P(X>4) = P(6) + P(5) = b(6;6,0.6) + b(5;6,0.6) = 0.6^6 + 6 0.4 0.6^5 ~= 0.23328

yes: b(4; 6, 0.6) = 6!/(4! 2!) ) 0.6^4 0.4^2 ~= 0.311042-P(X=4)

P(X=4)=b(4;6,0.6)

yes.3-mean

mean=n*p=3.6

Var = n p (1-p) = 6*0.6*0.4 ~= 1.444-variance

Var^2=n*p*(1-p)

Since X cannot take non-integer values P(X=3.6) = 0.5-P(X=mean)

Here since mean =3.6, I have problem to find P(X=mean)

my calculator does not give me a approximate value for b(3.6;4,0.6) !

How should I treat this question?

RonL