Answer
$P(X\gt100)=0.1841$
Since $P(X\gt100)\gt0.05$, it is not an unusual event.
Work Step by Step
$μ_{X ̅}=μ=90$
$σ_{X ̅}=\frac{σ}{\sqrt n}=\frac{35}{\sqrt {10}}=11.07$
Let's find the z-sore for 100:
$z=\frac{X-μ}{σ}=\frac{100-90}{11.07}=0.90$
According to Table V, the area of the standard normal curve to the left of z-score equal to 0.90 is 0.8159.
But, we want the area of the standard normal curve to the right of z-score equal to 0.90:
$1-0.8159=0.1841$
$P(X\gt100)=0.1841\gt0.05$
