APPENDIX D 779 e. The results suggest that the surveyed voters either lied or had defective memory of how they voted. 41. 0.0468 43. 0.000969 (Use A = 6, B = 43, n = 6, and x = 4.) Section 5-3 1. m = 0.470 arrivals; x = 2 arrivals; e ≈ 2.71828, which is a constant used in all applications of Formula 5-9. 3. The probability of exactly 2 Internet arrivals in one thousandth of a minute is P1x2 = 0.0690 (or 0.0691 if using the unrounded value of m). 5. a. P142 = 0.123 b. In 118 years, the expected number of years with 7 hurricanes is 14.5. c. The expected value of 14.5 years is very close to the actual value of 14 years, so in this case the Poisson distribution works quite well. 7. a. P132 = 0.113 b. In 118 years, the expected number of years with 3 hurricanes is 13.3. c. The expected value of 13.3 years is close to the actual value of 17 years, so in this case the Poisson distribution works well. 9. a. 16.3 births b. 0.0989 (or 0.0990 if using the unrounded mean) c. 0+ (or 0.0000000834 or 0.0000000851 if using the unrounded mean). Yes, 0 births in a day would be a significantly low number of births. 11. a. 62.2 b. 0.0155 (0.0156 using the rounded mean) 13. a. 0.170 b. The expected number is between 97.9 and 98.2, depending on rounding. c. The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2 hits. 15. a. m = 30.41176471 chocolate chips (or 30.4 rounded). b. P1262 = 0.0557 (or 0.0558 if using the rounded mean). c. Expected number: 1.9 cookies. The expected number of cookies is 1.9, and that is very close to the actual number of cookies with 26 chocolate chips, which is 2. 17. 0.0000172. No, it is highly unlikely that at least one jackpot win will occur in 50 years. You would need to play for more than 306,534 years to reach a 10% chance of winning at least one jackpot. Don’t hold your breath. Chapter 5: Quick Quiz 1. 0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible for event A to occur. 2. No, the sum of the probabilities is 1.5, which is greater than 1. 3. Yes. The random variable x has numerical values, the sum of the probabilities is 1, and each probability is between 0 and 1. 4. m = 3.0 5. The mean is a parameter. 6. Yes. The sum of the probabilities is 0.999, but because of rounding errors, it can be considered to be 1. 7. 0.827 or 0.828 8. m = 1.5 sleepwalkers 9. 1.0 sleepwalker2 7. Not binomial. Each response has more than two possible outcomes. 9. Not binomial. Because the senators are selected without replacement, the selections are not independent. (The 5% guideline for cumbersome calculations should not be used because the 40 selected senators constitute 40% of the population of 100 senators, and 40% exceeds 5%.) 11. Binomial. Although the events are not independent, they can be treated as being independent by applying the 5% guideline for cumbersome calculations. The sample size of 3600 is not more than 5% of the population of all households. 13. a. 0.128 b. WWC, WCW, CWW; 0.128 for each c. 0.384 15. 0.0413 (Table: 0.041) 17. 0.315 (Table: 0.316) 19. 0.0168 (Table: 0.017) 21. 0.238 23. 0.0121 25. 0.00000451. The result of 7 minorities is significantly low. The probability shows that it is very highly unlikely that a process of random selection would result in 7 or fewer minorities. (The Supreme Court rejected the claim that the process was random.) 27. a. 0.00128 b. Yes, because the probability of 12 or more is 0.00149, which is low (such as less than 0.05). c. 0.999 29. a. m = 18.0 girls; s = 3.0 girls b. Values of 12.0 girls or fewer are significantly low, values of 24.0 girls or greater are significantly high, and values between 12.0 girls and 24.0 girls are not significant. c. Yes, because the result of 26 girls is greater than or equal to 24.0 girls. A result of 26 girls would suggest that the XSORT method is effective. 31. a. m = 7.5 peas; s = 1.4 peas b. Values of 4.7 or less are significantly low, values of 10.3 or greater are significantly high, and values between 4.7 and 10.3 are not significant. c. No, because the result of 9 peas with green pods is not greater than or equal to 10.3. 33. With a probability of 0.158 that the combined sample tests positive, it is not unlikely because the probability of 0.158 is not very low. 35. 0.985. The high probability shows that a high percentage of ammo cans will be accepted, so it does not appear that there is a production problem. 37. a. 32.4 and 57.3 (or 32.5 and 57.3 if using the rounded s). The result of 36 brown M&Ms lies between those limits, so it is neither significantly low nor significantly high. b. 0.0241 c. The probability of 36 or fewer brown M&Ms is 0.0878. d. The probability from part (c) is relevant. The result of 36 brown M&Ms is not significantly low. e. The results do not provide strong evidence against the claim that 13% of M&Ms are brown. 39. a. 238.3 and 287.2 (or 287.1 if using the rounded s). Because 308 is greater than 287.2, the value of 308 is significantly high. b. 0.0000369 c. The probability of 308 or more is 0.000136. d. The probability from part (c) is relevant. The result of 308 voters who voted for the winner is significantly high.
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