If the variable takes the values 0,1,2,⋯,n with frequencies proportional to qn,nCrqn−1p,nC2qn−2p2,⋯,pn where p+q=1, then show that mean square deviation is n2p2+npq and variance is npq.
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Solution
μ′1=E(x)=0∑x=0x(nx)pxqn−x (nx)=nCx=nx(n−1x−1)=npn∑x−1(n−1x−1)px−1q(n−1)−(x−1) =np(p+q)n−1=np∴p+q=1 mean square deviation about x=0 is μ2=E(x2)=n∑x=0x2(nx)pxqn−x n∑x=0[x(x−1)+1](nx)pxqn−x n∑x=0x(x−1)(nxpxqn−x+n∑x=0)x(nx)pxqn−x n(n−1)p2n∑x=2(n−2n−2)px−2q(n−2)−(x−2)+np n(n−1)p2(q+p)n−2+npn(n−1)p2+np=n2p2+np(1−p) =n2p2+npq∴p+q=1variance =μ2=μ12−(μ′1)2 =n2p2+npq−(np)2=npq