Answer
$P(X\gt100)=0.3859$
Since $P(X\gt100)\gt0.05$, it is not an unusual event.
Work Step by Step
$μ=90$ and $σ=35$
Let's find the z-score for 100:
$z=\frac{X-μ}{σ}=\frac{100-90}{35}=0.29$
According to Table V, the area of the standard normal curve to the left of z-score equal to 0.29 is 0.6141.
But, we want the area of the standard normal curve to the right of z-score equal to 0.29:
$1-0.6141=0.3859$
$P(X\gt100)=0.3859\gt0.05$. It is not unusual.
