Answer
$P(X\lt60)=0.0062$
Work Step by Step
$μ_X=np=500\times0.16=80$
$σ_X=\sqrt {np(1-p)}=\sqrt {500\times0.16(1-0.16)}=8.20$
$P(X\lt60)=P(X\leq59)$
Let's find the z-scores for 59.5
$z=\frac{X-µ_X}{σ_X}=\frac{59.5-80}{8.20}=-2.5$
According to Table V, the area of the standard normal curve to the left of z-score equal to -2.5 is 0.0062.
