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Midterm Exam

Midterm Exam. Solution and discussion. 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.

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Midterm Exam

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  1. Midterm Exam Solution and discussion

  2. 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.

  3. 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.

  4. 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.

  5. 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}

  6. 5. The existence of covariance • Var(X1+X2)=Var(X1)+Var(X2)+2Cov(X1,X2) • Let X=X1=X2, then Var(X1)+Var(X2)+2Cov(X1,X2)=2Var(X)+2Var(X)=4Var(X)

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

  8. 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 Q1=the 6th number=32 • p=0.75, np=16.5, the Q3=the 17th number=59 • ∵Interquartile range=Q3-Q1=27, ∴the lower fence=-8.5, the upper fence=99.5 • Min=17>the lower fence, max=83<the upper fence, ∴ no significant outliers Q3 Q1 Q2 Max=83 Min=17 32 51 59

  9. 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

  10. 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

  11. 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

  12. 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)

  13. 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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