ODD ANSWERS A63 13. (a) 0.191 (b) 0.891 (c) 0.700 15. (a) 0.257 (b) 0.774 (c) 0.517 17. (a) x P(x) 0 0.002 1 0.022 2 0.114 3 0.293 4 0.376 5 0.193 (b) Number of adults Probability 0 1 2 3 4 5 Adults Who Have Read a Book in the Last Year P(x) x 0.10 0.20 0.30 0.40 0.05 0.15 0.25 0.35 Skewed left (c) The values 0 and 1 are unusual because they have a probability of less than or equal to 0.05. 19. m ≈ 1.0; s 2 ≈ 0.9; s ≈ 1.0. On average, 1 out of every 8 drivers is uninsured. The standard deviation is 1.0, so most samples of 8 drivers would differ from the mean by at most 1 driver. Unusual values would be any above 3. 21. (a) 0.104 (b) 0.166 (c) 0.405 23. (a) 0.154 (b) 0.217 (c) 0.011; unusual 25. (a) 0.073 (b) 0.458 (c) 0.384 Quiz for Chapter 4 (page 228) 1. (a) Discrete. The number of lightning strikes that occur in Wyoming during the month of June is a random variable that is countable. (b) Continuous. The fuel (in gallons) used by a jet during takeoff is a random variable that has an infinite number of possible outcomes and cannot be counted. (c) Discrete. The final score in a game of bowling is a random variable that is countable. 2. (a) x P(x) 0 0.238 1 0.405 2 0.209 3 0.090 4 0.040 5 0.019 (b) Number of wireless devices Probability 0.1 0 1 2 3 4 5 Wireless Devices per Household 0.2 0.3 0.4 0.5 P(x) x Skewed right (c) m ≈ 1.3; s 2 ≈ 1.4; s ≈ 1.2. The mean is 1.3, so the average number of wireless devices per household is 1.3. The standard deviation is 1.2, so most households will differ from the mean by no more than 1.2 wireless devices. (d) 0.058 3. (a) 0.273 (b) 0.860 (c) 0.040 4. (a) x P(x) 0 0.0002 1 0.004 2 0.033 3 0.132 4 0.297 5 0.356 6 0.178 (b) Number of patients Probability 0.35 0.30 0.40 0.25 0.20 0.15 0.10 0.05 1 2 3 5 6 0 4 5-Year Survival Rate of Liver Transplants P(x) x Skewed left (c) mean = 4.5; variance = 1.12; standard deviation = 1.06. On average 4.5 out of every 6 patients survive to five years. The standard deviation is 1.06, so most samples of 6 patients would differ from the mean by at most 1.06 survivors. 5. (a) 0.175 (b) 0.440 (c) 0.007 6. (a) 0.140 (b) 0.125 (c) 0.034 7. Event (c) in problem 5 and event (c) in problem 6 are both unusual because the probabilities are less than 0.05. Real Statistics—Real Decisions for Chapter 4 (page 230) 1. (a) Sample answer: Calculate the probability of obtaining 0 live births out of 6 randomly selected ART cycles. (b) Binomial. The distribution is discrete because the number of live births is countable. 2. n = 6, p = 0.52 x P(x) 0 0.012 1 0.079 2 0.215 3 0.311 4 0.253 5 0.109 6 0.020 Sample answer: Because P102 = 0.12, this event is unusual but not impossible. 3. (a) Not suspicious, because the probability is more than 0.05 (b) Suspicious, because the probability is less than 0.05 Chapter 5 Section 5.1 (page 242) 1. When it is a normal distribution, the mean is equal to the median. 3. Inflection points are where the curve changes concavity. Inflection points occur one standard deviation away from the mean on both the right and left. The x-values would be m - s and m + s.
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