CCF08182010_00016

# CCF08182010_00016 - — Solutions for Exam i— Morning...

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Unformatted text preview: — Solutions for Exam i— Morning Section 7 i Use Euler’s formula, elezcosﬁ-l-isinﬁ 6:7r/2 2'1 - e 2 =2 .' '1 1 Al zz=(622) :6 2 Answer is D. 18. A graph of the area is shown. Since y = \/ 4 — \$2 is the equation for the top half of a circle, the volume after a revolution will be a hemi— sphere. V = iVsphere = (i) (i) ”7‘3 = (i) (3)9913 : 16.76 (17) Answer is C. 19. p{at least 2}: p{2 or 3 or 4 or 5} = 1 —p{0 or 1} = l —p{0} -p{1} Use binomial distribution to ﬁnd the individual proba— bilities. p : probability of heads 2 0.5 q = probability of tails = 0.5 mm = (0.5)5 : 0.03125 p{1} = (i) (0.5)4(0.5)1 = 0.15625 p{at least 2} = 1 — 0.03125 — 0.15625 : 0.8125 (0.81) Answer is C . 1+4+7_ 4 3 20. mean = The standard deviation is (1—4)2+(4—34)2+(7—4)2 =¢é 0': = 2.45 (2.5) Notice that n — 1 would have been used if the sample standard deviation had been requested. Answer is A. number of orange balls 7 21. b b'l‘ = _— = A pro a 1 1ty total number of balls 17 = 0.4118 (0.41) Answer is D. 22. First calculate the probability of rolling no sixes. Then subtract that value from 1 to ﬁnd the probability of rolling at least one six. p{ﬁve dice, no sixes} = p{one die, no six}5 = 0.402 ﬁve dice, at 1) least one six } = 1 — p{ﬁve dice, no sixes} = 1 — 0.402 : 0.598 (0.60) Answer is C. 23. The mean is E_ :32. _ 2+7+9+12+34 :1. n 5 28 The sample standard deviation is (2 — 12.8)2 + (7 1 12.8)2 + (9 1 12.8)2 + (12 — 12.8)2 + (34 —12.8)2 5 — 1 = 12.397 (13) Answer is C . 24. The mean of the Poisson distribution, A, is 5 jams. The probability of 3 jams is 6—))? _ e_5(5)3 33' — 3' = 0.140 p{3} = Answer is B. # Professional Publications, Inc. ...
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