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HW2-Solutions.pdf - Bn‘osTaJt/OI 2.4 HM SoW‘cms H522...

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Unformatted text preview: Bn‘osTaJt/OI 2.4% HM) SoW‘cms H522: 3.5 33 3.8 3.10 a.) 6.8 6.? CHAPTER 3 My Exercise 6 a. The mean time to seizure is 241:1 3" 13 336.85 13 = 25.9 months. HI The median is the middle value, or 24 months. Since two different measurements each occur twice, the data set has two modes — 12 months and 24 months. The range is the largest value minus the smallest value, or 96 — 0.1 = 95.9 months. The interquartile range is the 75th percentile minus the 25th percentile, or 36 — 4 = 32 months. The standard deviation is 23:10:.- —~ i)” = 2,2,0..- - 25.9)2 13 -— 1 12 = v 749.2 = 27.4 months. b. Note that 13 Zn. — s) = (0.10 — 25.91) + (0.25 — 25.91) + (0.50 — 25.91) i=1 + (4 — 25.91) + (12 — 25.91) + (12 — 25.91) + (24 — 25.91) + (24 — 25.91) + (31 - 25.91) + (36 — 25.91) + (42 — 25.91) + (55 — 25.91) + (96 — 25.91) —-25.81 — 25.66 — 25.41 — 21.91 — 13.91 — 13.91 — 1.91 — 1.91 + 5.09 + 10.09 + 16.09 + 29.09 + 70.09 0.02 z 0. Because we use the mean rounded to two decimal places in the calculations, the sum is not exactly equal to 0. 1 E E E E E . Exercise 7 a. The mean calcium level is HI I = 3.14 mmol/l. The median is the average of the 4th and 5th values, or 2.99 + 3.17 2 = 3.08 mmol/l. The standard deviation is 8 — 1 23:,(2, — 3.14)2 7 0.2608 0.51 mmol/l. The range is the largest value minus the smallest value, or 3.84 — 2.37 1.47 mmol/l. b. The mean albumin level is 40.4 g/l. The median is 42 g/l. The standard deviation is 3.0 g/l. The range is 9 g/l. c. The patients sufl'ering from vitamin D intoxication all have albumin levels within the normal range. However, they do not have normal blood levels of calcium. Both the mean and the median lie above the upper limit of the normal range; overall, 6 of the 8 patients have calcium levels that are above normal. Exercise 8 a. The median daily caloric intake for the bulimic adolescents is 21.6 kcal/ kg, and the median for the healthy adolescents is 30.6 kcal/kg. b. The interquartile range for the bulimic adolescents is the 75th percentile minus the 25th percentile, or 25.2 — 18.1 = 7.1 kcal/kg. The interquartile range for the healthy adolescents is 36.6 — 23.8 = 12.8 kcal/kg. c. Daily caloric intake tends to be higher for the healthy adolescents. This group also exhibits a. greater amount of variability. E j 1. Exercise 10 8.. The grouped mean serum cotinine level for smokers is 3 J O 2L1 mifi is = 8 i=1 f7: 1 (—) [6.5(78) + 31.5(133) + 74.5(142) + 124.5(206) 1539 + 174.5(197) + 224.5(220) + 274.5(151) + 34o(412)] = 199.0 ng/ ml. The grouped variance for smokers is 2 232.1("12' " 5:9)2f.‘ 9 [XL-=1 ft] " 1 s (153; _ 1) [(—192.5)2(78) + (—167.5)2(133) + (—124.5)2(142) + (—74.5)2(206) + (-24.5)2(197) + (25.5)”(220) + (75.5)”(151)+ (141.0)”(412)] = 12, 535.3 (ng/ml)2, and the grouped standard deviation is s, = «12,5353 = 112.0 ng/ml. The grouped mean serum cotinine level for nonsmokers is a, = (374125) [6.5(3300) + 31.5(72) + 74.5(23) + 124.5(15) + 174.5(7) + 224.5(8) + 274.5(9) + 340(11)] = 10.6 ng/ml. The grouped variance for nonsmokers is 1 .992 = (3445 _ 1) [(—4.1)2(3300) + (20.9)”(72) + (63.9)”(23) + (113.9)”(15) + (163.9)”(7) + (213.9)”(8) + (263.9)2(9) + (329.3201)] = 798.4 (ng/ml)2, and the grouped standard deviation is 5,, = V7984 = 28.3 ng/ml. ”.5 A b. For smokers, the median serum ootinine level is the 769th measurement, which falls in the interval 200—249 ng/ml. For nonsmokers, the median is the 1722th measurement, which falls in the interval 0—13 ng/ml. c. The mean and median serum cotinine levels are much higher for smokers than for nonsmokers. There is also a lot more variability in values among the smokers. 6,? Exercise 7 ‘ a. A n B is the event that the individual is exposed to high levels of both carbon monom'de and nitrogen dioxide. b. A U B is the event that the individual is exposed to either carbon monoxide or nitrogen dioxide or both. c. Ac is the event that the individual is not exposed to high levels of carbon monoxide. d. The events A and B are not mutually exclusive. 6‘ 8 Exercise 8 b. Since P(A) x P(B) 0.142 x 0.051 0.0072 P(A n B) 0.031, ll‘iLll these two events are not independent. c. The probability that A or B or both occur is P(A u B) = P(A) + P(B) — P(A n B) 0.142 + 0.051 — 0.031 0.162. ll d. The probability that A occurs given that B occurs is P(A n B) P(B) 9-011 0.051 0.608. P(AlB) = II 6‘? Exercise 9 a. The probability that a woman who gave birth in 1992 was 24 years of age or younger is P(s 24) = P(< 15 or 15—19 or 20—24) P(< 15) + P(15—19) + P(20—24) 0.003 + 0.124 + 0.263 = 0.390. b. The probability that the woman was 40 years of age or older is P(Z 40) = P(40—44 or 45—49) = P(40—44) + P(45-49) = 0.014 + 0.001 0.015. c. Given that the woman was under 30 years of age, the probability that she was not yet 20 is P(< 20 and < 30) P(< 30) P(< 20) P(< 30) 0.003 + 0.124 0.003 + 0.124 + 0.263 + 0.290 = 0.187. P(< 20 |< 30) ll d. Given that the woman was 35 years of age or older, the probability that she was under 40 is P(< 40 and 2 35) H2 35) P(35 — -39) Hz 35) 0.085 0.085 + 0.014 + 0.001 0.850. P(< 40 I2 35) = Ii ...
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