Solution Sheet --- Statistics 101 Problem Set #31. Ex 4.24 page 250 a) S={UUUU, UUUD, UUDU, UDUU, DUUU, UUDD, DDUU, UDUD, DUDU, UDDU, DUUD, DDDU, DDUD, DUDD, UDDD, DDDD} b) S= {0,1,2,3,4}2. Ex 4.36 a) The height will be ½. b) P(Y<=1)=0.5*(1-0)=1/2. c) P(0.5<Y<1.3)=0.5*(1.3-0.5)=0.4. d) P(Y>=0.8)=0.5*(2-0.8)=0.6.3. Ex 4.54 a) DistributionsNumber of Persons (Household)0.100.300.50Probability Axis0 2 4 6 8Quantiles100.0% maximum 7.000099.5% 7.000097.5% 6.000090.0% 5.000075.0% quartile 4.000050.0% median 2.000025.0% quartile 1.250010.0% 1.00002.5% 1.00000.5% 1.00000.0% minimum 1.0000MomentsDistributionsNumber of Persons(Family)0.200.400.600.80Probability Axis1 2 3 4 5 6 7 8Mean 2.6Std Dev 1.42Std Err Mean 0.014upper 95% Mean 2.62lower 95% Mean 2.57N 10000Quantiles100.0% maximum 7.000099.5% 7.000097.5% 6.000090.0% 5.000075.0% quartile 4.000050.0% median 3.000025.0% quartile 2.000010.0% 2.00002.5% 2.00000.5% 2.00000.0% minimum 2.0000MomentsMean 3.14Std Dev 1.25Std Err Mean 0.012upper 95% Mean 3.16lower 95% Mean 3.11N 10000The average size of a household is 2.6 people and the standard deviation is 1.42. The average size of a family is 3.14 people and the standard deviation is 1.25. Both distributions are skewed to the right. The most obvious difference is that there are households that consist of only one person and there are no families of only one person.b) Ex 4.55 The distribution on number of rooms in owner-occupied units appears to be more spread out than number of rooms in renter-occupied units. Stdo=1.64, Stdr=1.31.4, Ex 5.40 a). P(X=12)=0.8^12=0.068; 1-P(X=12)= 0.93 b). E(Y)=12*0.2=2.4; Std(Y)=sqrt(12*0.2*0.8)=1.385 c). P(Y<2.4)=P(Y=0)+P(Y=1)+P(Y=2)= 0.00000452608 5, A wins game 2 B wins game 2A wins game 1 .6*.7=0.42 .6-0.42=.18 .6B wins game 1 .6-.42=.18 1-.78=.22 .4.6 .4 1 a) P (A wins at least one game)=1-0.22=0.78 b) A wins game 7 B wins game 7A wins game 6 0.49 .21 .7B wins game 6 0.3*(.18/.4)=.135 .165 .3.625 .375 1i) {{ A wins game 6}, { B wins game 6, A wins game 7}, { B wins game 6, B wins game 7}}. ii) P (A wins game 6) =0.7; P (B wins game 6, A wins game 7) =.135; P (B wins game 6, A wins game 7) =.165. iii) P (A wins the series)=1-.165=.835. iv) P (# of games=7 | A wins the series)= P(# of games=7, A wins the series)/P(A wins the series)=0.135/0.835=.1626, a) P1 p(x) P1p(x) P2 p(x) P2p(x) P3 p(x) P3p(x) 30 .4 12 50 .4 20 60 .4 24 90 .6 54 80 .6 48 60 .6 36 E(P1)= 66 E(P2)= 68 E(P3)=60 So P2 is preferred. b) Launch Not Launch + .4(3/4)=.3 .6(1/3)=.2 .5 - .4(1/4) =.1 .6(2/3)=.4 .5 .4 .6 P(Launch|+)=.3/.5 = .6 c) P1 p(x) P1p(x) P2 p(x) P2p(x) P3 p(x) P3p(x) 30 .6 18 50 .6 30 60 .6 36 90 .4 36 80 .4 32 60 .4 24 E(P1)= 54 E(P2)= 62 E(P3)=60 So P2 is preferred. d) If negative indication then probability it will be launched is .2 P1 p(x) P1p(x) P2 p(x) P2p(x) P3 p(x) P3p(x) 30 .2 6 50 .2 10 60 .2 12 90 .8 72 80 .8 64 60 .8 48 E(P1)= 78 E(P2)= 74 E(P3)=60 P1 preferred. Since you make 62 with a + indication and 78 with a negative indication and there is a 50% chance of each (see the above table), the expected profit is 70.7, Extra Credit Consider First Parent’s Eye Colors. P (M is Blue and F is Blue) =1/4*1/4=1/16 Since P (M is B) = ¼, P (M is Blue and F is not Blue)=1/4*3/4=3/16 P (M is not Blue and F is not Blue) = ¾*3/4=9/16 Therefore M is B & F is BM is B & F is not BM is not B & F is BM is not B & F is not BChild is B 1/16 1/16 1/16 1/16 4/16Child is not B 0 2/16 2/16 8/16 12/161/16 3/16 3/16 9/16 1a) P (M is B and F is B)=1/16+3/16+3/16=0.4375 b) P (M is not B or F is not B or C is not B)=1-P(M is B and F is B and C is B) =1-(1/16)=0.9375 c) P (C is B | F is not B and M is B) = (1/16)/(9/16)=1/9 d) P (M is B or F is B) | C is B) = (1/16+1/16+1/16)/(4/16) =