Answer
$P(\text{6-10}) =\frac{5}{24}\approx0.208$
It is not an unusual event.
Work Step by Step
The sample space is the caseload. So, N(S) = 24
There are 5 cases involving children between 6 and 10 years old. So, N(6-10) = 5.
P(6-10)$ = \frac{N(\text{6-10})}{N(s)}=\frac{5}{24}\approx0.208$
0.208 $\gt$ 0.05. So, it is not an unusual event.
