Answer
No, it is not unusual.
Work Step by Step
n = 400 and p = 18.4% = 0.184
Expected value (mean):
$E(X)=μ_X=np=400\times0.184=73.6$
Standard deviation:
$σ_X=\sqrt {np(1-p)}=\sqrt {400\times0.184(1-0.184)}=\sqrt {400\times0.184\times0.816}=7.75$
$np(1-p)=400\times0.184\times0.816=60.0576\gt10$. The probability histogram is bell shaped (see blue rectangle on page 343).
There is a probability of 95% that the number of successes is between $μ_X-2σ_X=73.6-2\times7.75=58.1$ and $μ_X+2σ_X=73.6+2\times7.75=89.1$. But, 86 is between 58.1 and 89.1. So, it is not unusual.
