Answer
$P(X\geq130)=0.0028$
Work Step by Step
55% = 0.55
$μ_X=np=200\times0.55=110$
$σ_X=\sqrt {np(1-p)}=\sqrt {200\times0.55(1-0.55)}=7.04$
Let's find the z-score for 129.5:
$z=\frac{X-μ}{σ}=\frac{129.5-110}{7.04}=2.77$
According to Table V, the area of the standard normal curve to the left of z-score equal to 2.77 is 0.9972.
But, we want the area of the standard normal curve to the right (at least 130) of z-score equal to 2.77:
$1-0.9972=0.0028$
