Answer
$P(sample~mean\geq19.825)=0.0351$
It is an unusual event.
Work Step by Step
From the previous item we found that $x ̅=19.825$. Let's find the z-score:
$z=\frac{x ̅-µ}{\frac{σ}{\sqrt n}}=\frac{19.825-19.1}{\frac{0.8}{\sqrt 4}}=1.81$
The area to the left of $z=1.81$, in the standard normal distribution, is 0.9649. That is: $P(z\lt1.81)=0.9649$
$P(sample~mean\geq19.825)=P(z\geq1.81)=1-P(z\lt1.81)=1-0.9649=0.0351\lt0.05$
So, it is an unusual event.
