Answer
$P(100\leq X\leq120)=0.7336$
If we randomly select 10,000 samples of 250 Americans 18 years old or older, we would expect to find about 7336 samples in which between 100 and 120, inclusive, read at least six books within the past year.
Work Step by Step
$μ_X=np=250\times0.46=115$
$σ_X=\sqrt {np(1-p)}=\sqrt {250\times0.46(1-0.46)}=7.88$
Let's find the z-scores for 99.5 and 120.5:
$z=\frac{X-μ}{σ}=\frac{99.5-115}{7.88}=-1.97$
$z=\frac{X-μ}{σ}=\frac{120.5-115}{7.88}=0.70$
According to Table V, the area to the left of the standard normal curve for a z-score equal to -1.97 is 0.0244.
According to Table V, the area to the left of the standard normal curve for a z-score equal to 0.70 is 0.7580.
The area of the standard normal curve between the z-scores -1.97 and 0.70 is the difference between the area to the left of the standard normal curve for a z-score equal to 0.70 and the area to the left of the standard normal curve for a z-score equal to -1.97:
$0.7580-0.0244=0.7336$
