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A balanced coin is tossed 3 times. Denote,  H: Occurrence of Head T: Occurrence of Tail Then, the probability of head P(H)=1//2=P(T) and Y = number of heads observed Clearly,  Y∼Binomial(n=3,p=0.5) and the pdf of Y is given by, f_(Y)(y)=([3],[y])(0.5)^(y)(1-0.5)^(3-y)=([3],[y])(0.5)^(3) a)  P[Y=0]=([3],[0])(0.5)^(3)=0.125P[Y=1]=([3],[1])(0.5)^(3)=0.375P[Y=2]=([3],[2])(0.5)^(3)=0.375P[Y=3]=([3],[3])(0.5)^(3)=0.125 b) t tt ttt ttty  ttt tttp(y) tt tt ttt0 ttt0.125 tt tt ttt ttt1 ttt ttt0.375 tt tt ttt2 ttt0.375 tt tt ttt3 ttt ttt0.125 ttt tt t   c) mu=E(Y) E(Y)=np=3xx0.5=1.5 Variance of Y is given by V(Y)=npq (where q=1-p=1-0.5=0.5 here) So, V(Y)=3xx0.5 xx0.5=0.75 Hence, the standard deviation of Y is given by,  So, sqrt(V(Y))=sigma=sqrt0.75=0.866 d) 1 sd of mean  implies the interval  (mu-sigma,mu+sigma)=(1.5-0.866,1.5+0.866)=(0.634,2.366) Then  fraction of population measurements lying within the interval ( 0.634, 2.366) is  P(Y in(0.634,2.366))=P(Y=1)+P(Y=2)=0.375+0.375=0.75 Tchebysheff's theorem says that at least 1-1//k^(2)& ... See the full answer