STAT111 - Homework 3 - Solutions

Transcription

1 STAT111 - Homework 3 - Solutions 30 points in total Problem 1 (IPS 4.39), 3 points Let A at least one of them is universal donor, then A c = all of them are not universal donor. By complement rule. P paq 1 P pa c q Further, let B i the i th donor is not universal donor, then P pa c q P pb 1 X B 2 X B 10 q Notice that B 1,, B 10 are independent, by multiplication rule P pa c q P pb 1 XB 2 XB 10 q P pb 1 qˆp pb 2 qˆ ˆP pb 1 0q p1 0.07q which implies P paq 1 P pa c q points total, for complement rule, for multiplication rule, and 1 for correct answer Alternative Solution: In the solution above, we basically assume sampling with replacement. But the real exact model should be without replacement which corresponds to a hyper-geometric model (not covered in class. So if you get same answer as above, ignore this part). Assume n is the number of Americans. When n is large, these two models are nearly the same because it is really hard to sample one element twice. In case any student treat the problem in this way, we give the solution below. The number of universal donors among all Americans is 0.07n and the rest 0.93n of them are not. ˆ0.93n P paq 1 P pa c q 1 10 ˆ n 10 n «320 millions. Substitute into the formula above, we get P paq If we keep enough digits (9 in this case), the two solutions will differ. The students might choose a different n but as long as the n is not very small, say n ą 10, 000, they will still get

2 Problem 2 (IPS 4.126), 4 points (a) See the table below. Bachelor s Master s Professional Doctorate Total Female Male Total , including the row and column Total is not necessary. (b) According to the table, the probability is (c) P pwoman Professionalq P pwoman and Professionalq P pprofessionalq (d) Not independent because P pwoman Professionalq P pwomanq (There are other ways to argue, e.g. P pwoman and Professionalq P pwomanq ˆ P pprofessionalq. Anything reasonable should be viewed as correct answer) Problem 3, 3 points The strategy is the same as Problem 1. Let L= Liara cannot be contacted, G= Garrus cannot be contacted and T= Tali cannot be contacted P pat least one can be contactedq 1 P pall cannot be contactedq 1 P pl X G X T q 1 P plq ˆ P pgq ˆ P pt q 1 p1 0.6q ˆ p1 0.45q ˆ p1 0.65q points total, for complement rule, for multiplication rule, and 1 for correct answer For those not comfortable with the shorthand L, G, T, P pl X G X T q P pliara, Garrus, Tali all cannot be contactedq and similar for P pgq, P pt q. P plq P pliara cannot be contactedq 2

3 Problem 4, 3 points (a) Let random variable X 1 be the time of first reaction and X 2 be the time of the second reaction and X be the time of the whole process. Therefore, X X 1 ` X 2 ` 3.5 and as a result, meanpxq meanpx 1 q ` meanpx 2 q ` ` 15 ` pminq (b) By the formula of standard deviation for the sum of two independent random variables sdpxq a sdpx 1 q 2 ` sdpx 2 q 2? ` pminq Problem 5, 5 points Let T= make the shot and F= miss the shot, then P pt q 2{3, P pf q 1{3 and the sample space S tt T T, T T F, T F T, F T T, T F F, F T F, F F T, F F F u Therefore, 3 ˆ2 P px 3q P pt T T q ˆ2 P px 2q P pt T F Y T F T Y F T T q 3 ˆ P px 1q P pt F F Y F T F Y F F T q 3 ˆ ˆ1 P px 0q P pf F F q ˆ for each scenario 3

4 Problem 6, 2 points (a) P pgame is tiedq P px 1q 2 9 (b) P pteam A winsq P px 2q ` P px 3q 4 ` Problem 7 (IPS 1.120), 3 points (a) -1.0, 2.1, -1.8, 0.4, 0.1, 2.6, -0.8, -1.7, 0.8, -0.1 for getting all numbers correct (b) We want to find t such that P pz ą tq Checking the normal table (or using R), t 1.04 (c) The second and sixth students earned an A. Problem 8 (IPS 1.133), Let S be the WAIS score. P ps ě 130q P p S where Z is standard normal distribution. for correct answer ě 2q P pz ě 2q Problem 9 (IPS 1.144), 3 points (a) HDL 55 P phdl ď 40q P p ď where Z is standard normal distribution. (b) HDL 55 P phdl ě 60q P p ě q P pz ď 0.97q q P pz ě 0.323q 0.373

5 where Z is standard normal distribution. (c) P p40 ă HDL ă 60q 1 P phdl ď 40q P phdl ě 60q 0.46 Problem 10, 3 points (a) Because normal distribution is symmetric, P piq ď 100q 1 2 For the proportion in the range [90, 110], we again reduce it to z-score IQ P p90 ď IQ ď 110q P p ď ď q P p ď Z ď 0.588q 1 2P pz ě 0.588q ď ˆ (b)we want to find t such that P piq ě tq 0.15, that is IQ 100 P p ě t 100 q 0.15 P pz ě t 100 q 0.15 Check the normal table, P pz ě 1.04q 0.15, therefore t which implies t 1.68 (1. 62 is the exact answer. Because of rounding, any solution nearby should be viewed as correct. ) 5