Answer
$P(Bryce)=\frac{119}{1009}\approx0.1179$
It is not an unusual event.
Work Step by Step
The sample space: 1009 cases. So, $N(S)=1009$
According to the marginal distribution (see page 235) of the third row: $N(Bryce)=119$
Using the Empirical Approach (page 258):
$P(Bryce)=\frac{N(Bryce)}{N(S)}=\frac{119}{1009}\approx0.1179$
$P(Bryce)\gt0.05$. So, it is not an unusual event.
