Anna Audler Stats 10: Section 3A UID: 304295761 TA: Luis Sosa Homework #5 6.2: A) Continuous B) Discrete 6.7: A) Winnings Probability $3 1/6 $0 1/2 -$4 1/3 B) 6.17: A) 0.8461 B) 1 – 0.8461 = 0.1539 6.19: A) 0.9608 B) 1 – 0.9608 = 0.0392 C) 0.1515 – 0.0968 = 0.0547 6.23: z = (675 – 500)/100 = 1.75; table shows value of 0.9599 Percentage of female college-bound students above 675 = 1 – 0.9599 = 0.0401 0"0.1"0.2"0.3"0.4"0.5"0.6"*5" *4" *3" *2" *1" 0" 1" 2" 3" 4"Probability*Winnings*(in*dollars)*6.24: z = (675 – 530)/100 = 1.45; table shows value of 0.9265 Percentage of male college-bound students above 675 = 1 – 0.9265 = 0.0735 6.27: z1 = (279 – 270)/9 = 1 z2 = (261 – 270)/9 = -1 Since the z scores are -1 and 1, we can use the empirical rule to determine the probability = 0.68 6.32: A) z = (39 – 38)/2 = 0.5; table shows value of 0.6915 Percentage of boys with height greater than 39in = 1 – 0.6915 = 0.3085 B) Since height is normally distributed with a mean of 38, we can expect to have .5000 of the boys to have a height greater than 38in; or z = (38 – 38)/2 = 0.00; table shows value of 0.5000 6.42: A) 0.53 B) 1.65 6.44: 96th percentile = .9600; the corresponding z-score is 1.75 1.75 = (x – 530)/100; x = 705 = SAT Score at 96th percentile 6.48: 20th percentile = .2000; the corresponding z-score is- 0.84 -0.84 = (x – 70)/3; x = 67.48in = height at 20th percentile 6.54: There are 2 outcomes: heads or tails; fixed number of trials: 3; same probability of getting heads or tails for each trial: 1/2; the trials are independent: the outcome of a coin flip does not depend on the previous coin flip 6.59: A) b(100, 0.54, 65) B) b(10, 0.90, 9) C) b(10, 0.25, 6) 6.61: b(6, 0.40, 2) = 0.311 6.65: A) (0.5)(0.5) = 0.25 B) (0.5)(0.5)(0.5) = 0.125 C) 1 – 0.125 = 0.875 D) Yes: the more children she has, the more likely will have at least one more boy 6.68: A) All cleared: 0.00034 B) None cleared: 0.0135 C) At least one cleared: 1-0.0135 = 0.987 D) About 4 or 5 would be cleared on average
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