Answer
P(10 or more people are waiting in line for lunch) = 0.029.
This is unusual.
Work Step by Step
P(10 or more people are waiting in line for lunch) = $P(X \geq 10)$
$P(X \geq 10) = P(X=10) + P(X=11) + P(X=12)
= 0.019 + 0.004 + 0.006
= 0.029
Here the probability is less than 0.05, hence the event is unusual.
