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Question

If the variable takes the values 0,1,2,,n with frequencies proportional to qn,nCrqn1p,nC2qn2p2,,pn where p+q=1, then show that mean square deviation is n2p2+npq and variance is npq.

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Solution

μ1=E(x)=0x=0x(nx)pxqnx
(nx)=nCx=nx(n1x1) =npnx1(n1x1)px1q(n1)(x1)
=np(p+q)n1=np p+q=1
mean square deviation about x=0 is
μ2=E(x2)=nx=0x2(nx)pxqnx
nx=0[x(x1)+1](nx)pxqnx
nx=0x(x1)(nxpxqnx+nx=0)x(nx)pxqnx
n(n1)p2nx=2(n2n2)px2q(n2)(x2)+np
n(n1)p2(q+p)n2+np n(n1)p2+np=n2p2+np(1p)
=n2p2+npq p+q=1variance =μ2=μ12(μ1)2
=n2p2+npq(np)2=npq

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