In order to calculate the minimum and maximum usual values, first let's calculate the mean and standard deviation of this distribution:
![\begin{gathered} \mu=n\cdot p=140\cdot0.5=70\\ \\ \sigma=√(np(1-p))=√(140\cdot0.5\cdot0.5)=5.92 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ty74hbyworbuc41avijsq3095m1hkyuoj8.png)
Now, calculating the minimum and maximum usual values, we have:
![\begin{gathered} minimum=\mu-2\sigma=70-11.84=58.16\\ \\ maximum=\mu+2\sigma=70+11.84=81.84 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/npxkg345ruc9r0hnhlicf6zw5e5v6oyef6.png)
Since the given result is 55, it is an unusual reslt, because it is less tahan the minimum usual value.
Correct option: third one.