Probability and Random Variables Multiple Choice Practice Problems for AP Statistics (page 2)
Review the following concepts if necessary:
- Probability for AP Statistics
- Random Variables for AP Statistics
- Normal Probabilities for AP Statistics
- Simulation and Random Number Generation for AP Statistics
- Combining Random Variables for AP Statistics
- 12/125, 28/125
- 12/63, 28/60
- 12/125, 28/63
- 12/125, 28/60
- 12/63, 28/63
- It turns out that 25 seniors at Fashionable High School took both the AP Statistics exam and the AP Spanish Language exam. The mean score on the Statistics exam for the 25 seniors was 2.4 with a standard deviation of 0.6 and the mean score on the Spanish Language exam was 2.65 with a standard deviation of 0.55. We want to combine the scores into a single score. What are the correct mean and standard deviation of the combined scores?
- 5.05; 1.15
- 5.05; 1.07
- 5.05; 0.66
- 5.05; 0.81
- 5.05; you cannot determine the standard deviation from this information.
- The GPA (grade point average) of students who take the AP Statistics exam are approximately normally distributed with a mean of 3.4 with a standard deviation of 0.3. Using Table A, what is the probability that a student selected at random from this group has a GPA lower than 3.0?
- The 2000 Census identified the ethnic breakdown of the state of California to be approximately as follows: White: 46%, Latino: 32%, Asian: 11%, Black: 7%, and Other: 4%. Assuming that these are mutually exclusive categories (this is not a realistic assumption, especially in California), what is the probability that a random selected person from the state of California is of Asian or Latino descent?
- The students in problem #4 above were normally distributed with a mean GPA of 3.4 and a standard deviation of 0.3. In order to qualify for the school honor society, a student must have a GPA in the top 5% of all GPAs. Accurate to two decimal places, what is the minimum GPA Norma must have in order to qualify for the honor society?
- The following are the probability distributions for two random variables, X and Y:
- The following table gives the probabilities of various outcomes for a gambling game.
- You can't answer this question since this is not a complete probability distribution.
- You own an unusual die. Three faces are marked with the letter "X," two faces with the letter "Y," and one face with the letter "Z." What is the probability that at least one of the first two rolls is a "Y"?
- You roll two dice. What is the probability that the sum is 6 given that one die shows a 4?
In the table above what are P(A and E ) and P(C | E )?
For the tree diagram pictured above, what is P(B | X )?
If X and Y are independent, what is P(X = 5 and Y = 4)?
What is the player's expected return on a bet of $1?
- The correct answer is (c). There are 12 values in the A and E cell and this is out of the total of 125. When we are given column E, the total is 63. Of those, 28 are C.
- The correct answer is (b).
- The correct answer is (e). If you knew that the variables "Score on Statistics Exam" and "Score on Spanish Language Exam" were independent, then the standard deviation would be given by
- The correct answer is (a).
- The correct answer is (d). Because ethnic group categories are assumed to be mutually exclusive, P(Asian or Latino) = P(Asian) + P(Latino) = 32% + 11% = 43%.
- The correct answer is (e). The situation is as pictured below:
- The correct answer is (c). P(Y = 4) = . Since they are independent,
- The correct answer is (c). The expected value is (–1)(0.6) + (1)(0.25) + (2)(0.15) = –0.05.
- The correct answer is (d). P(at least one of the first two rolls is "Y") = P(the first roll is "Y") + P(the second roll is "Y") – P(both rolls are "Y") = . Alternatively, P(at least one of the first two rolls is "Y") = 1 – P(neither roll is "Y") = .
- The correct answer is (b). The possible outcomes where one die shows a 4 are highlighted in the table of all possible sums:
(This problem is an example of what is known as Bayes's rule. It's still conditional probability, but sort of backwards. That is, rather than being given a path and finding the probability of going along that path—P(X | B) refers to the probability of first traveling along B and then along X—we are given the outcome and asked for the probability of having gone along a certain path to get there—P(B | X) refers to the probability of having gotten to X by first having traveled along B. You don't need to know Bayes's rule by name for the AP exam, but you may have to solve a problem like this one.)
However, you cannot assume that they are independent in this situation. In fact, they aren't because we have two scores on the same people. Hence, there is not enough information.
The calculator answer is normalcdf(-100,3,3.4,0.3) = 0.0912. Note that answer (d) makes no sense since probability values must be non-negative (and, of course, less than or equal to 1).
From Table A, zx = 1.645 (also, invNorm(0.95) = 1.645).
Hence, . Norma would need a minimum GPA of 3.89 in order to qualify for the honor society.
There are 11 cells for which one die is a 4 (be careful not to count the 8 twice), 2 of which are 6's.
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