Knowing that, the Mean #mu# of the Binomial Distribution with
the parameters #n=8 and p=0.4# is given by,
#mu=np=8xx0.4=3.2#.
Now, the Probability of #X=x# successes out of #n# trials, i.e.,
#P(X=x)=p(x)# is given by, #p(x)=""_nC_xp^xq^(n-x),#
where, #x=0,1,2,...,n, and q=1-p#.
Hence, #p(x)=""_8C_x(0.4)^x(0.6)^(8-x), x=0,1,...,8#.
#:."The Reqd. Prob.="P(X>mu)=P(X>3.2)=P(X>=4)#,
#=1-P(X<3)#,
#rArr"The Reqd. Prob.="1-[p(0)+p(1)+p(2)+p(3)]......(beta)#.
#p(0)=""_8C_0(0.4)^0(0.6)^8=0.6^8............(beta_0)#.
#p(1)=""_8C_1(0.4)^1(0.6)^7=8(0.4)(0.6)^7......(beta_1)#.
#p(2)=""_8C_2(0.4)^2(0.6)^6=28(0.4)^2(0.6)^6......(beta_2)#.
#p(3)=""_8C_3(0.4)^3(0.6)^5=56(0.4)^3(0.6)^5......(beta_3)#.
Utilising #(beta_0)" to "(beta_3)" in "(beta)#, we get,
#"The Reqd. Prob.="1-0.6^5{0.6^3+8xx0.4xx0.6^2+28xx0.4^2xx0.6+56xx0.4^3}#,
#=1-0.6^5{0.216+ 1.152+2.688+3.584}#,
#=1-0.07776(7.64)=1-0.5940864#,
#rArr"The Reqd. Prob.="0.4059136#.
