Running head: STATISTICSSTATISTICSName of the studentName of the university:Authors note:
1STATISTICSTable of ContentsAnswer 1:.........................................................................................................................................2Answer 2:.........................................................................................................................................2Answer 3:.........................................................................................................................................2References........................................................................................................................................3
2STATISTICSAnswer 1:(a) Given that the problem has geometric prior distribution with mean 100. The priordistribution is:P(N)=(1/100)(99/100)N-1Given that the far number 203 is being observed here. Hence, it can be expected to beat least 203 cars. Any of the cars has equal probability to be the number of the lastcar. The likelihood function can be evaluated here as:P(y/N) = (1/N), for each N>202,1, else where.The posterior distribution here will be:P(N/y) α P( N)*P(y/N)=(1/N)(0.01)(0.99)N-1α (1/N)(0.99)N(b) A probability distribution adds up to 1 and the posterior probability has to sum up to1. Therefore, the normalizing constants here are:P(X) = ∑(1/N)(0.99)N N=203(1) ∞Therefore, the posterior mean is:E[p(N/X)] = ∑N=20310000 N(1N)(99100)N−1p(X) = ∑N=20310000(99100)N−1p(X) = 279.0885And, the standard deviation is:
3STATISTICSsd[p(N/X)] = √∑(N−E[p(NX)])(1N)(99100)N−1p(X) = 79.96458N = 203:10000p.x. = sum((1/N)*(99/100)*(N-1))e.n. = sum(((99/100)*(N-1))/p.x)e.n.s.d.N = sqrt(sum((N-e.n.)^2*((1/N)*(99/100)*(N-1)/p.x.))s.d.N(c)Let the problem has constant prior distribution. The prior distribution is:P(N)=k.Given that the far number 203 is being observed here. Hence, it can be expected to beat least 203 cars. Any of the cars has equal probability to be the number of the lastcar. The likelihood function can be evaluated here as:P(y/N) = (1/N), for each N>202,2, else where.The posterior distribution here will be:P(N/y) α P( N)*P(y/N)=(1/N)*kA probability distribution adds up to 1 and the posterior probability has to sum up to 1.Therefore, the normalizing constants here are:
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