Answer
$P(X\lt120)=0.7157$
If we randomly select 10,000 samples of 250 Americans 18 years old or older, we would expect to find about 7157 samples in which fewer than 120 read at least six books within the past year.
Work Step by Step
$μ_X=np=250\times0.46=115$
$σ_X=\sqrt {np(1-p)}=\sqrt {250\times0.46(1-0.46)}=7.88$
Notice that; $P(X\lt120)=P(X\leq119)$
Let's find the z-score for 119.5:
$z=\frac{X-μ}{σ}=\frac{119.5-115}{7.88}=0.57$
According to Table V, the area to the left of the standard normal curve for a z-score equal to 0.57 is 0.7157.
