Question

Hello, I having trouble with finding the median. So I noticed that the median has two 3’s and I can’t remember how to do it.

Answer 1

The data given in ascending order is,

$1,2,2,3,3,3,3,4,4,4,5,9$

Let solve for Q1,Q2, and Q3.

$\begin{gathered} Q_1=\frac{1}{4}(N+1)^{th}term \\ where,N=12 \\ Q_1=\frac{1}{4}(12+1)=\frac{1}{4}(13)=3.25^{th}term=\frac{2+3}{2}=2.5 \end{gathered}$$Q_3=\frac{3}{4}(N+1)=\frac{3}{4}(12+1)=9.75^{th}term=4$$Q_2=Median=\frac{3+3}{2}=3$

Let us now find the Minimum and Maximum value

$\begin{gathered} Minimum\text{ value=1} \\ Maximum\text{ value=9} \end{gathered}$

The box plot is shown below

Then,

$\begin{gathered} Range=9-1=8 \\ \therefore Range=8 \\ \\ Interquartile\text{ range=}Q_3-Q_1=4-2.5=1.5 \\ \therefore Interquartile\text{ range=1.5} \end{gathered}$