Answer
$P(125)=0.0213$
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 124.5 and 125.5:
$z=\frac{X-μ}{σ}=\frac{124.5-115}{7.88}=1.21$
$z=\frac{X-μ}{σ}=\frac{125.5-115}{7.88}=1.33$
According to Table V, the area to the left of the standard normal curve for a z-score equal to 1.21 is 0.8869.
According to Table V, the area to the left of the standard normal curve for a z-score equal to 1.33 is 0.9082.
The area of the standard normal curve between the z-scores 1.21 and 1.33 is the difference between the area to the left of the standard normal curve for a z-score equal to 1.33 and the area to the left of the standard normal curve for a z-score equal to 1.21:
$0.9082-0.8869=0.0213$
