Question

From the following, determine lower quartile and upper quartile:

X

12

16

17

21

28

19

30

32

Answer 1

12, 16, 17, 19, 21, 28, 30, 32

N = 8

$$\begin{gathered} \mathbf{Lower}\mathbf{}\mathbf{Quartile}\mathbf{:} \\ {Q}_1=\mathrm{Size}\mathrm{of}{\left(\frac{\mathrm{N}+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},Q_1=\mathrm{Size}\mathrm{of}{\left(\frac{8+1}{4}\right)}^{\mathrm{th}}=2.{25}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},Q_1=\mathrm{Size}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{item}+\frac{1}{4}\left(\mathrm{Size}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{term}-\mathrm{Size}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{item}\right) \\ \mathrm{or},Q_1=16+\frac{1}{4}\left(17-16\right) \\ \Rightarrow Q_1=16.25 \\ \mathbf{Upper}\mathbf{}\mathbf{Quartile}\mathbf{:} \\ {Q}_3=\mathrm{Size}\mathrm{of}3{\left(\frac{\mathrm{N}+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},Q_3=\mathrm{Size}\mathrm{of}3{\left(\frac{8+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},Q_3=6.{75}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},Q_3=\mathrm{Size}\mathrm{of}{6}^{\mathrm{th}}\mathrm{item}+\frac{3}{4}(\mathrm{Size}\mathrm{of}{7}^{\mathrm{th}}\mathrm{term}-\mathrm{Size}\mathrm{of}{6}^{\mathrm{th}}\mathrm{item}) \\ \mathrm{or},Q_3=28+\frac{3}{4}\left(30-28\right) \\ \mathrm{or},Q_3=28+1.5 \\ \Rightarrow Q_3=29.5 \end{gathered}$$

Hence, lower and upper quartile are 16.25 and 29.5 respectively.