Answer:
[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{84.5}}}}}[/tex]
Step-by-step explanation:
Given data :
89 , 74 , 100 , 86 , 74 , 67 , 86 , 72 , 60 , 93 , 83 , 86
Arranging the data in ascending order, we have,
60 , 67 , 72 , 74 , 74 ,83 , 86 , 86 , 86 , 89 , 93 , 100
Here, n ( total no.of observation ) = 12
Now, finding the position of media :
[tex] \boxed{ \sf{position \: of \: median = ( \frac{n + 1}{2} ) ^{th} \: item}}[/tex]
⇒[tex] \sf{ (\frac{12 + 1}{2} ) ^{th} item}[/tex]
⇒[tex] \sf{( \frac{13}{2} ) ^{th} item}[/tex]
⇒[tex] \sf{ {6.5}^{th} item}[/tex]
6.5 th item is the average of 6 th and 7 th items.
Again,
[tex] \sf{median = \frac{ {6}^{th \: item} + {7}^{th} item}{2} }[/tex]
⇒[tex] \sf{ \frac{83 + 86}{2} }[/tex]
⇒[tex] \sf{ \frac{169}{2} }[/tex]
⇒[tex] \sf{84.5}[/tex]
Hope I helped!
Best regards!!
