Statistics Practice Test Classwork Answers

Chapter 5 AP Statistics Practice Test Section 1: Multiple Choice Question 15.1 Prompt: Dr. Stats plans to flip a coin 10,000 times in the hope that it will lead him to a deeper understanding of the laws of probability. Which of the following statements is true?   Options: (a) It is likely that Dr. Stats will get more than 5000 heads. (b) Whenever Dr. Stats gets a string of 15 tails in a row, it becomes more likely that the next toss will be heads.   (c) The fraction of tosses resulting in heads should be close to 1/2. (d) The chance that the 100th toss will be a head depends somewhat on the results of the first 99 tosses. (e) All of the above statements are true.   Note: A small text box on the right side states: "PROBABILITY ONLY TELLS US WHAT HAPPENS APPROXIMATELY IN THE LONG RUN, NOT WHAT WILL HAPPEN IN THE SHORT RUN." Question 15.2 Prompt: 1.2 billion people live in China. Marketers want to know which national brands they have heard of. A large study showed that 62% of Chinese adults have heard of Coca-Cola. You want to simulate choosing a Chinese adult at random and asking if he or she has heard of Coca-Cola. One correct way to assign random digits to answer this question is: Options: (a) One digit simulates one person's answer: odd means "Yes" and even means "No." (b) One digit simulates one person's answer: 0 to 6 means "Yes" and 7 to 9 means "No."   (c) One digit simulates one person's answer: 0 to 9 tells how many in the sample said "Yes." (d) Two digits simulate one person's answer: 00 to 61 means "Yes" and 62 to 99 means "No." (e) Two digits simulate one person's answer: 00 to 62 means "Yes" and 63 to 99 means "No." Question 15.3 Prompt: A housing company builds homes with two-car garages. What percent of households have more cars than the garage can hold? Table: | Number of cars | 0 | 1 | 2 | 3 | 4 | 5 | P(More than 2) = 0.13 + 0.05 + 0.02 = 0.20 | | Probability | 0.09 | 0.16 | 0.35 | 0.13 | 0.05 | 0.02 | Options: (a) 7% (b) 13% (c) 45% (d) 55% Question 15.4 Prompt: Computer voice recognition software is getting better. Some companies claim that their software can recognize 98% of all words spoken by a trained user. To simulate recognizing a single word when the probability of being correct is 0.98, let two digits simulate one word, 00 to 97 means "Correct." The program recognizes words independently. To simulate the program's performance on 10 words, I would randomly select 9 out of 10.   Prompt: The number of words scored correctly out of the 10 is: Options: (a) 10 (b) 9 (c) 8 (d) 7 (e) 6 Section 1: Multiple Choice Questions 15.5 to 15.7 Context: The data in the table below summarizes the results of a survey of one thousand students at a city high school. Our interest is in their GPA and whether or not they consistently skipped classes. Table: | GPA | Skipped Classes | |---|---| | 2.0 or Lower | Many | 90 | | 2.0 or Lower | Few | 25 | | Greater than 2.0 | Many | 175 | | Greater than 2.0 | Few | 710 | | Total | | 1000 | Question 15.5: What is the probability that a student has a GPA greater than 2.0? P(GPA > 2.0) = 255/1000 = 0.255 Question 15.6: What is the probability that a student has a GPA under 2.0 or has skipped many classes? P(GPA < 2.0 OR Skipped Many Classes) = (90 + 25 + 175)/1000 = 290/1000 = 0.290 Question 15.7: What is the probability that a student has a GPA under 2.0, given that he or she has skipped many classes? P(GPA < 2.0 | Skipped Many Classes) = 90/110 = 0.818 Question 15.8 For events A and B related to the same chance process, which of the following statements is true? (a) If A and B are mutually exclusive, then they must be independent. (b) If A and B are independent, then they must be mutually exclusive. (c) If A and B are not mutually exclusive, then they must be independent. (d) If A and B are not independent, then they must be mutually exclusive. (e) If A and B are not independent, then they cannot be mutually exclusive.   Explanation: If A and B are independent, then we don't know whether B has occurred if A occurred. But if A and B are mutually exclusive, then if B has occurred, then we know that A couldn't have occurred.   Question 15.9 Choose an American adult at random. The probability that the person is a woman is 0.52. The probability that the person has never married is 0.25. The probability that the person is a woman and has never married is 0.11. What is the probability that the person is a woman or has never married? P(Woman) = 0.52 P(Never Married) = 0.25 P(Woman and Never Married) = 0.11 P(Woman or Never Married) = 0.52 + 0.25 - 0.11 = 0.66 Question 15.10 A sleek card playing deck has 52 cards, of which 12 are face cards. If you shuffle the deck well and turn over the top 3 cards, one after the other, what's the probability that all 3 are face cards? P(1st face and 2nd face and 3rd face) = (12/52) * (11/51) * (10/50) = 1320/132600 = 0.009905