Given the data set 

62 , 65 , 68 , 70 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101, 

find 

a) the median, 

b) the first quartile, 

c) the third quartile, 

d) the interquartile range (IQR). 

Answer :
 

We have data : 62 , 65 , 68 , 70 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101


a )  We know Median  = $\frac{{\left({\displaystyle \frac{n}{2}}\right)}^{th} term +{\left(\frac{n}{2} +\hspace{0.17em}1 \right)}^{th} term }{2}$ , Here n = Number of terms  =  12
So,
Median of given data  = $\frac{{\left({\displaystyle \frac{12}{2}}\right)}^{th} term +{\left(\frac{12}{2} +\hspace{0.17em}1 \right)}^{th} term }{2} = \frac{{\left({\displaystyle 6}\right)}^{th} term +{\left(7 \right)}^{th} term }{2} = \frac{74 \hspace{0.17em}+ 76}{2} = \frac{150}{2} = \mathbf{ }\mathbf{75}\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{\hspace{0.17em}}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$
And
Quartiles are the values that divide a list of numbers into quarters
So , we divide our data , As  :


So,

b ) The first quartile Q₁ = $\frac{68 + 70}{2} = \frac{138}{2} = \mathbf{69}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$

c ) The third quartile Q₃ = $\frac{80 + 82}{2} = \frac{162}{2} = \mathbf{81}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$

d ) The Interquartile range ( IQR ) = Q₃ - Q₁ =  81  -  69  =  12                          ( Ans )