1个回答
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P(ξ=k)=1/n (k=1,2,3……,n)截到n吧
E(ξ)=1×1/n+2×1/n+3×1/n+.......+n×1/n)
=(1+2+3+........+n)/n
=(n+1)*n/2*1/n=(n+1)/2
D(ξ)=[1-(n+1)/2]²*1/n+[2-(n+1)/2]²*1/n+[3-(n+1)/2]²*1/n+.........+[n-(n+1)/2]²*1/n
={(1²+2²+........+n²)+n(n+1)²/4-[(n+1)+2(n+1)+3(n+1)+.......+n(n+1)]}/n
={(1²+2²+........+n²)+n(n+1)²/4-(n+1)(1+2+3+........+n)}/n
=[n(n+1)(2n+1)/6+n(n+1)²/4-n(n+1)²/2]/n
=[n(n+1)(2n+1)/6-n(n+1)²/4]/n
=1/12(n+1)[4n+2-3n-3]
=(n²-1)/12
E(ξ)=1×1/n+2×1/n+3×1/n+.......+n×1/n)
=(1+2+3+........+n)/n
=(n+1)*n/2*1/n=(n+1)/2
D(ξ)=[1-(n+1)/2]²*1/n+[2-(n+1)/2]²*1/n+[3-(n+1)/2]²*1/n+.........+[n-(n+1)/2]²*1/n
={(1²+2²+........+n²)+n(n+1)²/4-[(n+1)+2(n+1)+3(n+1)+.......+n(n+1)]}/n
={(1²+2²+........+n²)+n(n+1)²/4-(n+1)(1+2+3+........+n)}/n
=[n(n+1)(2n+1)/6+n(n+1)²/4-n(n+1)²/2]/n
=[n(n+1)(2n+1)/6-n(n+1)²/4]/n
=1/12(n+1)[4n+2-3n-3]
=(n²-1)/12
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