2005-Fall-Midterm_s

2005-Fall-Midterm_s - More PastPaper:...

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Unformatted text preview: More PastPaper: http://ihome.ust.hk/~cs_gxx ISMT111 F05 Midterm Exam Suggested solution Question 1 (a) One gets the right sequence by either the combination (B,R,R) or (R,R,R). The probability that a red ball is chosen on the 2nd and 3rd pick is then (3/8)(5/7)(4/6)+(5/8)(4/7)(3/6)=5/14 (b) (c) (d) (e) Mean is 100(5/14) and variance is 100(5/14)(9/14) Use P(Y100<=20.5) then apply the normal approximation. This is binomial with n=10, use the binomial exact formula. Let AR be the event that an extra red ball is added to a box. Then the probability of adding a red ball to a single box is P(AR)=P(AR|R)P(R)+P(AR|B)P(B)=1/2(5/8); This is a binomial problem and hence the expectation is 100(1/2) 5/8. Question 2 a) P(head)=1 x (1/2)+(1/2)(1/2)=3/4 b) P(coin 1 and head)=P(head|coin1)P(coin1)=(1/2)(1/2)=1/4 c) P(coin1|head)=[P(coin1, head)]/P(head)=1/3 P(coin2|head) )=[P(coin2, head)]/P(head)=2/3 d) P(head down|head up)=[P(head down and head up)]/[P(head)] = P(coin2)/P(head) =(1/2)/(3/4)=2/3 Or P(head down|head up)=P(coin2 |head)=2/3 om e. us t.h k/ ~c s_ gx x/ Question 3: Let X be the number of minutes need to complete the exam, X~N(100, 152). (a) P(X<120) = P(Z<1.33) = 0.9082 (b) Let Y be the number of students not able to complete their exam, Y~Bin(1000, 0.0918). Distribution of Y can be approximated by Y*~N(91.8, 9.132 ) P(Y>100) can be approximated by P(Y*>100.5) = P(Z>0.95) = 0.5-0.3289 = 0.1711 ht tp :// ih (c) E(Y) = 1000×0.0918 = 91.8. About 92 students More PastPaper: http://ihome.ust.hk/~cs_gxx (d) P(X<x0) = 0.98. or P(Z<(x0-100)/15) = 0.98 (x0-100)/15 = 2.05. x0 = 130.75. i.e., 2 hours and 11 minutes. (e) P(X<90|X<105) = P(X<90)/P(X<105) = P(Z<-0.67)/P(Z<0.33) = 0.2514/0.6293 = 0.3995 Question 4: (a) No. If a lot of applicants have taken more than one statistics courses, then the average can be larger than 0.5 even though more than half of them have taken no statistics course before. (b) Yes. If we add back the top 25% and bottom 25% of the expected salary data to the middle 50%, the median will be basically the same. (c) By Chebyshev inequality, the range, mean ± 2 × standard deviation , contains at least 50% of the observations. That is 20,000 ± 2828=(17172, 22828). (d) Approximately 50% of the expected salaries have absolute deviations no more than 1,500 from mean. The range (18500, 21500) contains about 50% of applicants. ht tp :// ih om e. us t.h k/ ~c s_ gx x/ Question 5: (a) E(X) = (-5)(0.3) + 3(0.4) + 8(0.3) = 2.1 E(X2)= 25(0.3) + 9(0.4) + 64(0.3) = 30.3 Var(X) = E(X2) - E(X) 2 = 30.3 – 2.12 = 25.89 (b) P(X = -5%, Y = 5%) = P(X = -5%) P(Y = 5%) = (0.3)(0.6) = 0.18, because of the independence of X and Y. (c) P(X + Y = 0) = P(X = -5%, Y = 5%) + P(X = -3%, Y = 3%) = 0.18 + 0.16 = 0.34 (d) Using similar techniques as in (c) to work out c -4% 0% 2.5% 4% 6.5% P(Z=c) 0.12 0.34 0.12 0.24 0.18 (e) E(Z) = (-4)(0.12) + 0(0.34) + 2.5(0.12) + 4(0.24) + 6.5(0.18) = 1.95 E(Z2)= (16)(0.12) + 0(0.34) + 6.25(0.12) + 16(0.24) + 42.25(0.18) = 14.115 Var(Z) = E(Z2) - E(Z) 2 = 14.115 – 1.872 = 10.3125 (f) If the employee prefers higher expected return, we recommend fund A. If the employee wants to expose less risk, we recommend the portfolio. ...
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This note was uploaded on 02/27/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Spring '09 term at HKUST.

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