See Answer

Add Answer +20 Points

Community Answer

See all the answers with 1 Unlock

Get 4 Free Unlocks by registration

Get 4 Free Unlocks by registration

Answer:- Given that: The following probability distributions are given, Distribution A : t tt tttxi tttP(X=xi) tt tt ttt0 ttt0.50 tt tt ttt1 ttt0.22 tt tt ttt2 ttt0.15 tt tt ttt3 ttt0.10 tt tt ttt4 ttt0.03 tt t Distribution B : t tt tttxi tttP(X=xi) tt tt ttt0 ttt0.03 tt tt ttt1 ttt0.10 tt tt ttt2 ttt0.15 tt tt ttt3 ttt0.22 tt tt ttt4 ttt0.50 tt t   (A) The expected value for each distribution is, Expected value for Distribution A : E(X)_(A)=sumx_(i)**P(X=x_(i))                  =0**0.50+1**0.22+2**0.15+3**0.10+4**0.03                  =0+0.22+0.3+0.3+0.12                  =0.94 Hence, the expected value for Distribution A = mu = 0.94   Expected value for Distribution B : E(X)_(B)=sumx_(i)**P(X=x_(i))                  =0**0.03+1**0.10+2**0.15+3**0.22+4**0.50                  =0+0.10+0.3+0.66+2                  =3.06 Hence, the expected value for Distribution B = mu = 3.06     (B) The standard deviation for each distribution is,   Standard deviation for Distribution A: SD_(A)=sqrt(E(X^(2))_(A)-[E(X)_(A)]^(2))   E(X^(2))_(A)=sumx_(i)^(2)**P(X=x_(i))                  =0^(2)**0.50+1^(2)**0.22+2^(2)**0.15+3^(2)**0.10+4^(2)**0.03 =0+0.22+0.6+0.9+0.48 =2.2                   SD_(A)=sqrt(E(X^(2))_(A)-[E(X)_(A)]^(2))                      =sqrt(2.2-[0.94]^(2))              =sqrt(2.2-0.8836)              ... See the full answer