Question 3.13: Examine the median of the r.v. X distributed as follows....

We have P(X\leq6)=16/32=0.50\geq0.50\ \ \mathrm{and}\ \ P(X\geq6)=17/32\gt 0.05\geq0.50, so that (47) is satisfied.
P(X\leq x_{0.50})\geq0.50\ \ \ \mathrm{and}\ \ \ \ P(X\geq x_{0.50})\geq0.50,\qquad\qquad (47)
Also,

P(X\leq7)=18/32\gt 0.50\geq0.50\quad\mathrm{and}\quad P(X\geq7)=16/32=0.50\geq0.50,

so that (47) is satisfied again. However, if we define the median as the point (6+7)/2=6.5, then P({{X}}\leq6.5)=P(X\geq6.5)=0.50, as (47) requires, and the median is uniquely defined.