91Ó°ÊÓ

Given data :

The interquartile range describes the spread of your distribution's middle half. Any distribution that is sorted from low to high is divided into four equal sections by quartiles.

The inter-quartile range is calculated as $$\begin{gathered} Q_1=\frac{N+1}{4}thitem \\ {Q}_1=\frac{14+1}{4}thitem \\ {Q}_1=\frac{15}{4}thitem \\ {Q}_1=3.75thitem \\ =\frac{1}{4}x3rditem+\frac{3}{4}x4thitem \\ =77.5 \\ Q1=\frac{1}{4}x76+\frac{3}{4}x78 \\ =77.5 \\ Q3=\frac{3(N+1)}{4}thitem \\ Q3=\frac{3(14+1)}{4}thitem \\ Q3=\frac{45}{4}thitem \\ Q3=11.25thitem=\frac{3}{4}x11thitem+\frac{1}{4}x12thitem \\ Q3=\frac{3}{4}x91+\frac{1}{4}x93 \\ =91.5 \\ IQR=Q_3−Q_1 \\ =91.5−77.5 \\ =14 \end{gathered}$$

The interquartile range is $14$

We can conclude that there are no outliers in Joe's first $14$ quiz grades in a marking session. If an observation falls more than $(1.5)$IQR above or below the first quartile, it is considered an outlier. $$\begin{gathered} IQR×1.5 \\ 14×1.5=21 \\ {Q}_1-21=77.5-21 \\ =56.5 \\ {Q}_3-21=91.5+21 \\ =112.5 \end{gathered}$$

There are no outliers because all of the items are within the range. Thus there are no outliers.