Let X represent the random variable with a binomial distribution.
X∼B(8,p)
P(X=3)=(83)p3(1−p)8−3=56p3(1−p)5∎
X∼B(8,0.35)
P(X≥4)=1−P(X≤3)=1−0.70640=0.29360=0.294∎ (to 3 s.f.)
[probability of X \geq 4]
E(X)=np=8(0.35)=2.8∎
Var(X)=np(1−p)=8(0.35)(1−0.35)=1.8200
σ=1.8200=1.3491=1.35∎ (to 3 s.f.)