Answer
$P(3\leq X\leq5)=0.036864+0.13824+0.27648=0.451584$
Work Step by Step
The solution is based on the binomial distribution: $P(X=x)=C(n,x)*p^{x}*q^{n-x}$ If $3\leq x\leq5$, this means that x is either 3, 4 or 5, therefore we have to add the probabilities of these $x$ values.
If $p=0.4$, $n=6$, then $q=1-0.4=0.6$ By substituting into the given variables, we get: $P(X=5)=C(6,5)*0.4^5*0.6^1=\frac{6!}{5!\times(6-5)!}*0.4^5*0.6^1=0.036864$ $P(X=4)=C(6,4)*0.4^4*0.6^2=\frac{6!}{4!\times(6-4)!}*0.4^4*0.6^2=0.13824$ $P(X=3)=C(6,3)*0.4^3*0.6^3=\frac{6!}{3!\times(6-3)!}*0.4^3*0.6^3=0.27648$
The sum of these values gives us: $P(3\leq X\leq5)=0.036864+0.13824+0.27648=0.451584$
