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hw7Spring08answers

hw7Spring08answers - 0,6 STAT 301T HW7 ANSWERS 10.8(3 Both...

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Unformatted text preview: 0,6 STAT 301T HW7 ANSWERS 10.8. (3) Both rules are satisfied: All probabilities are between 0 and 1 (0% and 100%), and the percents add to 100%. (We also should note that the four groups described are- nonoverlapping, and account for all possible classifications of potential customers.) (b) 35% (20% + 15%) are currently undergraduates. 10.42. (a) It is legitimate because every person must fall into exactly one category, and the probabilities add to 1. (b) 0.125 = 0.000 + 0.003 + 0.060 —l— 0.062 is the probability that a randomly chosen American is Hispanic. (c) 0.309 = 1 — 0.691 is the probability that a randomly chosen American is not a non-Hispanic white. 11.2. 68% is a parameter (related to the population of all registered voters in Indianapolis); 73% is a statistic (related to the sample of registered voters among those called). 11.8. (a) a/«/3 5 5.7735 mg. (b) Solve a/fi = 5: fl = 2, so 11 = 4. The average of several measurements is more likely than a single measurement to be close to the mean. 11.9. (a) If X has a N(300, 35) distribution, we have P(X > 300): 0. 5 and P(X > 335): P(Z > l)— — [58 2. b) The average of four independent NAEP scores has mean 1.1— - 300 and standard deviation a/«/__ — 17. 5. (Specifically, it has [approximately] a Normal distribution with that mean and standard deviation.) (c)_P(f > 300) = 0.5 and P(f > 335) = P(Z > 2) = = .012. 11.12. (a) Ramon’s score is about the 64th percentile: P (X 5 1100) éP P(_Z < 1——-10%091026)— - P(Z 5 0.35) = 0.6368 (W). (b) 56 = 1100 is about the 99.9th percentile: f is approximately Normal with mean 1026 and standard deviation 209/V7 70 4 24. 9803, so P(x < 1100)— _. P(Z_ < 2. 96): 0.9985. (c) The first answer jail-3% accurate: The distribution of an individual’ 5 score (like Ramon’ 3) might not be Normal, but the central limit theorem says that the distribution of x will be close to Normal. 11.13. State: What is the probability that the average loss will be no greater than $275? Formulate: Use the central limit theorem to approximate this probability. Solve: The central limit theorem says that, in spite of the skewness of the population distribution, the average loss among 10,000 policies will be approximately N ($250, U/m ) = N ($250, $10). Because $275 is 2.5 standard deviations above the mean, the probability of seeing an average loss of no more than $275 is about 0.9938. Conclude: We can be about 99.4% certain that average losses will not exceed $275 per policy. 11.36. Let X be Shelia’s measured glucose level. (a) P(X > 140) = P(Z > 1.5) = 0.0668. (b) If f is the mean of four measurements (assumed to be independent), then f has a N(125, IO/JZ) = N(125 mg/d1,5 mg/dl) distribution, and P(f > 140) z P(Z > 3) = 0.0013. 11.38. The mean of four, measurements has a N (125 mg/d1,5 mg/dl) distribution, and P(Z > 1.645) =0.05 if Z is N(O, 1), so L: 125 + 16.4!- 5: 133 .23 mg/dl. . 0R 115+165..5 =133.15 ...
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