Firstly, rearrange the data so that it's in ascending order: [tex] \{59,76,84,93,95,97,101,101,102,102,104,106\} [/tex]
Next, find the median:
[tex] \{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99 [/tex]
Now to find the lower quartile, find the "median" of the data set that's to the left of 99:
[tex] \{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5 [/tex]
Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:
[tex] \{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102 [/tex]
Now that we have the upper and lower quartile, subtract them:
[tex] 102-88.5=13.5 [/tex]
In short, the IQR of this data set is 13.5.
