Use the midpoint convention for a median

The mode is $4$ because $P(Y=4)=0.30$ is the largest probability. This is the Teacher completion.

$P(Y\le3)=0.1+0.2+0.2=0.5\ge0.5$ and $P(Y\ge4)=0.3+0.2=0.5\ge0.5$. The Teacher solution reports the midpoint $(3+4)/2=3.5$. Under the general median inequalities, every $m\in[3,4]$ is a median; $3.5$ is the source’s midpoint convention. This occurs despite five distinct listed values: do not apply a parity rule to the support.