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| Section | Objectives |
|---|---|
| Probability | - Probability rules
|
| Probability Distributions | - Discrete distributions
|
| Regression and Correlation | - Relationship analysis
|
| Descriptive Statistics | - Data visualization
|
| Statistical Inference | - Hypothesis testing
|
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NEW QUESTION # 92
There are 14 black, 16 white, and 10 red balls in a box.
What is the probability of selecting 1 black ball, then 1 red ball, and then 1 white ball, replacing the ball each time?
Answer: A
Explanation:
The box contains 14 black, 16 white, and 10 red balls, for a total of 40 balls. Because the ball is replaced after each selection, the probability of each color remains constant on every draw. This makes the three draws independent. The probability of selecting a black ball first is 14/40, which simplifies to 7/20. The probability of selecting a red ball second is 10/40, or 1/4. The probability of selecting a white ball third is 16/40, or 2/5.
Since the question asks for all three events in the stated order, multiply the probabilities: (7/20)(1/4)(2/5) = 14
/400 = 7/200. Replacement is the controlling condition; without replacement, the denominators would decrease after each draw. Here, the sample space remains 40 each time. References/topics from the Study Guide: probability, independent events, replacement, multiplication rule.
NEW QUESTION # 93
Standard deviation measures:
Answer: C
Explanation:
Standard deviation measures how spread out data values are around the mean. A smaller standard deviation indicates that values are clustered closely near the mean, while a larger standard deviation indicates greater variability. Standard deviation is based on deviations from the mean, squared deviations, variance, and then the square root of variance. It is expressed in the original units of measurement, making it easier to interpret than variance. Option B is incorrect because central tendency is measured by statistics such as the mean, median, and mode. Option C is incorrect because probability measures likelihood, not data dispersion. Option D refers to frequency or the number of observations, not their variability. For example, two classes could have the same mean test score but different standard deviations; the class with the larger standard deviation has more varied scores. Study Guide references/topics: standard deviation, variance, spread, descriptive statistics.
NEW QUESTION # 94
Uniform distribution 0-5: P(X < 3) = ?
Answer: C
Explanation:
For a continuous uniform distribution on the interval from 0 to 5, probability is proportional to interval length.
The total interval length is 5 # 0 = 5. The event X < 3 corresponds to the interval from 0 to 3, which has length 3. Therefore, P(X < 3) = 3/5 = 0.6. This works because the uniform distribution assigns constant density across the entire interval, so any subinterval's probability equals its length divided by the total length.
Option B would correspond to half the interval, such as X < 2.5. Option C would correspond to length 2 out of
5. Option D incorrectly treats the cutoff value 3 as if it were a percentage rather than an interval boundary.
The correct answer is 0.6, meaning 60% of the uniform distribution lies below 3. Study Guide references
/topics: uniform distribution, continuous probability, interval length, probability density.
NEW QUESTION # 95
Null hypothesis for a new drug shows no effect. True statement?
Answer: A
Explanation:
The null hypothesis represents the default claim or baseline assumption in hypothesis testing. For a new drug, a typical null hypothesis states that the drug has no effect, no difference from placebo, or no improvement over the existing standard. The testing process begins by assuming the null hypothesis is true, then evaluates whether the sample evidence is strong enough to reject it. Therefore, the correct statement is that researchers assume no effect until evidence suggests otherwise. This does not mean the drug is proven ineffective; it means the burden of evidence lies with demonstrating an effect. Option B is the alternative hypothesis, not the null. Option C overstates the conclusion because failing to reject the null does not prove the drug has no effect. Option D is false because this type of claim is testable using experimental data and statistical inference.
Study Guide references/topics: null hypothesis, alternative hypothesis, hypothesis testing, statistical evidence.
NEW QUESTION # 96
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: B
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 # 97
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