Answer
$P(100)=0.0018$
Work Step by Step
$μ_X=np=300\times0.267=80.1$
$σ_X=\sqrt {np(1-p)}=\sqrt {300\times0.267(1-0.267)}=7.66$
Let's find the z-scores for 99.5 and 100.5:
$z=\frac{X-μ}{σ}=\frac{99.5-80.1}{7.66}=2.53$
$z=\frac{X-μ}{σ}=\frac{100.5-80.1}{7.66}=2.66$
According to Table V, the area of the standard normal curve to the left of z-score equal to 2.53 is 0.9943.
According to Table V, the area the standard normal curve to the left of z-score equal to 2.66 is 0.9961.
The area of the standard normal curve between the z-scores 2.53 and 2.66 is the difference between the area of the standard normal curve to the left of z-score equal to 2.66 and the area of the standard normal curve to the left of z-score equal to 2.53:
$0.9961-0.9943=0.0018$
