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Midterm Exam Solution and discussion

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1.A random variable A variable is used in order to describe the quantities of interest that are determined by the outcomes of an experiment. A random variable is usually not a fixed parameter, but a variable composed of all possible experimental results. So the distribution of values of a random variable becomes the focus of experimentation.

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2. p.m.f. vs. p.d.f. The p.m.f is used to describe the relative probability for at most a countable number of a discrete random variable. In contrast, the p.d.f. is used for describing the relative interval probability of a continuous random variable.

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3. A competitive campaign A proper sampling frame should be exhaustively 1-to-1 mapping to the entire population. The members of Internet club are just belonging to a segment of total voters who may be more innovative, or more high-income, or possessing a special interest. The campaign manager must be careful to clarify the size of this special segment relative to the total voters.

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4. Exclusive vs. independent Suppose two events A, and B If A and B are mutually exclusive, then A ∩ B=φ, and P{A ∩ B}=0, i.e., P{A/B}=0 If A and B are independent, then P{A/B}=P{A}, or P{A ∩ B}=P{A} ×P{B}

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5. The existence of covariance Var(X 1 +X 2 )=Var(X 1 )+Var(X 2 )+2Cov(X 1,X 2 ) Let X=X 1 =X 2, then Var(X 1 )+Var(X 2 )+2Cov(X 1,X 2 )=2Var(X)+2 Var(X)=4Var(X)

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1. The stem-and-leaf plot 1 2 3 4 5 6 7 1 8 7,9 1,3,7 1,1,3,4,6,8,9 2,5 2,7 3 3,5,6 2,4

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2. The box-and-whisker plot ∵ N=22, p=0.5, np=11, the median=(the 11th+the 12th)/2=51 p=0.25, np=5.5, the Q 1 =the 6th number=32 p=0.75, np=16.5, the Q 3 =the 17th number=59 ∵ Interquartile range=Q 3 -Q 1 =27, ∴ the lower fence=-8.5, the upper fence=99.5 Min=17>the lower fence, max=83

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3. Marginal p.d.f. ∵ f(X,Y)< fx(X)× fy(Y), ∴ X is not independent from Y 此題因原 joint p.d.f. 有誤, 只要列式正確則一律給分

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4. The insurance risk Using the Bayes ’ rule P{accident}=P{g}×P{accident/g} +P{a}×P{accident/a}+P{b}×P{accident/b} =0.2×0.05+0.5×0.15+0.3×0.3 =0.175 P{no accidents}=1-0.175=0.825 P{g/no accident}=(P{g}×P{no accidents/g}) /P{no accidents} =(0.2×0.95)/0.825=0.23

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5. Operations of expectation E[X]=2, E[X2]=8, E[(2+4X)2]=E[4+16X+16X2] =4+16×2+16×8=164 E[X2+(X+1)2]= E[X2]+E[X2+2X+1] =8+8+4+1=21

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6. The circuit problem (a) The successful probability=1-the failure probability=[1-(1-p)2][1-(1-p)2] =[1-(1-p)2]2=4p2- 4p3+p4 (b) If the central bridge is closed, then the situation is the same as (a). But, we have to consider the additional situation that occurs when the central bridge is open. The failed upper/bottom circuit flow: 1-p2; The successful chance in the situation of open bridge: (1-p)[1-(1-p2)(1-p2)] ∴p(4p2-4p3+p4)+(1-p)[1-(1-p2)(1-p2)]=p2(2+2p-5p2+2p3)

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7. The birthday problem The probability of no two having the same birthday: (365/365)(364/365)(363/365) … [(365- n+1)/365]=(365)(364) … (365-n+1)/(365)n If n>23, the probability is less than 0.5, i.e., P{at least one pair having the same birthday}>0.5; Moreover, the number of students is usually more than 23. So, the needed gifts may cost too much. Don ’ t accept this marketing proposal.

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