3" has three favorable outcomes: 1, 2, and 3. The probability is therefore favorable outcomes divided by total outcomes: 3/6. This fraction simplifies to 1/2. Option B, 1/3, would correspond to two favorable outcomes out of six. Option C, 1/6, is the probability of rolling one specific number only. Option D, 2/3, would require four favorable outcomes out of six. Since exactly half of the die faces are 1, 2, or 3, the correct probability is one- half. This is a direct application of theoretical probability with equally likely outcomes. Study Guide references/topics: die probability, favorable outcomes, sample space, theoretical probability.
117. Frage
Binomial expected value = ?
Antwort: B
Begründung:
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. The binomial model applies when there is a fixed number of independent trials, each trial has two outcomes, and the probability of success remains constant. The expected value represents the long-run average number of successes across repeated binomial experiments. For example, if a test has 20 independent questions and the probability of success on each is 0.75, the expected number correct is 20 × 0.75 = 15. Option B, n/p, and option C, p/n, do not represent the binomial mean. Option D, n + p, incorrectly adds the number of trials and success probability, which are not comparable units. The formula np captures both how many opportunities there are and how likely success is on each opportunity. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.
118. Frage
In a standard normal distribution, z-score for the mean = ?
Antwort: A
Begründung:
A z-score measures how many standard deviations a value is above or below the mean. The formula is z = (x
# #) / #, where x is the observed value, # is the mean, and # is the standard deviation. If the observed value equals the mean, then x = #. Substituting into the formula gives z = (# # #) / # = 0 / # = 0. Therefore, the z- score corresponding to the mean is always 0 in any normal distribution, including the standard normal distribution. In the standard normal distribution specifically, the mean is 0 and the standard deviation is 1, so the mean lies exactly at z = 0. A z-score of 1 would indicate one standard deviation above the mean, and #1 would indicate one standard deviation below the mean. Study Guide references/topics: standard normal distribution, z-scores, mean, standard deviation.
119. Frage
The average lifespan of a dalmatian is 11.5 years, with a standard deviation of 1.5 years.
What are the two values between which 95% of the data falls for the average lifespan of a dalmatian, assuming a normal distribution?
Antwort: C
Begründung:
For a normal distribution, the empirical rule states that approximately 95% of observations lie within two standard deviations of the mean. The mean lifespan is 11.5 years, and the standard deviation is 1.5 years. Two standard deviations equal 2 × 1.5 = 3.0 years. Therefore, the interval containing approximately 95% of the data is found by subtracting and adding 3.0 years to the mean: 11.5 # 3.0 = 8.5 and 11.5 + 3.0 = 14.5. The resulting interval is 8.5 years to 14.5 years. Option C, 10.0 to 13.0, represents one standard deviation from the mean and would correspond to about 68% of the data. Options A and D are not centered correctly at the proper two-standard-deviation distance. The correct interval must be symmetric around the mean and use two standard deviations. References/topics from the Study Guide: normal distributions, empirical rule, standard deviation, measures of spread.
120. Frage
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