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Statistical Techniques for Transportation Engineering by Naresh Davergave, Anil Shah, G Rao, Kumar Molugaram

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Again we have

μ3=x=0nx3p(x)=x=0n[x(x1)(x2)+3x(x1)+x]p(x)(sincex3=x(x1)(x2)+3x(x1)+x)=x=0n[x(x1)(x2)+3x=0nx(x1)+p(x)+x=0nxP(x)=n(n1)(n2)p3(q+p)n3+3n(n1)p2(q+p)n2+np(q+p)n1=n(n1)(n2)p3+3n(n1)p2+np

image

Since μ3=μ33μ2μ1+2μ13image

We get

μ3=n(n1)(n2)p3+3n(n1)p2+np3(n2p2np2+np)(np)+2n3p3=npq(qp)

image

Similarly we get

μ4=n(n1)(n2)

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