In order to calculate the minimum and maximum usual values, first let's calculate the mean and standard deviation of this distribution:
[tex]\begin{gathered} \mu=n\cdot p=140\cdot0.5=70\\ \\ \sigma=\sqrt{np(1-p)}=\sqrt{140\cdot0.5\cdot0.5}=5.92 \end{gathered}[/tex]Now, calculating the minimum and maximum usual values, we have:
[tex]\begin{gathered} minimum=\mu-2\sigma=70-11.84=58.16\\ \\ maximum=\mu+2\sigma=70+11.84=81.84 \end{gathered}[/tex]Since the given result is 55, it is an unusual reslt, because it is less tahan the minimum usual value.
Correct option: third one.