Answer :
Give a fiven number data set: 1, 7, 12, 15, 20.
Minimum = 1
Maximum = 20
[tex]\begin{gathered} \text{First quartile = (}\frac{5+1}{4})^{th}\text{ obs.} \\ =1.5^{th\text{ obs.}} \\ =1^{st}obs.+0.5(2^{nd}obs.-1^{st}obs.) \\ =1+0.5(7-1) \\ =4 \end{gathered}[/tex][tex]\begin{gathered} Median=(\frac{5+1}{2})^{th}\text{ obs.} \\ =3^{rd}\text{ obs.} \\ =12 \end{gathered}[/tex][tex]\begin{gathered} \text{Third quartile = (}\frac{3(5+1)}{4})^{th}obs.\text{ } \\ =4.5^{th}obs. \\ =4^{th}obs.+0.5(5^{th}obs.-4^{th}obs.) \\ =15+0.5(20-15) \\ =15+0.5(5) \\ =17.5 \end{gathered}[/tex]Find the interquartile range.
[tex]\begin{gathered} \text{Interquartile range = Third quartile-First quartile} \\ =17.5-4 \\ =13.5 \end{gathered}[/tex]Now,
[tex]\begin{gathered} Q_1-1.5\text{ IQR=}4-1.5(13.5) \\ =-16.25 \end{gathered}[/tex][tex]\begin{gathered} Q_3+1.5IQR=17.5+1.5(13.5) \\ =37.75 \end{gathered}[/tex]a. An observation is considered an outlier if it is below -16.25.
b. An observation is considered an outlier if it is above 37.75